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8,000
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Which country is the University of Oxford located?
SELECT T2.country_name FROM university AS T1 INNER JOIN country AS T2 ON T1.country_id = T2.id WHERE university_name = 'University of Oxford'
University of Oxford refers to university_name = 'University of Oxford'; which country refers to country_name
[ "country.country_name", "country.id", "university.country_id" ]
8,001
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many times did the Yale University achieve a score of no less than 10 in the Quality of Education Rank?
SELECT COUNT(*) FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id INNER JOIN university AS T3 ON T3.id = T2.university_id WHERE T3.university_name = 'Yale University' AND T2.score >= 10 AND T1.criteria_name = 'Quality of Education Rank'
Yale University refers to university_name = 'Yale University'; a score of no less than 10 refers to score > = 10; in the Quality of Education Rank refers to criteria_name = 'Quality of Education Rank'
[ "ranking_criteria.criteria_name", "ranking_criteria.id", "university.id", "university.university_name", "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.university_id" ]
8,002
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What are the names of the criteria under Center for World University Rankings?
SELECT T2.criteria_name FROM ranking_system AS T1 INNER JOIN ranking_criteria AS T2 ON T1.id = T2.ranking_system_id WHERE T1.system_name = 'Center for World University Rankings'
names of the criteria refers to criteria_name; under Center for World University Rankings refers to system_name = 'Center for World University Rankings';
[ "ranking_criteria.criteria_name", "ranking_criteria.ranking_system_id", "ranking_system.id", "ranking_system.system_name" ]
8,003
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
List the names of all the universities that have no less than 50,000 students in the year 2012.
SELECT T2.university_name FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T1.num_students > 50000 AND T1.year = 2012
have no less than 50,000 students refers to num_students > 50000; name of university refers to university_name;
[ "university.id", "university.university_name", "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,004
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Between 2011 to 2016, in which countries can you find the universities where at least 50% of its students are international students?
SELECT DISTINCT T3.country_name FROM university AS T1 INNER JOIN university_year AS T2 ON T1.id = T2.university_id INNER JOIN country AS T3 ON T3.id = T1.country_id WHERE T2.pct_international_students > 50 AND T2.year BETWEEN 2011 AND 2016
Between 2011 to 2016 refers to year BETWEEN 2011 AND 2016; at least 50% of its students are international students refers to pct_international_students > 50; which country refers to country_name
[ "country.country_name", "country.id", "university.country_id", "university.id", "university_year.pct_international_students", "university_year.university_id", "university_year.year" ]
8,005
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many universities have no less than 20,000 female students in 2016? Identify how many of the said universities are located in the United States of America.
SELECT COUNT(*) , SUM(CASE WHEN T3.country_name = 'United States of America' THEN 1 ELSE 0 END) AS nums_in_usa FROM university AS T1 INNER JOIN university_year AS T2 ON T1.id = T2.university_id INNER JOIN country AS T3 ON T3.id = T1.country_id WHERE T2.year = 2016 AND T2.num_students * T2.pct_female_students / 100 > 20...
have no less than 20,000 female students refers to DIVIDE(MULTIPLY(pct_female_students, num_students), 100) > 20000; in 2016 refers to year = 2016; located in the United States of America refers to country_name = 'United States of America'
[ "country.country_name", "country.id", "university.country_id", "university.id", "university_year.num_students", "university_year.pct_female_students", "university_year.university_id", "university_year.year" ]
8,006
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What are the names of the top 5 universities with the highest number of international students?
SELECT DISTINCT T2.university_name FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id ORDER BY (CAST(T1.num_students * T1.pct_international_students AS REAL) / 100) DESC LIMIT 5
highest number of international students refers to MAX(DIVIDE(MULTIPLY(num_students, pct_international_students), 100)); name of university refers to university_name;
[ "university.id", "university.university_name", "university_year.num_students", "university_year.pct_international_students", "university_year.university_id" ]
8,007
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the university ID of the university with the largest student staff ratio?
SELECT university_id FROM university_year ORDER BY student_staff_ratio DESC LIMIT 1
the largest student staff ratio refers to max(student_staff_ratio)
[ "university_year.student_staff_ratio", "university_year.university_id" ]
8,008
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Give the year where a university had the lowest number of students.
SELECT year FROM university_year ORDER BY num_students ASC LIMIT 1
had the lowest number of students refers to MIN(num_students)
[ "university_year.num_students", "university_year.year" ]
8,009
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Compute the average percentage of female students.
SELECT AVG(pct_female_students) FROM university_year
average percentage of female students refers to avg(pct_female_students)
[ "university_year.pct_female_students" ]
8,010
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Provide the number of international students and number of students in 2013 in university ID 20.
SELECT pct_international_students * num_students, num_students FROM university_year WHERE year = 2013 AND university_id = 20
number of international students refers to DIVIDE(MULTIPLY(pct_international_students, num_students), 100); in 2013 refers to year = 2013
[ "university_year.num_students", "university_year.pct_international_students", "university_year.university_id", "university_year.year" ]
8,011
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the university ID of Harvard University?
SELECT id FROM university WHERE university_name = 'Harvard University'
of Harvard University refers to university_name = 'Harvard University';
[ "university.id", "university.university_name" ]
8,012
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
List the university ID of the university that scored 100 in 2011.
SELECT university_id FROM university_ranking_year WHERE score = 100 AND year = 2011
in 2011 refers to year = 2011; score = 100
[ "university_ranking_year.score", "university_ranking_year.university_id", "university_ranking_year.year" ]
8,013
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Provide the ranking system of the ranking criteria named Quality of Education Rank.
SELECT T1.system_name FROM ranking_system AS T1 INNER JOIN ranking_criteria AS T2 ON T1.id = T2.ranking_system_id WHERE T2.criteria_name = 'Quality of Education Rank'
criteria named Quality of Education Rank refers to criteria_name = 'Quality of Education Rank'; ranking system refers to system_name;
[ "ranking_criteria.criteria_name", "ranking_criteria.ranking_system_id", "ranking_system.id", "ranking_system.system_name" ]
8,014
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the student staff ratio of Harvard University in 2012?
SELECT T1.student_staff_ratio FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T2.university_name = 'Harvard University' AND T1.year = 2012
Harvard University refers to university_name = 'Harvard University'; in 2012 refers to year = 2012
[ "university.id", "university.university_name", "university_year.student_staff_ratio", "university_year.university_id", "university_year.year" ]
8,015
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Give the location of the university ID 112.
SELECT T2.country_name FROM university AS T1 INNER JOIN country AS T2 ON T1.country_id = T2.id WHERE T1.id = 112
location refers to country_name
[ "country.country_name", "country.id", "university.country_id", "university.id" ]
8,016
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Calculate the total number of students in universities located in Sweden.
SELECT SUM(T2.num_students) FROM university AS T1 INNER JOIN university_year AS T2 ON T1.id = T2.university_id INNER JOIN country AS T3 ON T3.id = T1.country_id WHERE T3.country_name = 'Sweden'
located in Sweden refers to country_name = 'Sweden'; number of students refers to num_students
[ "country.country_name", "country.id", "university.country_id", "university.id", "university_year.num_students", "university_year.university_id" ]
8,017
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the ranking criteria ID of Brown University in 2014?
SELECT T1.ranking_criteria_id FROM university_ranking_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T2.university_name = 'Brown University' AND T1.year = 2014
Brown University refers to university_name = 'Brown University'; in 2014 refers to year = 2014
[ "university.id", "university.university_name", "university_ranking_year.ranking_criteria_id", "university_ranking_year.university_id", "university_ranking_year.year" ]
8,018
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
List the name of universities located in Spain.
SELECT T1.university_name FROM university AS T1 INNER JOIN country AS T2 ON T1.country_id = T2.id WHERE T2.country_name = 'Spain'
name of universities refers to university_name; located in Spain refers to country_name = 'Spain';
[ "country.country_name", "country.id", "university.country_id", "university.university_name" ]
8,019
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the criteria name of the university ID 32 in 2015?
SELECT T1.criteria_name FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id WHERE T2.university_id = 32 AND T2.year = 2015
in 2015 refers to year = 2015
[ "ranking_criteria.criteria_name", "ranking_criteria.id", "university_ranking_year.ranking_criteria_id", "university_ranking_year.university_id", "university_ranking_year.year" ]
8,020
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Compute the average score of the university located in Brazil.
SELECT AVG(T2.score) FROM university AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.university_id INNER JOIN country AS T3 ON T3.id = T1.country_id WHERE T3.country_name = 'Brazil'
average score refers to avg(score); located in Brazil refers to country_name = 'Brazil';
[ "country.country_name", "country.id", "university.country_id", "university.id", "university_ranking_year.score", "university_ranking_year.university_id" ]
8,021
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
In which country does the most populated university in 2014 located ?
SELECT T2.country_id FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T1.year = 2014 ORDER BY T1.num_students DESC LIMIT 1
the most populated university refers to max(num_students); in 2014 refers to year = 2014
[ "university.country_id", "university.id", "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,022
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Give the score and number of international students in university ID 100 in 2015.
SELECT CAST(T1.num_students * T1.pct_international_students AS REAL) / 100, T2.score FROM university_year AS T1 INNER JOIN university_ranking_year AS T2 ON T1.university_id = T2.university_id WHERE T2.year = 2015 AND T1.university_id = 100
number of international students refers to DIVIDE(MULTIPLY(num_students, pct_international_students), 100); in 2015 refers to year = 2015
[ "university_ranking_year.score", "university_ranking_year.university_id", "university_ranking_year.year", "university_year.num_students", "university_year.pct_international_students", "university_year.university_id" ]
8,023
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the student population of the university that scored 98 in 2013?
SELECT SUM(T1.num_students) FROM university_year AS T1 INNER JOIN university_ranking_year AS T2 ON T1.university_id = T2.university_id WHERE T2.score = 98 AND T1.year = 2013
student population refers to num_students; in 2013 refers to year = 2013
[ "university_ranking_year.score", "university_ranking_year.university_id", "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,024
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
List the criteria names under the ranking system called Center for World University Ranking.
SELECT T2.criteria_name FROM ranking_system AS T1 INNER JOIN ranking_criteria AS T2 ON T1.id = T2.ranking_system_id WHERE T1.system_name = 'Center for World University Rankings'
ranking system called Center for World University Ranking refers to system_name = 'Center for World University Rankings';
[ "ranking_criteria.criteria_name", "ranking_criteria.ranking_system_id", "ranking_system.id", "ranking_system.system_name" ]
8,025
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Provide the country name of universities with the number of students greater than 98% of the average student population of all universities in 2013.
SELECT DISTINCT T3.country_name FROM university AS T1 INNER JOIN university_year AS T2 ON T1.id = T2.university_id INNER JOIN country AS T3 ON T3.id = T1.country_id WHERE T2.year = 2013 AND T2.num_students * 100 > ( SELECT AVG(num_students) FROM university_year ) * 98
number of students greater than 98% of the average student population of all universities refers to num_students >  MULTPLY(num_students, 0.98); in 2013 refers to year = 2013
[ "country.country_name", "country.id", "university.country_id", "university.id", "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,026
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Among universities that score below 80 in 2015, what is the percentage of international students?
SELECT SUM(CAST(T1.num_students * T1.pct_international_students AS REAL) / 100) / COUNT(*) * 100 FROM university_year AS T1 INNER JOIN university_ranking_year AS T2 ON T1.university_id = T2.university_id WHERE T2.score < 80 AND T1.year = 2015
score below 80 refers to score < 80; in 2015 refers to year 2015; percentage of international students refers to DIVIDE(SUM(DIVIDE(MULTIPLY(num_students, pct_international_students), 100)), SUM(num_students))
[ "university_ranking_year.score", "university_ranking_year.university_id", "university_year.num_students", "university_year.pct_international_students", "university_year.university_id", "university_year.year" ]
8,027
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many students attended universities were there in 2011?
SELECT SUM(num_students) FROM university_year WHERE year = 2011
in 2011 refers to year = 2011;
[ "university_year.num_students", "university_year.year" ]
8,028
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Among all universities, how many female students were there in 2011?
SELECT SUM(CAST(num_students * pct_female_students AS REAL) / 100) FROM university_year WHERE year = 2011
in 2011 refers to year = 2011; female students refers to SUM(DIVIDE(MULTIPLY(num_students, pct_female_students), 100))
[ "university_year.num_students", "university_year.pct_female_students", "university_year.year" ]
8,029
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the student staff ratio at the university with the greatest student staff ratio of all time?
SELECT MAX(student_staff_ratio) FROM university_year ORDER BY student_staff_ratio DESC LIMIT 1
greatest student staff ratio of all time refers to max(student_staff_ratio)
[ "university_year.student_staff_ratio" ]
8,030
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the university ID with the most students in 2011?
SELECT university_id FROM university_year WHERE year = 2011 ORDER BY num_students DESC LIMIT 1
most students refers to MAX(num_students), in 2011 refers to year = 2011
[ "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,031
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many institutions with over 50,000 students in 2011 had a percentage of oversea students of more than 10%?
SELECT COUNT(*) FROM university_year WHERE year = 2011 AND num_students > 50000 AND pct_international_students > 10
institutions with over 50,000 students refers to num_students > 50000; in 2011 refers to year = 2011; percentage of oversea students of more than 10% refers to pct_international_students > 10;
[ "university_year.num_students", "university_year.pct_international_students", "university_year.year" ]
8,032
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Provide the ID of the university with the highest percentage of female students in 2012.
SELECT university_id FROM university_year WHERE year = 2012 ORDER BY pct_female_students DESC LIMIT 1
in 2012 refers to year = 2012; highest percentage of female students  refers to MAX(pct_female_students); ID of the university refers to university_id
[ "university_year.pct_female_students", "university_year.university_id", "university_year.year" ]
8,033
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Which university had the highest reputation in 2012?
SELECT T2.university_name FROM university_ranking_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T1.year = 2012 ORDER BY T1.score DESC LIMIT 1
had the highest reputation refers to MAX(score), in 2012 refers to year = 2012; which university refers to university_name;
[ "university.id", "university.university_name", "university_ranking_year.score", "university_ranking_year.university_id", "university_ranking_year.year" ]
8,034
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Name the university that had the most students in 2011.
SELECT T2.university_name FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T1.year = 2011 ORDER BY T1.num_students DESC LIMIT 1
in 2011 refers to year = 2011; had the most students refers to MAX(num_students); name of university refers to university_name;
[ "university.id", "university.university_name", "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,035
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Indicate the university's name with the highest ranking score in Teaching.
SELECT T1.university_name FROM university AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.university_id INNER JOIN ranking_criteria AS T3 ON T3.id = T2.ranking_criteria_id WHERE T3.criteria_name = 'Teaching' ORDER BY T2.score DESC LIMIT 1
university's name refers to university_name; highest ranking score refers to MAX(score); in Teaching refers to criteria_name = 'Teaching'
[ "ranking_criteria.criteria_name", "ranking_criteria.id", "university.id", "university.university_name", "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.university_id" ]
8,036
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the percentage of Harvard university's international students in 2011?
SELECT T1.pct_international_students FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T1.year = 2011 AND T2.university_name = 'Harvard University'
Harvard university's refers to university_name = 'Harvard University'; in 2011 refers to year = 2011; percentage of Harvard university's international students refers to pct_international_students
[ "university.id", "university.university_name", "university_year.pct_international_students", "university_year.university_id", "university_year.year" ]
8,037
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many female students were there at Stanford University in 2011?
SELECT CAST(T1.num_students * T1.pct_female_students AS REAL) / 100 FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T1.year = 2011 AND T2.university_name = 'Stanford University'
in 2011 refers to year 2011; female students refers to DIVIDE(MULTIPLY(pct_female_students, num_students), 100); Stanford University refers to university_name = 'Stanford University';
[ "university.id", "university.university_name", "university_year.num_students", "university_year.pct_female_students", "university_year.university_id", "university_year.year" ]
8,038
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
In which nation is Harvard University located?
SELECT T2.country_name FROM university AS T1 INNER JOIN country AS T2 ON T1.country_id = T2.id WHERE T1.university_name = 'Harvard University'
Harvard University refers to university_name = 'Harvard University'; nation refers to country_name
[ "country.country_name", "country.id", "university.country_id", "university.university_name" ]
8,039
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the name of the ranking system for Teaching criteria?
SELECT T1.system_name FROM ranking_system AS T1 INNER JOIN ranking_criteria AS T2 ON T1.id = T2.ranking_system_id WHERE T2.criteria_name = 'Teaching'
Teaching criteria refers to criteria_name = 'Teaching'; name of the ranking system refers to system_name
[ "ranking_criteria.criteria_name", "ranking_criteria.ranking_system_id", "ranking_system.id", "ranking_system.system_name" ]
8,040
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Name the most famous university in Argentina.
SELECT T1.university_name FROM university AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.university_id INNER JOIN country AS T3 ON T3.id = T1.country_id WHERE T3.country_name = 'Argentina' GROUP BY T1.university_name ORDER BY SUM(T2.score) DESC LIMIT 1
in Argentina refers to country_name = 'Argentina';  most famous refers to MAX(SUM(score))
[ "country.country_name", "country.id", "university.country_id", "university.id", "university.university_name", "university_ranking_year.score", "university_ranking_year.university_id" ]
8,041
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
In Argentina, how many universities are there?
SELECT COUNT(*) FROM university AS T1 INNER JOIN country AS T2 ON T1.country_id = T2.id WHERE T2.country_name = 'Argentina'
In Argentina refers to country_name = 'Argentina';
[ "country.country_name", "country.id", "university.country_id" ]
8,042
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Which universities have more than 100,000 students in 2011?
SELECT T2.university_name FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T1.year = 2011 AND T1.num_students > 100000
in 2011 refers to year 2011; more than 100,000 students refers to num_students > 100000; which university refers to university_name;
[ "university.id", "university.university_name", "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,043
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many criteria are associated with ranking system Center for World University Rankings?
SELECT COUNT(T2.criteria_name) FROM ranking_system AS T1 INNER JOIN ranking_criteria AS T2 ON T1.id = T2.ranking_system_id WHERE T1.system_name = 'Center for World University Rankings'
ranking system Center for World University Rankings refers to system_name = 'Center for World University Rankings';
[ "ranking_criteria.criteria_name", "ranking_criteria.ranking_system_id", "ranking_system.id", "ranking_system.system_name" ]
8,044
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many students at the university earned a score of 90 in 2011?
SELECT COUNT(*) FROM university_year AS T1 INNER JOIN university_ranking_year AS T2 ON T1.university_id = T2.university_id WHERE T2.score = 90 AND T1.year = 2011
in 2011 refers to year 2011; earned a score of 90 refers to score = 90;
[ "university_ranking_year.score", "university_ranking_year.university_id", "university_year.university_id", "university_year.year" ]
8,045
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What are the top three universities with the most international students?
SELECT DISTINCT T2.university_name FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id GROUP BY T2.university_name ORDER BY SUM(T1.num_students * T1.pct_international_students / 100) DESC LIMIT 3
most international students refers to MAX(SUM(DIVIDE(MULTIPLE(pct_international_students, num_students), 100))); name of university refers to university_name;
[ "university.id", "university.university_name", "university_year.num_students", "university_year.pct_international_students", "university_year.university_id" ]
8,046
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the difference in overall student enrollment and international student enrollment at the Harvard university from 2011 to 2012?
SELECT SUM(T1.num_students) - SUM(CAST(T1.num_students * T1.pct_international_students AS REAL) / 100) FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T2.university_name = 'Harvard University' AND T1.year BETWEEN 2011 AND 2012
Harvard University refers to university_name = 'Harvard University'; difference in overall student enrollment and international student refers to SUBTRACT(SUM(num_students), SUM(DIVIDE(MULTIPLY(pct_international_students, num_students), 100))); from 2011 to 2012 refers to year BETWEEN 2011 AND 2012
[ "university.id", "university.university_name", "university_year.num_students", "university_year.pct_international_students", "university_year.university_id", "university_year.year" ]
8,047
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many universities had over 30000 students in 2011?
SELECT COUNT(*) FROM university_year WHERE year = 2011 AND num_students > 30000
in 2011 refers to year 2011; had over 30000 students refers to num_students > 30000;
[ "university_year.num_students", "university_year.year" ]
8,048
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the country ID of the University of Tokyo?
SELECT country_id FROM university WHERE university_name = 'University of Tokyo'
University of Tokyo refers to university_name = 'University of Tokyo';
[ "university.country_id", "university.university_name" ]
8,049
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Provide the ranking system ID of the Center for World University Rankings.
SELECT id FROM ranking_system WHERE system_name = 'Center for World University Rankings'
the Center for World University Rankings refers to system_name = 'Center for World University Rankings';
[ "ranking_system.id", "ranking_system.system_name" ]
8,050
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the ID of the Publications Rank criteria?
SELECT id FROM ranking_criteria WHERE criteria_name = 'Publications Rank'
Publications Rank criteria refers to criteria_name = 'Publications Rank';
[ "ranking_criteria.criteria_name", "ranking_criteria.id" ]
8,051
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many universities had above 30% of international students in 2013?
SELECT COUNT(*) FROM university_year WHERE pct_international_students > 30 AND year = 2013
had above 30% of international students refers to pct_international_students > 30; in 2013 refers to year = 2013
[ "university_year.pct_international_students", "university_year.year" ]
8,052
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many universities got less than 50 scores under ranking criteria ID 6 in 2011?
SELECT COUNT(*) FROM university_ranking_year WHERE ranking_criteria_id = 6 AND year = 2011 AND score < 50
in 2011 refers to year 2011; less than 50 scores refers to score < 50;
[ "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.year" ]
8,053
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Provide the number of students at Yale University in 2016.
SELECT T1.num_students FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T2.university_name = 'Yale University' AND T1.year = 2016
number of students refers to num_students; Yale University refers to university_name = 'Yale University'; in 2016 refers to year = 2016
[ "university.id", "university.university_name", "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,054
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
List the universities in Denmark.
SELECT T1.university_name FROM university AS T1 INNER JOIN country AS T2 ON T1.country_id = T2.id WHERE T2.country_name = 'Denmark'
in Denmark refers to country_name = 'Denmark'; name of university refers to university_name;
[ "country.country_name", "country.id", "university.country_id", "university.university_name" ]
8,055
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Provide the number of staff at the University of Auckland in 2015.
SELECT CAST(SUM(T1.num_students) AS REAL) / SUM(T1.student_staff_ratio) FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T2.university_name = 'University of Auckland' AND T1.year = 2015
University of Auckland refers to university_name = 'University of Auckland'; in 2015 refers to year = 2015; number of staff refers to DIVIDE(num_students, student_staff_ratio)
[ "university.id", "university.university_name", "university_year.num_students", "university_year.student_staff_ratio", "university_year.university_id", "university_year.year" ]
8,056
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Which country has the University of São Paulo?
SELECT T2.country_name FROM university AS T1 INNER JOIN country AS T2 ON T1.country_id = T2.id WHERE T1.university_name = 'University of São Paulo'
the University of São Paulo refers to university_name = 'University of São Paulo'; which country refers to country_name;
[ "country.country_name", "country.id", "university.country_id", "university.university_name" ]
8,057
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many international students attended Harvard University in 2012?
SELECT CAST(T2.num_students * T2.pct_international_students AS REAL) / 100 FROM university AS T1 INNER JOIN university_year AS T2 ON T1.id = T2.university_id WHERE T1.university_name = 'Harvard University' AND T2.year = 2012
Harvard University refers to university_name = 'Harvard University'; international students refers to DIVIDE(MULTIPLY(num_students, pct_international_students), 100); in 2012 refers to year = 2012
[ "university.id", "university.university_name", "university_year.num_students", "university_year.pct_international_students", "university_year.university_id", "university_year.year" ]
8,058
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Calculate the number of female students at Arizona State University in 2014.
SELECT CAST(T2.num_students * T2.pct_female_students AS REAL) / 100 FROM university AS T1 INNER JOIN university_year AS T2 ON T1.id = T2.university_id WHERE T1.university_name = 'Arizona State University' AND T2.year = 2014
female students refers to DIVIDE(MULTIPLY(pct_female_students, num_students), 100); at Arizona State University refers to university_name = 'Arizona State University'; in 2014 refers to year = 2014
[ "university.id", "university.university_name", "university_year.num_students", "university_year.pct_female_students", "university_year.university_id", "university_year.year" ]
8,059
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Provide the universities which got the highest scores.
SELECT T1.university_name FROM university AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.university_id GROUP BY T1.university_name ORDER BY SUM(T2.score) DESC LIMIT 1
got the highest scores refers to MAX(SUM(score))
[ "university.id", "university.university_name", "university_ranking_year.score", "university_ranking_year.university_id" ]
8,060
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
List the ranking criteria under the Shanghai Ranking system.
SELECT T2.criteria_name FROM ranking_system AS T1 INNER JOIN ranking_criteria AS T2 ON T1.id = T2.ranking_system_id WHERE T1.system_name = 'Shanghai Ranking'
Shanghai Ranking system refers to system_name = 'Shanghai Ranking'; ranking criteria refers to criteria_name
[ "ranking_criteria.criteria_name", "ranking_criteria.ranking_system_id", "ranking_system.id", "ranking_system.system_name" ]
8,061
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
In 2011, which university got the lowest score in teaching criteria?
SELECT T3.university_name FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id INNER JOIN university AS T3 ON T3.id = T2.university_id WHERE T1.criteria_name = 'Teaching' AND T2.year = 2011 ORDER BY T2.score ASC LIMIT 1
in 2011 refers to year 2011; got the lowest score refers to MIN(score), teaching criteria refers to criteria_name = 'Teaching'
[ "ranking_criteria.criteria_name", "ranking_criteria.id", "university.id", "university.university_name", "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.university_id", "university_ranking_year.year" ]
8,062
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Provide the ranking system name for the "Quality of Education Rank" criteria.
SELECT T1.system_name FROM ranking_system AS T1 INNER JOIN ranking_criteria AS T2 ON T1.id = T2.ranking_system_id WHERE T2.criteria_name = 'Quality of Education Rank'
the "Quality of Education Rank" criteria refers to criteria_name = 'Quality of Education Rank'; ranking system refers to system_name
[ "ranking_criteria.criteria_name", "ranking_criteria.ranking_system_id", "ranking_system.id", "ranking_system.system_name" ]
8,063
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many percent of universities got a score above 80 under International criteria in 2016? Among them, name the university which got the highest score.
SELECT CAST(SUM(CASE WHEN T2.score > 80 THEN 1 ELSE 0 END) AS REAL) / COUNT(*), ( SELECT T3.university_name FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id INNER JOIN university AS T3 ON T3.id = T2.university_id WHERE T1.criteria_name = 'International' AND T2.year ...
got a score above 80 refers to score > 80; under International criteria refers to criteria_name = 'International'; in 2016 refers to year = 2016; highest score refers to MAX(score)
[ "ranking_criteria.criteria_name", "ranking_criteria.id", "university.id", "university.university_name", "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.university_id", "university_ranking_year.year" ]
8,064
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Provide the ranking criteria and scores in 2005 that were received by Harvard University.
SELECT T1.criteria_name, T2.score FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id INNER JOIN university AS T3 ON T3.id = T2.university_id WHERE T3.university_name = 'Harvard University' AND T2.year = 2005
Harvard University refers to university_name = 'Harvard University'; in 2005 refers to year = 2005; ranking criteria refers to criteria_name;
[ "ranking_criteria.criteria_name", "ranking_criteria.id", "university.id", "university.university_name", "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.university_id", "university_ranking_year.year" ]
8,065
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Calculate the average score per university under Alumni criteria in 2008.
SELECT AVG(T2.score) FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id WHERE T1.criteria_name = 'Alumni' AND T2.year = 2008
under Alumni criteria refers to criteria_name = 'Alumni'; in 2008 refers to year = 2008; average score refers to DIVIDE(SUM(score), COUNT(university_id))
[ "ranking_criteria.criteria_name", "ranking_criteria.id", "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.year" ]
8,066
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Name the university and country which had the highest number of international students in 2015.
SELECT T1.university_name, T3.country_name FROM university AS T1 INNER JOIN university_year AS T2 ON T1.id = T2.university_id INNER JOIN country AS T3 ON T3.id = T1.country_id WHERE T2.year = 2015 ORDER BY T2.num_students DESC LIMIT 1
highest number of international students refers to MAX(DIVIDE(MULTIPLY(num_students, pct_international_students), 100)); in 2015 refers to year = 2015; name of university refers to university_name;
[ "country.country_name", "country.id", "university.country_id", "university.id", "university.university_name", "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,067
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many students were there in university ID 1 in 2011?
SELECT num_students FROM university_year WHERE year = 2011 AND university_id = 1
in 2011 refers to year 2011;
[ "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,068
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the ID of the university with the most students in 2011?
SELECT university_id FROM university_year WHERE year = 2011 ORDER BY num_students DESC LIMIT 1
in 2011 refers to year 2011; with the most students refers to MAX(num_students); ID of the university refers to university_id
[ "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,069
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Please list the IDs of the universities with a student staff ratio of over 15 in 2011.
SELECT university_id FROM university_year WHERE year = 2011 AND student_staff_ratio > 15
in 2011 refers to year 2011; student staff ratio of over 15 refers to student_staff_ratio > 15; ID of the university refers to university_id
[ "university_year.student_staff_ratio", "university_year.university_id", "university_year.year" ]
8,070
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Among the universities with over 20000 students in 2011, how many of them have an international students percentage of over 25% in the same year?
SELECT COUNT(*) FROM university_year WHERE year = 2011 AND pct_international_students > 25 AND num_students > 20000
in 2011 refers to year 2011; with over 20000 students refers to num_students > 20000; international students percentage of over 25% refers to pct_international_students > 25;
[ "university_year.num_students", "university_year.pct_international_students", "university_year.year" ]
8,071
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Please list the IDs of the universities with the top 3 female students percentage in 2011.
SELECT university_id FROM university_year WHERE year = 2011 ORDER BY pct_female_students DESC LIMIT 3
in 2011 refers to year 2011; top 3 female students percentage refers to MAX(pct_female_students) LIMIT 3; ID of the university refers to university_id
[ "university_year.pct_female_students", "university_year.university_id", "university_year.year" ]
8,072
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
In which year did university ID 1 have the most students?
SELECT year FROM university_year WHERE university_id = 1 ORDER BY num_students DESC LIMIT 1
have the most students refers to MAX(num_students)
[ "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,073
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many students did Harvard University have in 2011?
SELECT T1.num_students FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T2.university_name = 'Harvard University' AND T1.year = 2011
in 2011 refers to year 2011; Harvard University refers to university_name = 'Harvard University';
[ "university.id", "university.university_name", "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,074
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the name of the university with the most international students in 2011?
SELECT T2.university_name FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T1.year = 2011 ORDER BY T1.pct_international_students DESC LIMIT 1
in 2011 refers to year 2011; the most international students refers to MAX(DIVIDE(MULTIPLY(num_students,  pct_international_students), 100)); name of university refers to university_id
[ "university.id", "university.university_name", "university_year.pct_international_students", "university_year.university_id", "university_year.year" ]
8,075
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Please list the names of all the universities in Australia.
SELECT T1.university_name FROM university AS T1 INNER JOIN country AS T2 ON T1.country_id = T2.id WHERE T2.country_name = 'Australia'
in Australia refers to country_name = 'Australia'; name of university refers to university_name
[ "country.country_name", "country.id", "university.country_id", "university.university_name" ]
8,076
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Among the universities in Australia, how many of them have more than 15000 students in 2011?
SELECT COUNT(*) FROM university AS T1 INNER JOIN university_year AS T2 ON T1.id = T2.university_id INNER JOIN country AS T3 ON T3.id = T1.country_id WHERE T3.country_name = 'Australia' AND T2.year = 2011 AND T2.num_students > 15000
in 2011 refers to year 2011; have more than 15000 students refers to num_students > 15000; in Australia refers to country_name = 'Australia';
[ "country.country_name", "country.id", "university.country_id", "university.id", "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,077
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Which country is Harvard University in?
SELECT T2.country_name FROM university AS T1 INNER JOIN country AS T2 ON T1.country_id = T2.id WHERE T1.university_name = 'Harvard University'
Harvard University refers to university_name = 'Harvard University'; which country refers to country_name
[ "country.country_name", "country.id", "university.country_id", "university.university_name" ]
8,078
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the name of the university with the highest score in teaching in the year 2011?
SELECT T3.university_name FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id INNER JOIN university AS T3 ON T3.id = T2.university_id WHERE T1.criteria_name = 'Teaching' AND T2.year = 2011 ORDER BY T2.score DESC LIMIT 1
with the highest score refers to MAX(score); in teaching refers to criteria_name = 'Teaching'; name of university refers to university_name;
[ "ranking_criteria.criteria_name", "ranking_criteria.id", "university.id", "university.university_name", "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.university_id", "university_ranking_year.year" ]
8,079
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Please list the names of the universities with a score in teaching of over 90 in 2011.
SELECT T3.university_name FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id INNER JOIN university AS T3 ON T3.id = T2.university_id WHERE T1.criteria_name = 'Teaching' AND T2.year = 2011 AND T2.score > 90
in 2011 refers to year 2011; in teaching refers to  criteria_name = 'Teaching'; score in teaching of over 90 refers to score > 90; name of university refers to university_name;
[ "ranking_criteria.criteria_name", "ranking_criteria.id", "university.id", "university.university_name", "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.university_id", "university_ranking_year.year" ]
8,080
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Among the universities with a score in teaching of over 90 in 2011, how many of them are in the United States of America?
SELECT COUNT(*) FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id INNER JOIN university AS T3 ON T3.id = T2.university_id WHERE T1.criteria_name = 'Teaching' AND T2.year = 2011 AND T2.score > 90
in 2011 refers to year 2011; in teaching refers to  criteria_name = 'Teaching'; score in teaching of over 90 refers to score > 90; in the United States of America refers to country_name = 'United States of America';
[ "ranking_criteria.criteria_name", "ranking_criteria.id", "university.id", "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.university_id", "university_ranking_year.year" ]
8,081
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Please list the names of all the ranking criteria of Harvard University in 2011.
SELECT T1.criteria_name FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id INNER JOIN university AS T3 ON T3.id = T2.university_id WHERE T3.university_name = 'Harvard University' AND T2.year = 2011
in 2011 refers to year 2011; Harvard University refers to university_name = 'Harvard University'; names of all the ranking criteria refers to criteria_name
[ "ranking_criteria.criteria_name", "ranking_criteria.id", "university.id", "university.university_name", "university_ranking_year.ranking_criteria_id", "university_ranking_year.university_id", "university_ranking_year.year" ]
8,082
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What are the names of the universities that got 98 in teaching in 2011?
SELECT T3.university_name FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id INNER JOIN university AS T3 ON T3.id = T2.university_id WHERE T1.criteria_name = 'Teaching' AND T2.year = 2011 AND T2.score = 98
in 2011 refers to year 2011; that got 98 refers to score = 98; in teaching refers to criteria_name = 'Teaching'; name of university refers to university_name
[ "ranking_criteria.criteria_name", "ranking_criteria.id", "university.id", "university.university_name", "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.university_id", "university_ranking_year.year" ]
8,083
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Please list the names of all the universities that scored under 60 in teaching in 2011 and are in the United States of America.
SELECT T3.university_name FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id INNER JOIN university AS T3 ON T3.id = T2.university_id INNER JOIN country AS T4 ON T4.id = T3.country_id WHERE T4.country_name = 'United States of America' AND T2.year = 2011 AND T2.score < ...
scored under 60 refers to score < 60; in 2011 refers to year 2011; in teaching refers to criteria_name = 'Teaching'; in the United States of America refers to country_name = 'United States of America';
[ "country.country_name", "country.id", "ranking_criteria.criteria_name", "ranking_criteria.id", "university.country_id", "university.id", "university.university_name", "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.university_id", "univers...
8,084
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Among the universities in Australia, how many of them have a student staff ratio of over 15 in 2011?
SELECT COUNT(*) FROM university AS T1 INNER JOIN university_year AS T2 ON T1.id = T2.university_id INNER JOIN country AS T3 ON T3.id = T1.country_id WHERE T3.country_name = 'Australia' AND T2.student_staff_ratio > 15 AND T2.year = 2011
in 2011 refers to year 2011; in Australia refers to country_name = 'Australia'; student staff ratio of over 15 refers to student_staff_ratio > 15
[ "country.country_name", "country.id", "university.country_id", "university.id", "university_year.student_staff_ratio", "university_year.university_id", "university_year.year" ]
8,085
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many female students did Stanford University have in 2011?
SELECT CAST(T1.num_students * T1.pct_female_students AS REAL) / 100 FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T1.year = 2011 AND T2.university_name = 'Stanford University'
in 2011 refers to year 2011; female students refers to DIVIDE(MULTIPLY(pct_female_students, num_students), 100); Stanford University refers to university_name = 'Stanford University';
[ "university.id", "university.university_name", "university_year.num_students", "university_year.pct_female_students", "university_year.university_id", "university_year.year" ]
8,086
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Among the universities with a score in teaching of over 90 in 2011, what is the percentage of those in the United States of America?
SELECT CAST(SUM(CASE WHEN T4.country_name = 'United States of America' THEN 1 ELSE 0 END) AS REAL) * 100 / COUNT(*) AS per FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id INNER JOIN university AS T3 ON T3.id = T2.university_id INNER JOIN country AS T4 ON T4.id = T3...
in 2011 refers to year 2011; in teaching refers to  criteria_name = 'Teaching'; score in teaching of over 90 refers to score > 90; in the United States of America refers to country_name = 'United States of America'; percentage refers to DIVIDE(COUNT(country_name = 'United States of America'), COUNT(id))
[ "country.country_name", "country.id", "ranking_criteria.criteria_name", "ranking_criteria.id", "university.country_id", "university.id", "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.university_id", "university_ranking_year.year" ]
8,087
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Give the id of "Center for World University Rankings".
SELECT id FROM ranking_system WHERE system_name = 'Center for World University Rankings'
"Center for World University Rankings" refers to system_name = 'Center for World University Rankings';
[ "ranking_system.id", "ranking_system.system_name" ]
8,088
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Which country is University of Veterinary Medicine Vienna located in? Give its country id.
SELECT country_id FROM university WHERE university_name = 'University of Veterinary Medicine Vienna'
University of Veterinary Medicine Vienna refers to university_name = 'University of Veterinary Medicine Vienna';
[ "university.country_id", "university.university_name" ]
8,089
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the id of the criteria "Citations Rank"?
SELECT id FROM ranking_criteria WHERE criteria_name = 'Citations Rank'
criteria "Citations Rank" refers to criteria_name = 'Citations Rank';
[ "ranking_criteria.criteria_name", "ranking_criteria.id" ]
8,090
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Show the id of University of Orléans.
SELECT id FROM university WHERE university_name = 'University of Orléans'
University of Orléans refers to university_name = 'University of Orléans';
[ "university.id", "university.university_name" ]
8,091
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
For the university id 268, show its number of students in 2013.
SELECT num_students FROM university_year WHERE university_id = 268 AND year = 2013
number of students refers to num_students; in 2013 refers to year = 2013
[ "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,092
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Show the name of country id 66.
SELECT country_name FROM country WHERE id = 66
name of country refers to country_name
[ "country.country_name", "country.id" ]
8,093
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Which country is McMaster University located in?
SELECT T2.country_name FROM university AS T1 INNER JOIN country AS T2 ON T1.country_id = T2.id WHERE T1.university_name = 'McMaster University'
McMaster University refers to university_name = 'McMaster University'; which country refers to country_name
[ "country.country_name", "country.id", "university.country_id", "university.university_name" ]
8,094
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many Turkish universities are there in the database?
SELECT COUNT(*) FROM university AS T1 INNER JOIN country AS T2 ON T1.country_id = T2.id WHERE T2.country_name = 'Turkey'
Turkish universities refers to country_name = 'Turkey';
[ "country.country_name", "country.id", "university.country_id" ]
8,095
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
Which university had the most students in 2011? Show its name.
SELECT T2.university_name FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T1.year = 2011 ORDER BY T1.num_students DESC LIMIT 1
in 2011 refers to year 2011; the most students refers to MAX(num_students); which university refers to university_name;
[ "university.id", "university.university_name", "university_year.num_students", "university_year.university_id", "university_year.year" ]
8,096
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
How many students were there in University of Michigan in 2011?
SELECT COUNT(*) FROM university AS T1 INNER JOIN university_year AS T2 ON T1.id = T2.university_id WHERE T1.university_name = 'University of Michigan' AND T2.year = 2011
in 2011 refers to year 2011; in University of Michigan refers to university_name = 'University of Michigan';
[ "university.id", "university.university_name", "university_year.university_id", "university_year.year" ]
8,097
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
For Chosun University, what was its score on "Influence Rank" in 2015?
SELECT T2.score FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id INNER JOIN university AS T3 ON T3.id = T2.university_id WHERE T3.university_name = 'Chosun University' AND T1.criteria_name = 'Influence Rank' AND T2.year = 2015
Chosun University refers to university_name = 'Chosun University'; in 2015 refers to year = 2015; on "Influence Rank" refers to criteria_name = 'Influence Rank';
[ "ranking_criteria.criteria_name", "ranking_criteria.id", "university.id", "university.university_name", "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.university_id", "university_ranking_year.year" ]
8,098
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
What is the percentage of the international students in University of Oslo in 2015?
SELECT T2.pct_international_students FROM university AS T1 INNER JOIN university_year AS T2 ON T1.id = T2.university_id WHERE T1.university_name = 'University of Oslo' AND T2.year = 2015
percentage of the international students refers to pct_international_students; in 2015 refers to year = 2015; in University of Oslo refers to university_name = 'University of Oslo';
[ "university.id", "university.university_name", "university_year.pct_international_students", "university_year.university_id", "university_year.year" ]
8,099
university
CREATE TABLE country ( id INTEGER not null primary key, country_name TEXT default NULL ); CREATE TABLE ranking_system ( id INTEGER not null primary key, system_name TEXT default NULL ); CREATE TABLE ranking_criteria ( id INTEGER not null ...
For the University of Southampton in 2015, on which criteria did it score the best?
SELECT T1.criteria_name FROM ranking_criteria AS T1 INNER JOIN university_ranking_year AS T2 ON T1.id = T2.ranking_criteria_id INNER JOIN university AS T3 ON T3.id = T2.university_id WHERE T3.university_name = 'University of Southampton' AND T2.year = 2015 ORDER BY T2.score DESC LIMIT 1
University of Southampton refers to university_name = 'University of Southampton'; in 2015 refers to year = 2015; score the best refers to MAX(score); which criteria refers to criteria_name
[ "ranking_criteria.criteria_name", "ranking_criteria.id", "university.id", "university.university_name", "university_ranking_year.ranking_criteria_id", "university_ranking_year.score", "university_ranking_year.university_id", "university_ranking_year.year" ]