markdown stringlengths 0 37k | code stringlengths 1 33.3k | path stringlengths 8 215 | repo_name stringlengths 6 77 | license stringclasses 15
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Now let's use the above arrangement of axes as a template for the figurefirst:figure objects that we have drawn in an svg file. Below is a rendering of the svg file showing where we would like the templated sets of axes to go. | display(SVG('mpl_fig_for_templates.svg')) | examples/mpl_to_svg_layout/mpl_fig_to_figurefirst_svg.ipynb | FlyRanch/figurefirst | mit |
Now we call add_mpl_fig_to_figurefirst_svg with the necessary arguments to add a layer to the above svg file. This new layer is similar to the layer created previously using mpl_fig_to_figurefirst_svg, except that instead of placing this layer into an empty svg, we are adding it to an existing svg.
The one trick being ... | fifi_svg_filename = 'mpl_fig_for_templates.svg'
mpl_fig = fig
output_filename = 'mpl_fig_for_templates_output.svg'
layout = fifi.mpl_fig_to_figurefirst_svg.add_mpl_fig_to_figurefirst_svg( fifi_svg_filename,
mpl_fig,
... | examples/mpl_to_svg_layout/mpl_fig_to_figurefirst_svg.ipynb | FlyRanch/figurefirst | mit |
Now we can load the layout that has the added layer generated from the Matplotlib figure, and make a sample plot. | layout = fifi.svg_to_axes.FigureLayout('mpl_fig_for_templates_output.svg', make_mplfigures=True)
# collect the group and axis names
# Note: you could also do this by noting your figurefirst:figure names from the layout file
import numpy as np
groups_and_axes = np.unique(np.ravel(layout.axes.keys()))
groups = [g fo... | examples/mpl_to_svg_layout/mpl_fig_to_figurefirst_svg.ipynb | FlyRanch/figurefirst | mit |
The second line of this code uses the keyword if to tell Python that we want to make a choice. If the test that follows the if statement is true, the body of the if (i.e., the lines indented underneath it) are executed. If the test is false, the body of the else is executed instead. Only one or the other is ever execut... | num = 53
print('before conditional...')
if num > 100:
print('53 is greater than 100')
print('...after conditional') | 02-Python1/02-Python-1-Logic_Instructor.ipynb | OpenAstronomy/workshop_sunpy_astropy | mit |
We can also chain several tests together using elif, which is short for “else if”. The following Python code uses elif to print the sign of a number. | num = -3
if num > 0:
print(num, "is positive")
elif num == 0:
print(num, "is zero")
else:
print(num, "is negative") | 02-Python1/02-Python-1-Logic_Instructor.ipynb | OpenAstronomy/workshop_sunpy_astropy | mit |
One important thing to notice in the code above is that we use a double equals sign == to test for equality rather than a single equals sign because the latter is used to mean assignment.
We can also combine tests using and and or. and is only true if both parts are true: | if (1 > 0) and (-1 > 0):
print('both parts are true')
else:
print('at least one part is false') | 02-Python1/02-Python-1-Logic_Instructor.ipynb | OpenAstronomy/workshop_sunpy_astropy | mit |
Checking out Data
Now that we’ve seen how conditionals work, we can use them to check for the suspicious features we saw in our inflammation data. In the first couple of plots, the maximum inflammation per day seemed to rise like a straight line, one unit per day.
We can check for this inside the for loop we wrote wit... | import numpy as np
data = np.loadtxt(fname='data/inflammation-01.csv', delimiter=',')
if data.max(axis=0)[0] == 0 and data.max(axis=0)[20] == 20:
print('Suspicious looking maxima!')
elif data.min(axis=0).sum() == 0:
print('Minima add up to zero!')
else:
print('Seems OK!')
data = numpy.loadtxt(fname='infla... | 02-Python1/02-Python-1-Logic_Instructor.ipynb | OpenAstronomy/workshop_sunpy_astropy | mit |
<a id='strings'></a>
Strings
In Python, human language text gets represented as a string. These contain sequential sets of characters and they are offset by quotation marks, either double (") or single (').
We will explore different kinds of operations in Python that are specific to human language objects, but it is us... | # The iconic string
print("Hello, World!")
# Assign these strings to variables
a = "Hello"
b = 'World'
# Try out arithmetic operations.
# When we add strings we call it 'concatenation'
print(a+" "+b)
print(a*5)
# Unlike a number that consists of a single value, a string is an ordered
# sequence of characters. We ... | 01-IntroToPython/.ipynb_checkpoints/00-PythonBasics-checkpoint.ipynb | lknelson/text-analysis-2017 | bsd-3-clause |
<a id='lists'></a>
Lists
The numbers and strings we have just looked at are the two basic data types that we will focus our attention on in this workshop. (In a few days, we will look at a third data type, boolean, which consists of just True/False values.) When we are working with just a few numbers or strings, it is ... | # Let's assign a couple lists to variables
list1 = ['Call', 'me', 'Ishmael']
list2 = ['In', 'the', 'beginning']
## Q. Predict what will happen when we perform the following operations
print(list1+list2)
print(list1*5)
# As with a string, we can find out the length of a list
len(list1)
# Sometimes we just want a s... | 01-IntroToPython/.ipynb_checkpoints/00-PythonBasics-checkpoint.ipynb | lknelson/text-analysis-2017 | bsd-3-clause |
<a id='tricks'></a>
A couple of useful tricks
<a id='string_methods'></a>
String Methods
The creators of Python recognize that human language has many important yet idiosyncratic features, so they have tried to make it easy for us to identify and manipulate them. For example, in the demonstration at the very beginning ... | # Let's assign a variable to perform methods upon
greeting = "Hello, World!"
# We saw the 'endswith' method at the very beginning
# Note the type of output that gets printed
greeting.startswith('H'), greeting.endswith('d')
# We can check whether the string is a letter or a number
this_string = 'f'
this_string.isa... | 01-IntroToPython/.ipynb_checkpoints/00-PythonBasics-checkpoint.ipynb | lknelson/text-analysis-2017 | bsd-3-clause |
<a id='list_methods'></a>
List Comprehension
List comprehensions are a fairly advanced programming technique that we will spend more time talking about tomorrow. For now, you can think of them as list filters. Often, we don't need every value in a list, just a few that fulfill certain criteria. | # 'list1' had contained three words, two of which were in title case.
# We can automatically return those words using a list comprehension
[word for word in list1 if word.istitle()]
# Or we can include all the words in the list but just take their first letters
[word[0] for word in list1]
for word in list1:
prin... | 01-IntroToPython/.ipynb_checkpoints/00-PythonBasics-checkpoint.ipynb | lknelson/text-analysis-2017 | bsd-3-clause |
which uses pickle internally, optionally mmap‘ing the model’s internal large NumPy matrices into virtual memory directly from disk files, for inter-process memory sharing.
In addition, you can load models created by the original C tool, both using its text and binary formats:
model = gensim.models.Word2Vec.load_word2ve... | model = gensim.models.Word2Vec.load(temp_path)
more_sentences = [['Advanced', 'users', 'can', 'load', 'a', 'model', 'and', 'continue',
'training', 'it', 'with', 'more', 'sentences']]
model.build_vocab(more_sentences, update=True)
model.train(more_sentences, )
# cleaning up temp
os.close(fs)
os.remov... | docs/notebooks/word2vec.ipynb | isohyt/gensim | lgpl-2.1 |
Your task is to provide a drop-down so the user of the program can select one of the 5 mailboxes. Upon running the interaction the program will:
read the selected mailbox file a line at a time
find any lines beginning with From:.
extract out the email address from the From: line.
use the isEmail() function (provided b... | # Step 2: Write code here
| content/lessons/07-Files/HW-Files.ipynb | IST256/learn-python | mit |
Part 3: Questions
Did you write your own user-defined function? For what purpose?
--== Double-Click and Write Your Answer Below This Line ==--
Explain how you might re-write this program to create one large file from all the mailboxes. No code, just explain it.
--== Double-Click and Write Your Answer Below This L... | # run this code to turn in your work!
from coursetools.submission import Submission
Submission().submit() | content/lessons/07-Files/HW-Files.ipynb | IST256/learn-python | mit |
From the above info(),We can see columns Age, Cabin and Embarked have missing values.
Handling the missing values:
Ignore the rows with missing data,
Exclude the variable at all or we might substite it with mean or median.
Age 80% of the data is available,which seems a important variable so not to exclude.
Port of ... | #exculding some coloumns
for a in ['Ticket','Cabin','Embarked','Name','PassengerId','Fare']:
if a in data.columns:
del data[a]
print "Age median values by Age and Sex:"
#we are grouping by gender and class and taking median of age so we can replace with corrresponding values instead of NaN
print data.grou... | P6 data_viz/data/.ipynb_checkpoints/P2-checkpoint.ipynb | gangadhara691/gangadhara691.github.io | mit |
Data Exploration and Visualization | data_s=data
survival_group = data_s.groupby('Survived')
survival_group.describe() | P6 data_viz/data/.ipynb_checkpoints/P2-checkpoint.ipynb | gangadhara691/gangadhara691.github.io | mit |
From the above statistics
- Youngest to survive: 0.42
- Youngest to die: 1.0
- Oldest to survive: 80.0
- Oldest to die: 74.0 | import matplotlib.pyplot as plt
import seaborn as sns
# Set style for all graphs
#sns.set_style("light")
#sns.set_style("whitegrid")
sns.set_style("ticks", {"xtick.major.size": 8, "ytick.major.size": 8}) | P6 data_viz/data/.ipynb_checkpoints/P2-checkpoint.ipynb | gangadhara691/gangadhara691.github.io | mit |
From the above plot we can see how female individuals are given 1st preference and based on class.
Social-economic standing was a factor in survival rate of passengers by gender
Class 1 - female survival rate: 96.81%
Class 1 - male survival rate: 36.89%
Class 2 - female survival rate: 92.11%
Class 2 - male surviv... | def group(d,v):
if (d == 'female') and (v >= 18):
return 'Woman'
elif v < 18:
return 'child'
elif (d == 'male') and (v >= 18):
return 'Man'
data['Category'] = data.apply(lambda row:group(row['Sex'], row['Age']), axis=1)
data.head(5)
# We are dividing the ... | P6 data_viz/data/.ipynb_checkpoints/P2-checkpoint.ipynb | gangadhara691/gangadhara691.github.io | mit |
Load some example scores (output of a classifier)
The example data (from binary classification), presented next, contains:
- Column 1: scores or probas (output of predict_proba()) in the range [0, 1]
- Column 2: target or actual outcome (y truth) | df_results = pd.read_csv('../data/classifier_prediction_scores.csv')
print('Number of rows:', df_results.shape[0])
df_results.head() | units/11-validation-metrics/examples/01 - Validation Metrics for Classification.ipynb | LDSSA/learning-units | mit |
Let's take a look at the scores distribution. As an output of the predict_proba(), the scores range is [0, 1]. | df_results['scores'].hist(bins=50)
plt.ylabel('Frequency')
plt.xlabel('Scores')
plt.title('Distribution of Scores')
plt.xlim(0, 1)
plt.show() | units/11-validation-metrics/examples/01 - Validation Metrics for Classification.ipynb | LDSSA/learning-units | mit |
Classification Metrics
Accuracy score
The accuracy_score is the fraction (default) or the count (normalize=False) of correct predictions. It is given by:
$$ A = \frac{TP + TN}{TP + TN + FP + FN} $$
Where, TP is the True Positives, TN the True Negatives, FP the False Positives, and False Negative.
Disavantages:
- Not ... | # Specifying the threshold above which the predicted label is considered 1:
threshold = 0.50
# Generate the predicted labels (above threshold = 1, below = 0)
predicted_outcome = [0 if k <= threshold else 1 for k in df_results['scores']]
print('Accuracy = %2.3f' % accuracy_score(df_results['target'], predicted_outcome)... | units/11-validation-metrics/examples/01 - Validation Metrics for Classification.ipynb | LDSSA/learning-units | mit |
Confusion Matrix
The confusion_matrix C provides several performance indicators:
- C(0,0) - TN count
- C(1,0) - FN count
- C(0,1) - FP count
- C(1,1) - TP count | # Get the confusion matrix:
confmat = confusion_matrix(y_true=df_results['target'], y_pred=predicted_outcome)
# Plot the confusion matrix
fig, ax = plt.subplots(figsize=(5, 5))
ax.matshow(confmat, cmap=plt.cm.Blues, alpha=0.4)
for i in range(confmat.shape[0]):
for j in range(confmat.shape[1]):
ax.text(x=j,... | units/11-validation-metrics/examples/01 - Validation Metrics for Classification.ipynb | LDSSA/learning-units | mit |
As can be seen, the number of False Negatives is very high, which, depending on the business could be harmful.
Precision, Recall and F1-score
Precision is the ability of the classifier not to label as positive a sample that is negative (i.e., a measure of result relevancy).
$$ P = \frac{T_P}{T_P+F_P} $$
Recall is... | df_results['target'].value_counts(normalize=True) | units/11-validation-metrics/examples/01 - Validation Metrics for Classification.ipynb | LDSSA/learning-units | mit |
Rather imbalanced! Approximately 83% of the labels are 0. Let's take a look at the other metrics more appropriate for this type of datasets: | print('Precision score = %1.3f' % precision_score(df_results['target'], predicted_outcome))
print('Recall score = %1.3f' % recall_score(df_results['target'], predicted_outcome))
print('F1 score = %1.3f' % f1_score(df_results['target'], predicted_outcome)) | units/11-validation-metrics/examples/01 - Validation Metrics for Classification.ipynb | LDSSA/learning-units | mit |
As you can see, the results actually not so good as the accuracy metric would show us.
Receiver Operating Characteristic (ROC) and Area Under the ROC (AUROC)
The ROC is very common for binary classification problems. It is created by plotting the fraction of true positives out of the positives (TPR = true positive rate... | # Data to compute the ROC curve (FPR and TPR):
fpr, tpr, thresholds = roc_curve(df_results['target'], df_results['scores'])
# The Area Under the ROC curve:
roc_auc = roc_auc_score(df_results['target'], df_results['scores'])
# Plot ROC Curve
plt.figure(figsize=(8,6))
lw = 2
plt.plot(fpr, tpr, color='orange', lw=lw, lab... | units/11-validation-metrics/examples/01 - Validation Metrics for Classification.ipynb | LDSSA/learning-units | mit |
Time to build the network
Below you'll build your network. We've built out the structure and the backwards pass. You'll implement the forward pass through the network. You'll also set the hyperparameters: the learning rate, the number of hidden units, and the number of training passes.
<img src="assets/neural_network.p... | class NeuralNetwork(object):
def __init__(self, input_nodes, hidden_nodes, output_nodes, learning_rate):
# Set number of nodes in input, hidden and output layers.
self.input_nodes = input_nodes
self.hidden_nodes = hidden_nodes
self.output_nodes = output_nodes
# Initialize we... | first-neural-network/Your_first_neural_network.ipynb | gururajl/deep-learning | mit |
Training the network
Here you'll set the hyperparameters for the network. The strategy here is to find hyperparameters such that the error on the training set is low, but you're not overfitting to the data. If you train the network too long or have too many hidden nodes, it can become overly specific to the training se... | import sys
### Set the hyperparameters here ###
iterations = 5500
learning_rate = 0.4
hidden_nodes = 30
output_nodes = 1
N_i = train_features.shape[1]
network = NeuralNetwork(N_i, hidden_nodes, output_nodes, learning_rate)
losses = {'train':[], 'validation':[]}
for ii in range(iterations):
# Go through a random ... | first-neural-network/Your_first_neural_network.ipynb | gururajl/deep-learning | mit |
Exercises
Exercise: In NSFG Cycles 6 and 7, the variable cmdivorcx contains the date of divorce for the respondent’s first marriage, if applicable, encoded in century-months.
Compute the duration of marriages that have ended in divorce, and the duration, so far, of marriages that are ongoing. Estimate the hazard and... | def CleanData(resp):
"""Cleans respondent data.
resp: DataFrame
"""
resp.cmdivorcx.replace([9998, 9999], np.nan, inplace=True)
resp["notdivorced"] = resp.cmdivorcx.isnull().astype(int)
resp["duration"] = (resp.cmdivorcx - resp.cmmarrhx) / 12.0
resp["durationsofar"] = (resp.cmintvw - resp.c... | solutions/chap13soln.ipynb | AllenDowney/ThinkStats2 | gpl-3.0 |
In this equation:
$\epsilon$ is the single particle energy.
$\mu$ is the chemical potential, which is related to the total number of particles.
$k$ is the Boltzmann constant.
$T$ is the temperature in Kelvin.
In the cell below, typeset this equation using LaTeX:
YOUR ANSWER HERE
Define a function fermidist(energy, mu... | def fermidist(energy, mu, kT):
"""Compute the Fermi distribution at energy, mu and kT."""
# YOUR CODE HERE
f=1/(np.exp((energy-mu)/kT)+1)
return f
assert np.allclose(fermidist(0.5, 1.0, 10.0), 0.51249739648421033)
assert np.allclose(fermidist(np.linspace(0.0,1.0,10), 1.0, 10.0),
np.array([ 0.5249... | assignments/midterm/InteractEx06.ipynb | JackDi/phys202-2015-work | mit |
Write a function plot_fermidist(mu, kT) that plots the Fermi distribution $F(\epsilon)$ as a function of $\epsilon$ as a line plot for the parameters mu and kT.
Use enegies over the range $[0,10.0]$ and a suitable number of points.
Choose an appropriate x and y limit for your visualization.
Label your x and y axis and... | def plot_fermidist(mu, kT):
# YOUR CODE HERE
f=plt.figure(figsize=(9,6))
energy=np.linspace(0,10,100)
plt.plot(energy,fermidist(energy,mu,kT),'k')
plt.xlim(0,10)
plt.ylim(0,1)
plt.tick_params(direction='out')
plt.title('Fermi Dirac Distridution over a range of Energy Levels for Constant ... | assignments/midterm/InteractEx06.ipynb | JackDi/phys202-2015-work | mit |
Use interact with plot_fermidist to explore the distribution:
For mu use a floating point slider over the range $[0.0,5.0]$.
for kT use a floating point slider over the range $[0.1,10.0]$. | # YOUR CODE HERE
interact(plot_fermidist,mu=(0,5,0.1),kT=(0.1,10,0.1)) | assignments/midterm/InteractEx06.ipynb | JackDi/phys202-2015-work | mit |
Make population dynamic model
Basic parameters | pop_size = 60
seq_length = 100
alphabet = ['A', 'T', 'G', 'C']
base_haplotype = "AAAAAAAAAA" | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Setup a population of sequences
Store this as a lightweight Dictionary that maps a string to a count. All the sequences together will have count N. | pop = {}
pop["AAAAAAAAAA"] = 40
pop["AAATAAAAAA"] = 30
pop["AATTTAAAAA"] = 30
pop["AAATAAAAAA"] | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Add mutation
Mutations occur each generation in each individual in every basepair. | mutation_rate = 0.0001 # per gen per individual per site | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Walk through population and mutate basepairs. Use Poisson splitting to speed this up (you may be familiar with Poisson splitting from its use in the Gillespie algorithm).
In naive scenario A: take each element and check for each if event occurs. For example, 100 elements, each with 1% chance. This requires 100 random... | def get_mutation_count():
mean = mutation_rate * pop_size * seq_length
return np.random.poisson(mean) | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Here we use Numpy's Poisson random number. | get_mutation_count() | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
We need to get random haplotype from the population. | pop.keys()
[x/float(pop_size) for x in pop.values()]
def get_random_haplotype():
haplotypes = pop.keys()
frequencies = [x/float(pop_size) for x in pop.values()]
total = sum(frequencies)
frequencies = [x / total for x in frequencies]
return np.random.choice(haplotypes, p=frequencies) | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Here we use Numpy's weighted random choice. | get_random_haplotype() | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Here, we take a supplied haplotype and mutate a site at random. | def get_mutant(haplotype):
site = np.random.randint(seq_length)
possible_mutations = list(alphabet)
possible_mutations.remove(haplotype[site])
mutation = np.random.choice(possible_mutations)
new_haplotype = haplotype[:site] + mutation + haplotype[site+1:]
return new_haplotype
get_mutant("AAAAAA... | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Putting things together, in a single mutation event, we grab a random haplotype from the population, mutate it, decrement its count, and then check if the mutant already exists in the population. If it does, increment this mutant haplotype; if it doesn't create a new haplotype of count 1. | def mutation_event():
haplotype = get_random_haplotype()
if pop[haplotype] > 1:
pop[haplotype] -= 1
new_haplotype = get_mutant(haplotype)
if new_haplotype in pop:
pop[new_haplotype] += 1
else:
pop[new_haplotype] = 1
mutation_event()
pop | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
To create all the mutations that occur in a single generation, we draw the total count of mutations and then iteratively add mutation events. | def mutation_step():
mutation_count = get_mutation_count()
for i in range(mutation_count):
mutation_event()
mutation_step()
pop | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Add genetic drift
Given a list of haplotype frequencies currently in the population, we can take a multinomial draw to get haplotype counts in the following generation. | def get_offspring_counts():
haplotypes = pop.keys()
frequencies = [x/float(pop_size) for x in pop.values()]
return list(np.random.multinomial(pop_size, frequencies)) | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Here we use Numpy's multinomial random sample. | get_offspring_counts() | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
We then need to assign this new list of haplotype counts to the pop dictionary. To save memory and computation, if a haplotype goes to 0, we remove it entirely from the pop dictionary. | def offspring_step():
counts = get_offspring_counts()
for (haplotype, count) in zip(pop.keys(), counts):
if (count > 0):
pop[haplotype] = count
else:
del pop[haplotype]
offspring_step()
pop | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Combine and iterate
Each generation is simply a mutation step where a random number of mutations are thrown down, and an offspring step where haplotype counts are updated. | def time_step():
mutation_step()
offspring_step() | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Can iterate this over a number of generations. | generations = 500
def simulate():
for i in range(generations):
time_step()
simulate()
pop | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Record
We want to keep a record of past population frequencies to understand dynamics through time. At each step in the simulation, we append to a history object. | pop = {"AAAAAAAAAA": pop_size}
history = []
def simulate():
clone_pop = dict(pop)
history.append(clone_pop)
for i in range(generations):
time_step()
clone_pop = dict(pop)
history.append(clone_pop)
simulate()
pop
history[0]
history[1]
history[2] | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Analyze trajectories
Calculate diversity
Here, diversity in population genetics is usually shorthand for the statistic π, which measures pairwise differences between random individuals in the population. π is usually measured as substitutions per site. | pop | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
First, we need to calculate the number of differences per site between two arbitrary sequences. | def get_distance(seq_a, seq_b):
diffs = 0
length = len(seq_a)
assert len(seq_a) == len(seq_b)
for chr_a, chr_b in zip(seq_a, seq_b):
if chr_a != chr_b:
diffs += 1
return diffs / float(length)
get_distance("AAAAAAAAAA", "AAAAAAAAAB") | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
We calculate diversity as a weighted average between all pairs of haplotypes, weighted by pairwise haplotype frequency. | def get_diversity(population):
haplotypes = population.keys()
haplotype_count = len(haplotypes)
diversity = 0
for i in range(haplotype_count):
for j in range(haplotype_count):
haplotype_a = haplotypes[i]
haplotype_b = haplotypes[j]
frequency_a = population[hap... | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Plot diversity
Here, we use matplotlib for all Python plotting. | %matplotlib inline
import matplotlib.pyplot as plt
import matplotlib as mpl | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Here, we make a simple line plot using matplotlib's plot function. | plt.plot(get_diversity_trajectory()) | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Here, we style the plot a bit with x and y axes labels. | def diversity_plot():
mpl.rcParams['font.size']=14
trajectory = get_diversity_trajectory()
plt.plot(trajectory, "#447CCD")
plt.ylabel("diversity")
plt.xlabel("generation")
diversity_plot() | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Analyze and plot divergence
In population genetics, divergence is generally the number of substitutions away from a reference sequence. In this case, we can measure the average distance of the population to the starting haplotype. Again, this will be measured in terms of substitutions per site. | def get_divergence(population):
haplotypes = population.keys()
divergence = 0
for haplotype in haplotypes:
frequency = population[haplotype] / float(pop_size)
divergence += frequency * get_distance(base_haplotype, haplotype)
return divergence
def get_divergence_trajectory():
traject... | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Plot haplotype trajectories
We also want to directly look at haplotype frequencies through time. | def get_frequency(haplotype, generation):
pop_at_generation = history[generation]
if haplotype in pop_at_generation:
return pop_at_generation[haplotype]/float(pop_size)
else:
return 0
get_frequency("AAAAAAAAAA", 4)
def get_trajectory(haplotype):
trajectory = [get_frequency(haplotype, g... | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
We want to plot all haplotypes seen during the simulation. | def get_all_haplotypes():
haplotypes = set()
for generation in history:
for haplotype in generation:
haplotypes.add(haplotype)
return haplotypes
get_all_haplotypes() | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Here is a simple plot of their overall frequencies. | haplotypes = get_all_haplotypes()
for haplotype in haplotypes:
plt.plot(get_trajectory(haplotype))
plt.show()
colors = ["#781C86", "#571EA2", "#462EB9", "#3F47C9", "#3F63CF", "#447CCD", "#4C90C0", "#56A0AE", "#63AC9A", "#72B485", "#83BA70", "#96BD60", "#AABD52", "#BDBB48", "#CEB541", "#DCAB3C", "#E49938", "#E68133... | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
We can use stackplot to stack these trajectoies on top of each other to get a better picture of what's going on. | def stacked_trajectory_plot(xlabel="generation"):
mpl.rcParams['font.size']=18
haplotypes = get_all_haplotypes()
trajectories = [get_trajectory(haplotype) for haplotype in haplotypes]
plt.stackplot(range(generations), trajectories, colors=colors_lighter)
plt.ylim(0, 1)
plt.ylabel("frequency")
... | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Plot SNP trajectories | def get_snp_frequency(site, generation):
minor_allele_frequency = 0.0
pop_at_generation = history[generation]
for haplotype in pop_at_generation.keys():
allele = haplotype[site]
frequency = pop_at_generation[haplotype] / float(pop_size)
if allele != "A":
minor_allele_freq... | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Find all variable sites. | def get_all_snps():
snps = set()
for generation in history:
for haplotype in generation:
for site in range(seq_length):
if haplotype[site] != "A":
snps.add(site)
return snps
def snp_trajectory_plot(xlabel="generation"):
mpl.rcParams['font.size'... | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Scale up
Here, we scale up to more interesting parameter values. | pop_size = 50
seq_length = 100
generations = 500
mutation_rate = 0.0001 # per gen per individual per site | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
And the population genetic parameter $\theta$, which equals $2N\mu$, is 1. | 2 * pop_size * seq_length * mutation_rate
base_haplotype = ''.join(["A" for i in range(seq_length)])
pop.clear()
del history[:]
pop[base_haplotype] = pop_size
simulate()
plt.figure(num=None, figsize=(14, 14), dpi=80, facecolor='w', edgecolor='k')
plt.subplot2grid((3,2), (0,0), colspan=2)
stacked_trajectory_plot(xlab... | code/mutation-drift.ipynb | alvason/probability-insighter | gpl-2.0 |
Create histograms
h1, h2, ... have their alternatives in physt.dask_compat. They should work similarly. Although, they are not complete and unexpected errors may occur. | from physt.compat.dask import h1 as d1
from physt.compat.dask import h2 as d2
# Use chunks to create a 1D histogram
ha = d1(chunked2, "fixed_width", bin_width=0.2)
check_ha = h1(million2, "fixed_width", bin_width=0.2)
ok = (ha == check_ha)
print("Check: ", ok)
ha.plot()
ha
# Use chunks to create a 2D histogram
hb = d... | doc/dask.ipynb | janpipek/physt | mit |
Some timings
Your results may vary substantially. These numbers are just for illustration, on 4-core (8-thread) machine. The real gain comes when we have data that don't fit into memory.
Efficiency | # Standard
%time h1(million2, "fixed_width", bin_width=0.2)
# Same array, but using dask
%time d1(million2, "fixed_width", bin_width=0.2)
# Most efficient: dask with already chunked data
%time d1(chunked2, "fixed_width", bin_width=0.2) | doc/dask.ipynb | janpipek/physt | mit |
Different scheduling | %time d1(chunked2, "fixed_width", bin_width=0.2)
%%time
# Hyper-threading or not?
graph, name = d1(chunked2, "fixed_width", bin_width=0.2, compute=False)
dask.threaded.get(graph, name, num_workers=4)
# Multiprocessing not so efficient for small arrays?
%time d1(chunked2, "fixed_width", bin_width=0.2, dask_method=dask... | doc/dask.ipynb | janpipek/physt | mit |
High level API
We recommend using tf.keras as a high-level API for building neural networks. That said, most TensorFlow APIs are usable with eager execution.
Layers: common sets of useful operations
Most of the time when writing code for machine learning models you want to operate at a higher level of abstraction than ... | # In the tf.keras.layers package, layers are objects. To construct a layer,
# simply construct the object. Most layers take as a first argument the number
# of output dimensions / channels.
layer = tf.keras.layers.Dense(100)
# The number of input dimensions is often unnecessary, as it can be inferred
# the first time t... | tensorflow/contrib/eager/python/examples/notebooks/4_high_level.ipynb | gojira/tensorflow | apache-2.0 |
Heu de seleccionar un valor intermedi per a distingir entre blanc i negre. Serà el llindar que usareu:
si la lectura està per damunt, el sensor està detectant blanc
si està per baix, negre.
Seguiment de línia, versió 1.0
<img src="img/line_circuit.jpg" width=240 align="right">
Per a seguir la línia, la idea és senzi... | from functions import left, right
try:
while True:
if ___:
___
else:
___
except KeyboardInterrupt:
stop() | task/light.ipynb | ecervera/mindstorms-nb | mit |
Versió 2.0
Si el robot se n'ix per la curva, cal fer un gir més pronunciat. Les funcions que hem utilitzat fins ara giren una roda mentre l'altra està parada. Es pot girar més si, en lloc d'estar parada, l'altra roda gira en sentit contrari. Per a fer-ho, useu les següents funcions: | from functions import left_sharp, right_sharp
try:
while True:
if ___:
___
else:
___
except KeyboardInterrupt:
stop() | task/light.ipynb | ecervera/mindstorms-nb | mit |
Versió 3.0
En general, pot haver curves a esquerra i dreta. Aleshores, el mètode anterior no és suficient. Una solució seria usar dos sensors, un per cada costat de la línia, però només en tenim un per robot.
Una altra solució és fer que el sensor vaja per la vora de la línia, de manera que la superfície serà meitat ne... | try:
while True:
if ___:
___
else:
if ___:
___
else:
___
except KeyboardInterrupt:
stop() | task/light.ipynb | ecervera/mindstorms-nb | mit |
Versió 4.0
Si el robot oscil·la massa, fins i tot pot travessar la línia i desorientar-se completament.
La solució és disminuir la velocitat, indicant-ho a les funcions de moviment, per exemple:
python
forward(speed=65)
El valor per defecte és 100, el màxim, i el mínim és 0. | try:
while True:
if ___:
___
else:
if ___:
___
else:
___
except KeyboardInterrupt:
stop() | task/light.ipynb | ecervera/mindstorms-nb | mit |
Recapitulem
En aquesta pàgina no hem vist conceptes nous, però hem aplicat el que ja sabíem al problema de seguir la línia.
Passem a vore el darrer sensor.
Abans de continuar, desconnecteu el robot: | from functions import disconnect, next_notebook
disconnect()
next_notebook('ultrasonic') | task/light.ipynb | ecervera/mindstorms-nb | mit |
2. Calculate the translational partition function of a CO molecule in the bottle at 298 K. What is the unit of the partition function?
For $T \gg \theta_{trans}$, $\Lambda \ll L$, $q_{trans} = V/\Lambda^3$. $\Lambda = h \left(\frac{\beta}{2\pi m}\right)^{1/2}$. | Lamda = h*(1/(kB*298*2*np.pi*m))**0.5
print(Lamda)
q_trans = V/Lamda**3
print('The translational partition function of a CO molecule in the bottle at 298 K is {:.4E}.'.format(q_trans))
print('It is dimensionless.') | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
3. Plot the rotational and vibrational partition functions of a CO molecule in the bottle from $T$ = 200 to 2000 K (assume the CO remains a gas over the whole range). Hint: Use your answer to Problem 1 to simplify calculating the rotational partition function.
$q_{rot} = \frac{1}{\sigma}\frac{T}{\theta_{rot}} = \frac... | T = np.linspace(200,2000,1000) # r = R/a_0
q_rot = T/T_rot
q_vib = 1/(1-np.exp(-T_vib/T))
plt.plot(T,q_rot)
plt.xlabel('T (K)')
plt.ylabel('$q_{rot}$')
plt.title('The rotational partition function of a CO molecule')
plt.show()
plt.plot(T,q_vib)
plt.xlabel('T (K)')
plt.ylabel('$q_{vib}$')
plt.title('The vibrational part... | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
4. Plot the total translational, rotational, and vibrational energies of CO in the bottle from $T =$ 200 to 2000 K (assume the CO remains a gas over the whole range). Which (if any) of the three types of motions dominate the total energy?
$U_{trans} = \frac{3}{2}RT$
$U_{rot} = RT$
$U_{vib} = R\frac{\theta_{vib}}{e^{\th... | R = 8.31447 # J/(mol*K)
U_trans = 1.5*R*T
U_rot = R*T
U_vib = R*T_vib/(np.exp(T_vib/T)-1)
plt.plot(T,U_trans,label='U_trans')
plt.plot(T,U_rot,label='U_rot')
plt.plot(T,U_vib,label='U_vib')
plt.legend()
plt.xlabel('T (K)')
plt.ylabel('Internal Energy (J/mol)')
plt.title('Internal energies of CO in the bottle')
plt.show... | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
5. Plot the total translational, rotational, and vibrational constant volume molar heat capacities of CO in the bottle from $T =$ 200 to 2000 K. Which (if any) of the three types of motions dominate the heat capacity?
$C_{V,trans} = \frac{3}{2}R$
$C_{V,rot} = R$
$C_{V,vib} = R\left(\frac{\theta_{vib}}{T}\frac{e^{\theta... | Cv_trans = np.linspace(1.5*R,1.5*R,1000)
Cv_rot = np.linspace(R,R,1000)
Cv_vib = R*(T_vib/T*np.exp(T_vib/2./T)/(np.exp(T_vib/T)-1))**2
plt.plot(T,Cv_trans,label='Cv_trans')
plt.plot(T,Cv_rot,label='Cv_rot')
plt.plot(T,Cv_vib,label='Cv_vib')
plt.legend()
plt.xlabel('T (K)')
plt.ylabel('Heat Capacity (J/mol/K)')
plt.titl... | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
6. Plot the total translational, rotational, and vibrational Helmholtz energies of CO in the bottle from $T =$ 200 to 2000 K. Which (if any) of the three types of motions dominate the Helmholtz energy?
$A = U - TS$
$S_{trans} = Rln\left(\frac{e^{5/2}V}{N\Lambda^3}\right)$
$S_{rot} = R(1-ln(\theta_{rot}/T))$
$S_{vib} = ... | NA = 6.022e23
S_trans = R*np.log(np.exp(2.5)*V/NA/Lamda**3)
S_rot = R*(1-np.log(T_rot/T))
S_vib = R*(T_vib/T/(np.exp(T_vib/T)-1)-np.log(1-np.exp(-T_vib/T)))
A_trans = U_trans-T*S_trans
A_rot = U_rot-T*S_rot
A_vib = U_vib-T*S_vib
plt.plot(T,A_trans,label='A_trans')
plt.plot(T,A_rot,label='A_rot')
plt.plot(T,A_vib,label=... | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
7. Use your formulas to calculate $\Delta P$, $\Delta U$, $\Delta A$, and $\Delta S$ associated with isothermally expanding the gas from 20 dm$^3$ to 40 dm$^3$.
T = 298 K. $\Delta U=0$.
$\Delta P = \frac{RT}{V_2} - \frac{RT}{V_1}$.
$\Delta S = S_{trans,2} - S_{trans,1}$.
$A = U-TS$, so, $\Delta A = -T\Delta S$. | V2 = 0.04 # m^3
deltaP = R*298*(1/V2-1/V)
deltaS = R*np.log(np.exp(2.5)*V2/NA/Lamda**3) - R*np.log(np.exp(2.5)*V/NA/Lamda**3)
deltaA = -deltaS*298
print('Delta P = {0:.3f} Pa, Delta U = 0, Delta A = {1:.3f} J/mol, and Delta S = {2:.3f} J/mol/K.'.format(deltaP,deltaA,deltaS)) | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
Reactions from scratch
In 1996, Schneider and co-workers used quantum chemistry to compute the reaction pathway for unimolecular decomposition of trifluoromethanol, a reaction of relevance to the atmospheric degradation of hydrofluorocarbon refrigerants (J. Phys. Chem. 1996, 100, 6097- 6103, doi:10.1021/jp952703m): $$\... | import numpy as np
T = 298 # K
k = 1.38065e-23 # J/K
R = 8.31447 # J/(mol*K)
Na = 6.0221e23 # 1/mol
c = 6.0221e26 # 1/m^3, conversion factor of 1mol/L = 6.02e26 particles/m^3
autokJ = 2625.50
Eelec = [-412.90047 ,-312.57028 ,-100.31885 ] # kJ/mol
ZPE = [0.02889 ,0.01422 ,0.00925 ] # kJ/mol
dE0 = ((Eelec[1] + ZPE[1] ... | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
9. Using the data provided, determine $\Delta A^{\circ}$(298 K) in kJ mol$^{-1}$, assuming ideal behavior and 1 M standard state. Recall that $A^\circ=E^\text{elec} + \text{ZPE}-RT\ln(q^\circ)-RT$ and that $q^\circ =(q^\text{trans}/V)q^\text{rot}q^\text{vib}/c^\circ$ in units corresponding with the standard state.
$\... | q_trans = [7.72e32/c,1.59e32/c,8.65e31/c] # change translational partition functions from 1/m3 to mol/l std state
q_rot = [61830,679,9.59] # unitless
q_vib = [2.33,1.16,1] # unitless
Q = (q_trans[1]*q_rot[1]*q_vib[1])*(q_trans[2]*q_rot[2]*q_vib[2])/(q_trans[0]*q_rot[0]*q_vib[0]) # total partition
dA = dE0 + (-R*T*... | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
10. Determine $\Delta G^\circ$(298 K). Recall that $G = A + PV = A + RT$ for an ideal ga.
$\Delta{G^{\circ}} = \Delta{A^{\circ}} + \Delta(PV)$
$\quad = \quad \Delta{A^{\circ}} + RT$ | dG = dA + R*T/1000 #kJ/mol
print("delta_G = %.2f kJ/mol"%dG) | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
11. Determine $\Delta S^{\circ}$(298 K), in J mol$^{-1}$ K$^{-1}$ , assuming a 1 M standard state. Recall that $S = (U - A)/T$.
$\Delta A^{\circ} = \Delta U^{\circ} - T\Delta S^{\circ}$
$\Delta S^{\circ} = \frac{\Delta U^{\circ} - \Delta A^{\circ}}{T}$ | dS = 1000*(dU - dA)/T
print("delta_S = %.2f J/mol K"%dS) | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
12. Using the data provided, determine $K_c$ (298 K), assuming a 1 M standard state. You may either determine from partition functions of from the relationship between $K_c$ and $\Delta G^\circ$.
$\mathrm{A} \rightarrow \mathrm{B + C}$$$K_c(T) = \frac{\frac{q_{\mathrm{B}}}{V}\frac{q_{\mathrm{C}}}{V}}{\frac{q_{\mathrm{... | Kc = Q *np.exp(-dE0*1000/(R*T)) # Kc equation from lecture notes
print('Kc = %.3f (unitless). '%(Kc))
Kc = np.exp(-dG*1000/(R*T))
print('Kc = %.3f (unitless). '%(Kc)) | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
13. 1 mole of CF$_3$OH is generated in a 20 L vessel at 298 K and left long enough to come to equilibrium with respect to its decomposition reaction. What is the composition of the gas (concentrations of all the components) at equilibrium (in mol/L)?
1 mol/ 20 L = 0.05 mol/L
$\mathrm{A} \rightarrow \mathrm{B + C}$
$K_c... | from sympy import *
x = symbols('x',positive=True)
c = solve(x**2-(0.05-x)*Kc,x)
print('At equilibrium, CF3OH = %.2E mol/L, COF2 = %.5f mol/L, HF = %.5f mol/L.'%(0.05-c[0],c[0],c[0]))
print('At equilibrium, CF3OH = %.2E mol, COF2 = %.5f mol, HF = %.5f mol.'%((0.05-c[0])*20,c[0]*20,c[0]*20)) | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
14. How, directionally, would your answer to Question 13 change if the vessel was at a higher temperature? Use the van'T Hoff relationship to determine the equilibrium constant and equilibrium concentrations at 273 and 323 K. How good was your guess?
From question #8, we know that at 298K, $\Delta U$ = 18.64 kJ/mol.
... | dn = 2-1
R = 8.314/1000 #kJ/mol K
T = 298 #K
dH = dU+dn*R*T #kJ/mol
print("dH =",round(dH,3),"kJ/mol") | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
Since $\Delta H$ is positive, we expect $K \uparrow$ as $T \uparrow$
The Van't Hoff equation states:
$ln\frac{K(T_2)}{K(T_1)}=\frac{-\Delta H^o}{R}(\frac{1}{T_2}-\frac{1}{T_1})$
We know from question 12, K = 2.926 at 298 K. | K1 = 2.926
T1 = 298 #K
T2 = 273 #K
K2 = K1*np.exp(-dH/R*(1/T2-1/T1))
print('K=', round(K2,4), 'at 273 K.')
x = symbols('x',positive=True)
c = solve(x**2-(0.05-x)*K2,x)
print('At equilibrium, CF3OH = %.2E mol/L, COF2 = %.5f mol/L, HF = %.5f mol/L.'%(0.05-c[0],c[0],c[0]))
K1 = 2.926
T1 = 298 #K
T2 = 323 #K
K2 = K1*np.ex... | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
Therefore, the at higher temperatures, the reaction shifts towards the products.
15. How, directionally, would your answer to Question 13 change if the vessel had a volume of 5 L? Redo the calculation at this volume to verify your guess.
If V = 5L, the initial concentration = $\frac{1 mol}{5 L}=.2 M$
$K_c = \frac{x^2}{... | T = 298 #K
R = 8.314
Kc = np.exp(-dG*1000/(R*T))
print('At 298 K, Kc = %.3f (unitless). '%(Kc))
x = symbols('x',positive=True)
c = solve(x**2-(0.2-x)*Kc,x)
print('At equilibrium, CF3OH = %.2E mol/L, COF2 = %.5f mol/L, HF = %.5f mol/L.'%(0.2-c[0],c[0],c[0]))
print('At equilibrium, CF3OH = %.2E mol, COF2 = %.5f mol, HF =... | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
16. Consult a thermodynamics source (e.g. https://webbook.nist.gov/chemistry/) to determine $\Delta H^\circ$(298 K), $\Delta S^\circ$(298 K), and $\Delta G^\circ$(298 K) for the homologous reaction CH$_3$OH (g)$\rightarrow$ H$_2$O (g) + H$_2$CO (g). Does the substitution of F by H make the reaction more or less favora... | T = 298 #K
#All values were taken from NIST
#Methanol
Hm = -205 #kJ/mol
Sm = .2399 #kJ/mol K
Gm = Hm - T*Sm #kJ/mol
#Hydrogen
Hh = 0
Sh = .13068 #J/mol K
Gh = Hh - T*Sh #kJ/mol
#Formaldehyde
Hf = -108.6 #kJ/mol
Sf = .21895 #kJ/mol K
Gf = Hf - T*Sf #kJ/mol
delta_H = Hf+Hh-Hm #kJ/mol
delta_S = Sf+Sh-Sm #kJ/mol K
del... | Homework/HW10-soln.ipynb | wmfschneider/CHE30324 | gpl-3.0 |
This Notebool is Mainly to Check the Buffer built from Entrances .shp (previously) and from geojson (.json) in most recent Dumbo-Can-Run scripts.
Load | from shapely.geometry import Point
import pyproj
import geopandas as gpd
proj = pyproj.Proj(init='epsg:2263', preserve_units=True)
entr_points = sqlContext.read.load('../why_yellow_taxi/Data/2016_(May)_New_York_City_Subway_Station_Entrances.json', \
format='json', header=True, inferSche... | Buffer/Use_Geojson2build_Buffer_Not_Shapefile.ipynb | djfan/why_yellow_taxi | mit |
List | entr_geo.head(2)
shp.head(2) | Buffer/Use_Geojson2build_Buffer_Not_Shapefile.ipynb | djfan/why_yellow_taxi | mit |
Identical or Not? | entr_geo.head(2).geometry[1] == shp.head(2).geometry[1] | Buffer/Use_Geojson2build_Buffer_Not_Shapefile.ipynb | djfan/why_yellow_taxi | mit |
Detail | shp.head(2).geometry[0].centroid.x
shp.head(2).geometry[0].centroid.y
entr_geo.head(2).geometry[0].centroid.x
entr_geo.head(2).geometry[0].centroid.y | Buffer/Use_Geojson2build_Buffer_Not_Shapefile.ipynb | djfan/why_yellow_taxi | mit |
Unit Test | %run ../utils/results.py
# %load test_dfs.py
from nose.tools import assert_equal
class TestDfs(object):
def __init__(self):
self.results = Results()
def test_dfs(self):
node = Node(5)
insert(node, 2)
insert(node, 8)
insert(node, 1)
insert(node, 3)
in... | interactive-coding-challenges/graphs_trees/tree_dfs/dfs_challenge.ipynb | saashimi/code_guild | mit |
因为需要全市场回测所以本章无法使用沙盒数据，《量化交易之路》中的原始示例使用的是美股市场，这里的示例改为使用A股市场。
本节建议对照阅读abu量化文档第20-23节内容
本节的基础是在abu量化文档中第20节内容完成运行后有A股训练集交易和A股测试集交易数据之后
abu量化系统github地址 (您的star是我的动力！)
abu量化文档教程ipython notebook
第11章 量化系统-机器学习•ABU | from abupy import AbuFactorAtrNStop, AbuFactorPreAtrNStop, AbuFactorCloseAtrNStop, AbuFactorBuyBreak
from abupy import abu, EMarketTargetType, AbuMetricsBase, ABuMarketDrawing, ABuProgress, ABuSymbolPd
from abupy import EMarketTargetType, EDataCacheType, EMarketSourceType, EMarketDataFetchMode, EStoreAbu, AbuUmpMainMul... | ipython/第十一章-量化系统——机器学习•ABU.ipynb | bbfamily/abu | gpl-3.0 |
11.1 搜索引擎与量化交易
本节建议对照abu量化文档第15节内容进行阅读 | orders_pd_train = abu_result_tuple_train.orders_pd
# 选择失败的前20笔交易绘制交易快照
# 这里只是示例，实战中根据需要挑选，rank或者其他方式
plot_simple = orders_pd_train[orders_pd_train.profit_cg < 0][:20]
# save=True保存在本地， 文件保存在~/abu/data/save_png/中
ABuMarketDrawing.plot_candle_from_order(plot_simple, save=True) | ipython/第十一章-量化系统——机器学习•ABU.ipynb | bbfamily/abu | gpl-3.0 |
11.2 主裁
11.2.1 角度主裁
请对照阅读ABU量化系统使用文档 ：第15节 中相关内容 | from abupy import AbuUmpMainDeg
# 参数为orders_pd
ump_deg = AbuUmpMainDeg(orders_pd_train)
# df即由之前ump_main_make_xy生成的类df，表11-1所示
ump_deg.fiter.df.head() | ipython/第十一章-量化系统——机器学习•ABU.ipynb | bbfamily/abu | gpl-3.0 |
耗时操作，快的电脑大概几分钟，具体根据电脑性能，cpu数量，启动多进程进行训练： | _ = ump_deg.fit(brust_min=False)
ump_deg.cprs
max_failed_cluster = ump_deg.cprs.loc[ump_deg.cprs.lrs.argmax()]
print('失败概率最大的分类簇{0}, 失败率为{1:.2f}%, 簇交易总数{2}, ' \
'簇平均交易获利{3:.2f}%'.format(ump_deg.cprs.lrs.argmax(),
max_failed_cluster.lrs * 100,
max_fai... | ipython/第十一章-量化系统——机器学习•ABU.ipynb | bbfamily/abu | gpl-3.0 |
由于不是同一份沙盒数据，所以下面结果内容与书中分析内容不符，需要按照实际情况分析:
比如下面的特征即是42日和60日的deg格外大，21和252相对训练集平均值也很大： | from abupy import ml
ml.show_orders_hist(max_failed_cluster_orders, ['buy_deg_ang21', 'buy_deg_ang42', 'buy_deg_ang60','buy_deg_ang252'])
print('分类簇中deg_ang60平均值为{0:.2f}'.format(
max_failed_cluster_orders.buy_deg_ang60.mean()))
print('分类簇中deg_ang21平均值为{0:.2f}'.format(
max_failed_cluster_orders.buy_deg_ang21.m... | ipython/第十一章-量化系统——机器学习•ABU.ipynb | bbfamily/abu | gpl-3.0 |
交易快照文件保存在~/abu/data/save_png/中, 下面打开对应目录：save_png | if abupy.env.g_is_mac_os:
!open $abupy.env.g_project_data_dir
else:
!echo $abupy.env.g_project_data_dir | ipython/第十一章-量化系统——机器学习•ABU.ipynb | bbfamily/abu | gpl-3.0 |
11.2.2 使用全局最优对分类簇集合进行筛选 | brust_min = ump_deg.brust_min()
brust_min
llps = ump_deg.cprs[(ump_deg.cprs['lps'] <= brust_min[0]) & (ump_deg.cprs['lms'] <= brust_min[1] )& (ump_deg.cprs['lrs'] >=brust_min[2])]
llps
ump_deg.choose_cprs_component(llps)
ump_deg.dump_clf(llps) | ipython/第十一章-量化系统——机器学习•ABU.ipynb | bbfamily/abu | gpl-3.0 |
11.2.3 跳空主裁
请对照阅读ABU量化系统使用文档 ：第16节 UMP主裁交易决策 中相关内容 | from abupy import AbuUmpMainJump
# 耗时操作，大概需要10几分钟，具体根据电脑性能，cpu情况
ump_jump = AbuUmpMainJump.ump_main_clf_dump(orders_pd_train, save_order=False)
ump_jump.fiter.df.head() | ipython/第十一章-量化系统——机器学习•ABU.ipynb | bbfamily/abu | gpl-3.0 |
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