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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
9033 | 1 | null | null | 2 | 136 | Going through some old notes of mine, I found a formula whose purpose seemed statistical, but whose provenance and use I have dumbly forgotten to write. I am asking if anybody might remember this one:
\begin{align*}
-N-\frac1{N}\sum_{j=1}^N (2j-1)(A_j+B_j),
\end{align*}
where
\begin{align*}
A_j=\ln\left(\frac{1}{2}+\in... | Identifying a statistical formula | CC BY-SA 2.5 | null | 2011-04-01T09:19:56.027 | 2011-04-01T18:11:38.237 | 2011-04-01T11:46:09.357 | null | null | [
"normal-distribution"
] |
9034 | 2 | null | 7166 | 4 | null | Predicting hourly data has become my main interest. This problem arises normally in Call Center Forecasting. One needs to be concerned with hourly patterns within the day , different daily patterns across the week and seasonal patterns across the year ( Monthly indicators/Weekly Indicators. In addition there can be and... | null | CC BY-SA 2.5 | null | 2011-04-01T10:57:40.617 | 2011-04-01T10:57:40.617 | null | null | 3382 | null |
9035 | 2 | null | 8800 | 2 | null | Ok, as far as I see your central question can be summarized in the following way:
>
I transform one feature F in my
dataset, but only for those rows where
the label is "flower", the rest of the
rows remain unchanged. I evaluate
NB,C4.5 and Logistic Regression before
and after the transformation. The results ... | null | CC BY-SA 2.5 | null | 2011-04-01T11:26:39.950 | 2011-04-01T11:26:39.950 | null | null | 264 | null |
9036 | 1 | 9037 | null | 4 | 2440 | Using R, I noticed that I get different results if I use `fisher.test()` with raw data (as factors) versus a contingency table. The documentation states that `fisher.test(x,y)` will compute a contingency table from `x, y` by treating them as factors.
My example is a baseline statistics exercise: comparing the gender s... | In R, fisher.test returns different results if I use vectors vs contingency table | CC BY-SA 3.0 | null | 2011-04-01T11:31:34.533 | 2013-07-22T16:41:54.237 | 2013-07-22T16:41:54.237 | 7290 | 1138 | [
"r",
"hypothesis-testing",
"p-value",
"contingency-tables",
"fishers-exact-test"
] |
9037 | 2 | null | 9036 | 5 | null | The first test should read
```
> fisher.test(expData[,1], expData[,2])
Fisher's Exact Test for Count Data
data: expData[, 1] and expData[, 2]
p-value = 0.4857
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
0.001607888 4.722931239
sample estimates:
odds ratio
0.15... | null | CC BY-SA 2.5 | null | 2011-04-01T11:39:40.613 | 2011-04-01T11:39:40.613 | null | null | 930 | null |
9038 | 1 | 9051 | null | 2 | 155 | I'm using [ENSO dataset](http://www.itl.nist.gov/div898/strd/nls/data/LINKS/DATA/ENSO.dat) from the NIST Statistical Reference Datasets as a test for nonlinear regression code.
The data are monthly averaged atmospheric pressure differences between Easter Island and Darwin, called [Southern Oscillation](http://en.wikip... | The source of ENSO data from NIST | CC BY-SA 2.5 | null | 2011-04-01T11:40:34.837 | 2011-04-01T14:50:47.687 | 2011-04-01T11:44:21.793 | null | 3981 | [
"dataset",
"nonlinear-regression"
] |
9039 | 2 | null | 8567 | -2 | null | From wiki "Given a data set, several candidate models may be ranked according to their AIC values. From the AIC values one may also infer that e.g. the top two models are roughly in a tie and the rest are far worse. Thus, AIC provides a means for comparison among models—a tool for model selection. AIC does not provide ... | null | CC BY-SA 2.5 | null | 2011-04-01T11:45:03.477 | 2011-04-01T11:45:03.477 | null | null | 3382 | null |
9040 | 1 | 9048 | null | 7 | 1061 | First off, I'm a programmer but my experience with true statistics ended at A-Level so I'm looking to all of you for help with a little side project I've been tinkering with.
At home I use Plex Media Center to display all of my movies. I built an export tool for this to generate a HTML file containing information on yo... | Visualize movie/actor relationships | CC BY-SA 2.5 | null | 2011-04-01T11:54:19.400 | 2011-05-04T13:50:39.353 | null | null | 3988 | [
"data-visualization"
] |
9041 | 2 | null | 9040 | 3 | null | Graphviz can optimise the layout, see something similar [here](http://www.graphviz.org/content/siblings).
| null | CC BY-SA 2.5 | null | 2011-04-01T12:10:43.893 | 2011-04-01T12:10:43.893 | null | null | 3911 | null |
9042 | 2 | null | 9020 | 2 | null | Hi
Take a look at the varclus function in the Frank Harrel's Hmisc package.
```
require(Hmisc)
? varclus
```
| null | CC BY-SA 2.5 | null | 2011-04-01T12:48:36.933 | 2011-04-01T12:48:36.933 | null | null | 2028 | null |
9043 | 1 | 9044 | null | 1 | 387 | I'm helping my father to translate a text from my own native Latvian language to English. Even though I'm a fair English speaker and have a bit of mathematics in my past, I never did have a firm grasp of statistics. :(
The term I'm looking for has something to do with error calculation. A GPS device is measuring coordi... | What is the English name for a statistics term that I'm looking for? | CC BY-SA 2.5 | null | 2011-04-01T08:53:26.107 | 2011-04-01T13:10:24.590 | null | null | 2590 | [
"terminology"
] |
9044 | 2 | null | 9043 | 3 | null | Google translate gives ["rms error of assessment"](http://translate.google.co.uk/?hl=en&tab=wT#lv%7Cen%7Cvid%C4%93jo%20kvadr%C4%81tisko%20k%C4%BC%C5%ABdu%20nov%C4%93rt%C4%93jums) which (as root mean square error) seems to be the intended meaning.
The difference between mean square error and root
mean square error is... | null | CC BY-SA 2.5 | null | 2011-04-01T09:51:37.740 | 2011-04-01T09:51:37.740 | null | null | 2958 | null |
9045 | 2 | null | 9043 | 0 | null | I would guess, it is a Least Mean Squared Error (LMSE) estimator. [publication1](http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=5090171) [publication2](http://web.cecs.pdx.edu/~ssp/Reports/2006/Monaghan.pdf) [wikipedia](http://en.wikipedia.org/wiki/Mean_squared_error)
the GPS is trying to estimate coordinates,... | null | CC BY-SA 2.5 | null | 2011-04-01T12:21:40.590 | 2011-04-01T12:29:20.880 | null | null | 4060 | null |
9048 | 2 | null | 9040 | 6 | null | N.B.: This was previously a (long) comment that I've converted to an answer. Hopefully I'll be able to post an example of what I describe below within a day or two.
Why not try something like a heatmap? Have movies as rows and actors as columns. Maybe sort each of them in terms of the number of actors in the movie and... | null | CC BY-SA 2.5 | null | 2011-04-01T13:58:06.120 | 2011-04-01T13:58:06.120 | null | null | 2970 | null |
9049 | 2 | null | 9040 | 1 | null | I wouldn't know how you'd go about constructing this but I liked the method using hyperbolic geometry
[http://www.newscientist.com/data/images/ns/cms/dn19420/dn19420-1_800.jpg](http://www.newscientist.com/data/images/ns/cms/dn19420/dn19420-1_800.jpg)
[http://www.newscientist.com/article/dn19420-escherlike-internet-map-... | null | CC BY-SA 2.5 | null | 2011-04-01T14:30:18.650 | 2011-04-01T14:30:18.650 | null | null | null | null |
9050 | 1 | null | null | 11 | 4650 | I'd like to obtain a graphic representation of the correlations in articles I have gathered so far to easily explore the relationships between variables. I used to draw a (messy) graph but I have too much data now.
Basically, I have a table with:
- [0]: name of variable 1
- [1]: name of variable 2
- [2]: correlatio... | How to display a matrix of correlations with missing entries? | CC BY-SA 2.5 | null | 2011-04-01T14:40:54.433 | 2011-04-01T17:02:00.770 | 2011-04-01T16:27:54.967 | 3827 | 3827 | [
"r",
"data-visualization",
"correlation"
] |
9051 | 2 | null | 9038 | 3 | null | You would not expect your two sets of numbers to be the same: the NIST numbers seem to be have been published to [test statistical software](http://www.itl.nist.gov/div898/strd/general/bkground.html) while the Australian BOM numbers are designed to [identify whether the ENSO cycle is above or below average](http://www.... | null | CC BY-SA 2.5 | null | 2011-04-01T14:50:47.687 | 2011-04-01T14:50:47.687 | null | null | 2958 | null |
9052 | 2 | null | 9050 | 5 | null | Your data may be like
```
name1 name2 correlation
1 V1 V2 0.2
2 V2 V3 0.4
```
You can rearrange your long table into a wide one with the following R code
```
d = structure(list(name1 = c("V1", "V2"), name2 = c("V2", "V3"),
correlation = c(0.2, 0.4)), .Names = c("name1", "name2",
... | null | CC BY-SA 2.5 | null | 2011-04-01T15:10:47.950 | 2011-04-01T15:10:47.950 | null | null | 3911 | null |
9053 | 1 | 9055 | null | 45 | 54369 | Why does a cross-validation procedure overcome the problem of overfitting a model?
| How does cross-validation overcome the overfitting problem? | CC BY-SA 2.5 | null | 2011-04-01T16:26:57.173 | 2020-07-19T23:07:44.370 | 2020-07-19T23:07:44.370 | 12359 | 3269 | [
"regression",
"cross-validation",
"model-selection",
"overfitting"
] |
9054 | 2 | null | 9050 | 12 | null | Building upon @GaBorgulya's response, I would suggest trying fluctuation or level plot (aka heatmap displays).
For example, using [ggplot2](http://had.co.nz/ggplot2/):
```
library(ggplot2, quietly=TRUE)
k <- 100
rvals <- sample(seq(-1,1,by=.001), k, replace=TRUE)
rvals[sample(1:k, 10)] <- NA
cc <- matrix(rvals, nr=10)
... | null | CC BY-SA 2.5 | null | 2011-04-01T16:38:04.397 | 2011-04-01T17:02:00.770 | 2011-04-01T17:02:00.770 | 930 | 930 | null |
9055 | 2 | null | 9053 | 35 | null | I can't think of a sufficiently clear explanation just at the moment, so I'll leave that to someone else; however cross-validation does not completely overcome the over-fitting problem in model selection, it just reduces it. The cross-validation error does not have a negligible variance, especially if the size of the ... | null | CC BY-SA 3.0 | null | 2011-04-01T16:51:53.760 | 2015-12-10T17:36:21.880 | 2015-12-10T17:36:21.880 | 887 | 887 | null |
9056 | 2 | null | 9050 | 3 | null | The `corrplot` package is a useful function for visualizing correlation matrices. It accepts a correlation matrix as the input object and has several options for displaying the matrix itself. A nice feature is that it can reorder your variables using hierarchical clustering or PCA methods.
See the accepted answer in [t... | null | CC BY-SA 2.5 | null | 2011-04-01T16:52:54.727 | 2011-04-01T16:52:54.727 | 2017-04-13T12:44:25.283 | -1 | 3309 | null |
9057 | 2 | null | 9029 | 6 | null | Agresti 2007 discusses them. They're in chapter 9 and 10. The 2002 edition probably discusses them too, as @suncoolsu mentioned.
Agresti refers to the group of response variables as a cluster and discusses according analysis with marginal models, conditional models and generalized estimating equations.
| null | CC BY-SA 3.0 | null | 2011-04-01T16:57:26.190 | 2014-10-18T17:16:56.540 | 2014-10-18T17:16:56.540 | 3874 | 3874 | null |
9058 | 2 | null | 9033 | 4 | null | When correctly rewritten (as indicated in a comment), it will be the [Anderson-Darling](http://en.wikipedia.org/wiki/Anderson%E2%80%93Darling_test) statistic (for a normality test).
| null | CC BY-SA 2.5 | null | 2011-04-01T18:11:38.237 | 2011-04-01T18:11:38.237 | null | null | 919 | null |
9059 | 2 | null | 9053 | 18 | null | My answer is more intuitive than rigorous, but maybe it will help...
As I understand it, overfitting is the result of model selection based on training and testing using the same data, where you have a flexible fitting mechanism: you fit your sample of data so closely that you're fitting the noise, outliers, and all th... | null | CC BY-SA 2.5 | null | 2011-04-01T18:23:29.910 | 2011-04-01T18:23:29.910 | null | null | 1764 | null |
9060 | 2 | null | 9030 | 2 | null | Some people have started to look at this issue in the chemometrics literature. For instance, about 20 years ago Robert Gibbons started to do statistical analyses suggesting instrument responses (for low-level measurement of chemicals) were nonlinear, heteroscedastic, and had non-normal (perhaps lognormal) error distri... | null | CC BY-SA 4.0 | null | 2011-04-01T20:39:18.377 | 2022-06-20T12:23:38.553 | 2022-06-20T12:23:38.553 | 919 | 919 | null |
9061 | 2 | null | 8911 | 0 | null | R would be my first vote. Another free option would be gretl. If you happen to know BUGS, JAGS makes sense and is free. And I really don't like its syntax, but if you have some knowledge of Matlab, the free alternative Octave runs on MacOS X as well.
| null | CC BY-SA 2.5 | null | 2011-04-01T21:35:45.077 | 2011-04-01T21:35:45.077 | null | null | 1764 | null |
9062 | 1 | 9096 | null | 6 | 8032 | I have a 2x3 contingency table - the row variable is a factor, the column variable is an ordered factor (ordinal level). I'd like to apply either symmetrical or asymmetrical association technique. What do you recommend me to do? Which technique do you find the most appropriate?
| Measure of association for 2x3 contingency table | CC BY-SA 2.5 | null | 2011-04-01T21:45:15.740 | 2011-04-02T16:46:40.420 | null | null | 1356 | [
"correlation",
"categorical-data",
"contingency-tables",
"association-measure"
] |
9063 | 2 | null | 9062 | 3 | null | On a 2x3 contingency table where the three-level factor is ordered you may use rank correlation (Spearman or Kendall) to assess association between the two variables.
You may also think about the data as an ordered variable observed in two groups. A corresponding significance test could be the Mann-Whitney test (with m... | null | CC BY-SA 2.5 | null | 2011-04-01T22:39:37.960 | 2011-04-01T22:39:37.960 | null | null | 3911 | null |
9064 | 1 | 9084 | null | 6 | 2376 | I've heard a bit about the 'kernel trick' for support vector machines, and I was wondering:
- How do you identify problems that might benefit from the kernel trick?
- How to implement it in R?
Thank you
| Implementing the 'kernel trick' for a support vector machine in R | CC BY-SA 2.5 | null | 2011-04-01T23:15:45.397 | 2015-04-19T20:48:43.750 | 2015-04-19T20:48:43.750 | 9964 | 2817 | [
"r",
"machine-learning",
"svm",
"kernel-trick"
] |
9065 | 5 | null | null | 0 | null | Overview
From The Discipline of Machine Learning by Tom Mitchell:
>
The field of Machine Learning seeks to answer the question "How can we build computer systems that automatically improve with experience, and what are the fundamental laws that govern all learning processes?" This question covers a broad range of lear... | null | CC BY-SA 3.0 | null | 2011-04-01T23:43:56.183 | 2017-08-30T17:14:06.607 | 2017-08-30T17:14:06.607 | 171895 | 7365 | null |
9066 | 4 | null | null | 0 | null | Machine learning algorithms build a model of the training data. The term "machine learning" is vaguely defined; it includes what is also called statistical learning, reinforcement learning, unsupervised learning, etc. ALWAYS ADD A MORE SPECIFIC TAG. | null | CC BY-SA 4.0 | null | 2011-04-01T23:43:56.183 | 2019-04-08T14:17:29.407 | 2019-04-08T14:17:29.407 | 28666 | 7365 | null |
9067 | 1 | 9088 | null | 1 | 9404 | I am taking a class in data mining and I am working on a term project using the [BRFSS](http://www.cdc.gov/brfss/) dataset. I have a huge dataset with 405 columns and 12,000 rows. There are many columns which are completely empty. I was trying to remove empty columns using SAS, R or Excel but it doesn't work. Could you... | Removing empty columns from a dataset | CC BY-SA 2.5 | null | 2011-04-02T00:07:35.167 | 2012-12-14T15:19:22.137 | 2011-04-02T13:09:16.883 | null | 3897 | [
"excel"
] |
9068 | 1 | 9090 | null | 12 | 13951 | I am having a problem computing the pearson correlation coefficient of data sets with possibly zero standard deviation (i.e. all data has the same value).
Suppose that I have the following two data sets:
```
float x[] = {2, 2, 2, 3, 2};
float y[] = {2, 2, 2, 2, 2};
```
The correlation coefficient "r", would be compute... | Pearson correlation of data sets with possibly zero standard deviation? | CC BY-SA 2.5 | null | 2011-04-01T14:57:10.287 | 2011-04-29T01:08:20.040 | 2011-04-29T01:08:20.040 | 3911 | 3993 | [
"correlation"
] |
9069 | 2 | null | 9062 | 2 | null | One way to incorporate the ordering of the column factor into your analysis is to use the cumulative frequencies instead of the cell frequencies. So in your table you have:
$$f_{ij}=\frac{n_{ij}}{n_{\bullet\bullet}}\;\;\;\; i=1,2\;\;j=1,2,3$$
where a "$\bullet$" indicates sum over that index. So I suggesting modeling... | null | CC BY-SA 2.5 | null | 2011-04-02T00:25:13.240 | 2011-04-02T00:25:13.240 | null | null | 2392 | null |
9070 | 2 | null | 9062 | 1 | null | You could use the Jonckheere Terpstra test. In SAS, you can get this in PROC FREQ with the /JT option on the tables statement. I didn't see a function for it in R, but there may be one out there.
| null | CC BY-SA 2.5 | null | 2011-04-02T01:14:14.767 | 2011-04-02T01:14:14.767 | null | null | 686 | null |
9071 | 1 | 9073 | null | 18 | 6714 | If I have two normally distributed independent random variables $X$ and $Y$ with means $\mu_X$ and $\mu_Y$ and standard deviations $\sigma_X$ and $\sigma_Y$ and I discover that $X+Y=c$, then (assuming I have not made any errors) the conditional distribution of $X$ and $Y$ given $c$ are also normally distributed with me... | Intuitive explanation of contribution to sum of two normally distributed random variables | CC BY-SA 2.5 | null | 2011-04-02T01:28:12.757 | 2011-04-02T03:05:20.290 | 2017-04-13T12:19:38.800 | -1 | 2958 | [
"normal-distribution",
"conditional-probability"
] |
9072 | 2 | null | 9068 | 0 | null | The correlation is undefined in that case. If you must define it, I would define it as 0, but consider a simple mean absolute difference instead.
| null | CC BY-SA 2.5 | null | 2011-04-02T02:07:54.087 | 2011-04-02T02:07:54.087 | null | null | 2456 | null |
9073 | 2 | null | 9071 | 17 | null | The question readily reduces to the case $\mu_X = \mu_Y = 0$ by looking at $X-\mu_X$ and $Y-\mu_Y$.
Clearly the conditional distributions are Normal. Thus, the mean, median, and mode of each are coincident. The modes will occur at the coordinates of a local maximum of the bivariate PDF of $X$ and $Y$ constrained to t... | null | CC BY-SA 2.5 | null | 2011-04-02T03:05:20.290 | 2011-04-02T03:05:20.290 | null | null | 919 | null |
9074 | 1 | 9082 | null | 7 | 11660 | I'm wondering how to implement two-way clustering, as explained in [Statistica documentation](http://www.statsoft.com/textbook/cluster-analysis/#twotwo) in R. Any help in this regard will be highly appreciated. Thanks
| Two-way clustering in R | CC BY-SA 2.5 | null | 2011-04-02T05:39:09.150 | 2011-04-02T13:07:14.437 | 2011-04-02T13:07:14.437 | null | 3903 | [
"r",
"clustering",
"multivariate-analysis"
] |
9075 | 2 | null | 9064 | 3 | null | you should take a look at kernlab R package. They even have a very nice [vignette](http://cran.r-project.org/web/packages/kernlab/vignettes/kernlab.pdf).
| null | CC BY-SA 2.5 | null | 2011-04-02T06:08:42.183 | 2011-04-02T06:08:42.183 | null | null | 223 | null |
9076 | 1 | 9078 | null | 2 | 547 | I am trying to programmatically identify an ARIMA model for a series of data and forecast values.
Currently the problem i am facing is to find a way to evaluate partial autocorrelation. I have been looking for methods to calculate PACF for quite a long time now but in vain.
Please provide some online resources which ... | Methods for evaluating partial autocorrelation for identification of ARIMA models | CC BY-SA 2.5 | null | 2011-04-02T06:29:34.960 | 2011-04-04T18:01:02.420 | 2011-04-02T10:22:43.337 | 930 | 3972 | [
"autocorrelation",
"arima"
] |
9077 | 2 | null | 9067 | 2 | null | If they really are just completely blank columns then in R...
```
read.table( 'myBigFile', strip.white = TRUE)
```
might do what you want. You will have to set other arguments of the read.table() command as needed. It's best to use this when specifying the specific column delimiter you have.
| null | CC BY-SA 2.5 | null | 2011-04-02T07:11:25.833 | 2011-04-02T07:24:39.017 | 2011-04-02T07:24:39.017 | 601 | 601 | null |
9078 | 2 | null | 9076 | 3 | null | Use the Durbin-Levinson algorithm. It will be explained in any good time series book such as [Brockwell and Davis](http://rads.stackoverflow.com/amzn/click/0387953515). Here are some online explanations:
- http://www.stat.ufl.edu/~berg/sta4853/files/sta4853-4.pdf
- http://amath.colorado.edu/courses/4540/2008Spr/HandO... | null | CC BY-SA 2.5 | null | 2011-04-02T07:12:52.580 | 2011-04-02T07:12:52.580 | null | null | 159 | null |
9079 | 2 | null | 9068 | 6 | null | I agree with sesqu that the correlation is undefined in this case. Depending on your type of application you could e.g. calculate the Gower Similarity between both vectors, which is:
$gower(v1,v2)=\frac{\sum_{i=1}^{n}\delta(v1_i,v2_i)}{n}$ where $\delta$ represents the [kronecker-delta](http://en.wikipedia.org/wiki/Kro... | null | CC BY-SA 2.5 | null | 2011-04-02T08:31:46.503 | 2011-04-02T08:31:46.503 | null | null | 264 | null |
9080 | 1 | null | null | 1 | 304 | The objective of this research was to investigate the long-term effects of irrigation with treated waste water on some chemical soil properties.
The investigation was carried out by comparison of soil properties in two different fields: one irrigated with the effluent from Parkan Waste water Treatment Plant over a per... | How to do a linear model in R? | CC BY-SA 2.5 | null | 2011-04-02T09:09:33.990 | 2011-04-02T12:45:38.977 | 2011-04-02T10:36:18.717 | 930 | 3996 | [
"r",
"regression"
] |
9081 | 2 | null | 9080 | 5 | null | The syntax is actually absurdly simple:
```
mymodel <- lm(dependentvariable ~ continuousvariable + categoricalvariable,
data=yourdata).
```
You can then call `summary()` to get the coefficients, and `plot()` to examine the residuals.
HTH.
| null | CC BY-SA 2.5 | null | 2011-04-02T09:16:40.097 | 2011-04-02T10:37:25.003 | 2011-04-02T10:37:25.003 | 930 | 656 | null |
9082 | 2 | null | 9074 | 9 | null | Generally speaking, you should always find useful pointers by looking at the relevant CRAN TAsk Views, in this case the one that deals with [Cluster](http://cran.r-project.org/web/views/Cluster.html) packages, or maybe [Quick-R](http://www.statmethods.net/).
It's not clear to me whether the link you gave referenced sta... | null | CC BY-SA 2.5 | null | 2011-04-02T10:20:17.737 | 2011-04-02T10:27:54.797 | 2017-04-13T12:44:21.160 | -1 | 930 | null |
9083 | 2 | null | 9080 | 2 | null | First you have to enter your data into R, see [this class note](http://www.ats.ucla.edu/stat/r/notes/entering.htm). You can follow the steps of [this tutorial](http://data.princeton.edu/R/linearModels.html) in the analysis, section 4.4 has a very similar example. In visualization you could do something similar as the `... | null | CC BY-SA 2.5 | null | 2011-04-02T11:24:37.767 | 2011-04-02T12:45:38.977 | 2011-04-02T12:45:38.977 | 3911 | 3911 | null |
9084 | 2 | null | 9064 | 9 | null |
- Basically anything what is not separable with a line (ok, hyperplane), for instance 2D data like this:
kernel trick will effectively project this situation into a (higher-dim) space in which linear separation is possible; see this movie for an effect of a gaussian kernel on similar data.
- Look for a kernel argume... | null | CC BY-SA 2.5 | null | 2011-04-02T13:21:19.140 | 2011-04-02T13:21:19.140 | null | null | null | null |
9085 | 1 | 9146 | null | 20 | 3156 | In my attempts to fight spreadsheet mayhem, I am often evangelical in pushing for more robust tools such as true statistics software (R, Stata, and the like). Recently, I was challenged on this view by someone who stated flat out that they simply will not learn to program. I would like to provide them with data analy... | Software for easy-yet-robust data exploration | CC BY-SA 2.5 | null | 2011-04-02T13:32:36.553 | 2015-07-06T00:51:14.617 | null | null | 3488 | [
"data-visualization",
"software"
] |
9086 | 2 | null | 9085 | 3 | null | A new software system that looks promising for this purpose is [Deducer](http://www.r-bloggers.com/r-ready-to-deduce-you/), built on top of R. Unfortunately, being new, I suspect it does not yet cover the breadth of questions that people might ask, but it does meet the toe-in-the-water criterion of leading people towa... | null | CC BY-SA 3.0 | null | 2011-04-02T13:35:47.410 | 2012-07-29T00:03:30.923 | 2012-07-29T00:03:30.923 | 3488 | 3488 | null |
9087 | 1 | 9091 | null | 7 | 277 | Let $X = N(0,\frac{1}{\alpha})$, $Y = 2X + 8 + N_{y}$, and $N_{y}$ be a noise $N_{y} = N(0,1)$. Then, $P(y|x) = \frac{1}{\sqrt{2\pi}}exp\{ -\frac{1}{2}(y - 2x - 8)^{2} \}$
and $P(x) = \sqrt{\frac{\alpha}{2\pi}}exp\{-\frac{\alpha x^{2}}{2}\} $.
The mean vector is:
$$\mathbf{\mu} = \left( \begin{array}{c}
\mu_{x}\\
... | Inference with Gaussian Random Variable | CC BY-SA 2.5 | null | 2011-04-02T13:49:55.227 | 2011-04-29T00:42:48.390 | 2011-04-29T00:42:48.390 | 3911 | 1371 | [
"normal-distribution",
"variance",
"random-variable",
"inference"
] |
9088 | 2 | null | 9067 | 3 | null | Empty columns contain `NA`s only or `""`s only, they have has no variability. This code removes all columns without variability (which is probably a plus in this case).
```
d=data.frame(r=seq(1, 5), a=rep('a', 5), n=rep(NA, 5), n1=c(NA, NA, 3, 3, 3))
homogenous = apply(d, 2, function(var) length(unique(var)) == 1)
d[,... | null | CC BY-SA 2.5 | null | 2011-04-02T14:05:43.993 | 2011-04-02T14:05:43.993 | null | null | 3911 | null |
9089 | 2 | null | 9087 | 6 | null | Solution to this homework is straightforward application of simple algebra and independence of $X$ and $N_y$: $\mathbb{E} (2 X + N_y)^2 = 4 \mathbb{E} X^2 + 4 \mathbb{E} X \mathbb{E} N_y + \mathbb{E} N_y^2 = 4 Var X + 0 + Var N_y = \frac{4}{\alpha} + 1$.
| null | CC BY-SA 2.5 | null | 2011-04-02T14:05:58.993 | 2011-04-02T16:02:38.683 | 2011-04-02T16:02:38.683 | 2645 | 2645 | null |
9090 | 2 | null | 9068 | 9 | null | The "sampling theory" people will tell you that no such estimate exists. But you can get one, you just need to be reasonable about your prior information, and do a lot harder mathematical work.
If you specified a Bayesian method of estimation, and the posterior is the same as the prior, then you can say the data say n... | null | CC BY-SA 2.5 | null | 2011-04-02T14:06:14.687 | 2011-04-02T14:06:14.687 | null | null | 2392 | null |
9091 | 2 | null | 9087 | 8 | null | The law of iterated expectations can help here. We have:
$$Var[Y]=E(Var[Y|X])+Var[E(Y|X)]$$
Now conditional on $X$ the expected value of $Y$ is $2X+8$, and its variance is $1$. So we have:
$$Var[Y]=E(1)+Var[2X+8]=1+4 Var[X]=1+\frac{4}{\alpha}$$
| null | CC BY-SA 2.5 | null | 2011-04-02T14:25:22.247 | 2011-04-02T14:25:22.247 | null | null | 2392 | null |
9092 | 2 | null | 9068 | 0 | null | This question is coming from programmers, so I'd suggest plugging in zero. There's no evidence of a correlation, and the null hypothesis would be zero (no correlation). There might be other context knowledge that would provide a "typical" correlation in one context, but the code might be re-used in another context.
| null | CC BY-SA 2.5 | null | 2011-04-02T15:04:59.187 | 2011-04-02T15:04:59.187 | null | null | 3919 | null |
9093 | 2 | null | 9085 | 8 | null | Some people think of programming as simply entering a command line statement. At that point then perhaps you are a bit lost in encouraging them. However, if they are using spreadsheets already then they already have to enter formulas. These are akin to command line statements. If they really mean they don't want to... | null | CC BY-SA 2.5 | null | 2011-04-02T15:11:16.043 | 2011-04-04T17:09:56.883 | 2011-04-04T17:09:56.883 | 601 | 601 | null |
9094 | 2 | null | 9085 | 8 | null | As far as exploratory (possibly interactive) data analysis is concerned, I would suggest to take a look at:
- Weka, originally targets data-mining applications, but can be used for data summaries.
- Mondrian, for interactive data visualization.
- KNIME, which relies on the idea of building data flows and is compatib... | null | CC BY-SA 2.5 | null | 2011-04-02T15:11:34.827 | 2011-04-02T15:11:34.827 | null | null | 930 | null |
9096 | 2 | null | 9062 | 5 | null | Linear or monotonic trend tests--$M^2$ association measure, WMW test cited by @GaBorgulya, or the Cochran-Armitage trend test--can also be used, and they are well explained in Agresti ([CDA](http://www.stat.ufl.edu/~aa/cda/cda.html), 2002, §3.4.6, p. 90).
The latter is actually equivalent to a score test for testing $... | null | CC BY-SA 2.5 | null | 2011-04-02T16:46:40.420 | 2011-04-02T16:46:40.420 | 2017-04-13T12:44:46.083 | -1 | 930 | null |
9097 | 2 | null | 5690 | 6 | null | One thing to keep in mind with the Kaplan-Meier survival curve is that it is basically descriptive and not inferential. It is just a function of the data, with an incredibly flexible model that lies behind it. This is a strength because this means there is virtually no assumptions that might be broken, but a weakness... | null | CC BY-SA 2.5 | null | 2011-04-02T16:47:55.793 | 2011-04-02T23:46:01.627 | 2011-04-02T23:46:01.627 | 2392 | 2392 | null |
9098 | 2 | null | 5690 | 3 | null | First I would visualize the data: calculate confidence intervals and standard errors for the median survivals in each state and show CIs on a forest plot, medians and their SEs using a funnel plot.
The “mean median survival all across the country” is a quantity that is estimated from the data and thus has uncertainty ... | null | CC BY-SA 2.5 | null | 2011-04-02T17:07:50.450 | 2011-04-03T01:21:43.877 | 2011-04-03T01:21:43.877 | 3911 | 3911 | null |
9099 | 1 | 52350 | null | 5 | 4778 | Diallel Analysis using the Griffing and Hayman approach is so common in plant breeding and genetics. I'm wondering if someone can share R worked example on Diallel Analysis. Is there any good referenced book which covered worked examples? Thanks
References:
Griffing B (1956) Concept of general and specific combining ab... | How to perform diallel analysis in R? | CC BY-SA 3.0 | null | 2011-04-02T17:18:59.177 | 2013-12-03T09:22:09.447 | 2011-08-21T07:02:46.460 | 5862 | 3903 | [
"r",
"experiment-design",
"genetics"
] |
9100 | 2 | null | 9053 | 4 | null | From a Bayesian perspective, I'm not so sure that cross validation does anything that a "proper" Bayesian analysis doesn't do for comparing models. But I am not 100% certain that it does.
This is because if you are comparing models in a Bayesian way, then you are essentially already doing cross validation. This is be... | null | CC BY-SA 2.5 | null | 2011-04-02T17:40:55.537 | 2011-04-02T17:40:55.537 | null | null | 2392 | null |
9101 | 2 | null | 8567 | 1 | null | I do not know the answer, so I can only offer some thoughts hoping that someone else can throw light on the issue.
It seems to me that there is no problem in computing the likelihood. To compute the value of the AIC criterion (which may or may not make sense in this context), what would be required is the number of fit... | null | CC BY-SA 2.5 | null | 2011-04-02T17:53:53.967 | 2011-04-02T17:53:53.967 | null | null | 892 | null |
9102 | 2 | null | 8898 | 5 | null | As it is described in the original post, the experiment is a randomized block.
- Pathologist (4 levels) is a blocking factor; the experiment is repeated within each pathologist.
- Instrument (3 levels) and the true result (2 levels) of the test are the two treatments, which I assume were assigned randomly.
- Conside... | null | CC BY-SA 3.0 | null | 2011-04-02T19:09:55.037 | 2011-04-24T17:14:47.580 | 2011-04-24T17:14:47.580 | 3874 | 3874 | null |
9103 | 2 | null | 9067 | 0 | null | Hard for me to think of this as a huge dataset, but these things are relative ;)
To do this in excel (a version like 2007 or 2010 which has enough columns), you could insert two rows at the top. In the first row, just have consecutive integers: 1 in column A, 2 in column B, etc. You can do this with a function [I thi... | null | CC BY-SA 2.5 | null | 2011-04-02T20:36:24.113 | 2011-04-02T20:36:24.113 | null | null | 3919 | null |
9104 | 1 | 9161 | null | 12 | 19326 | I have the following question for a course I'm working on:
>
Conduct a Monte Carlo study to estimate the coverage probabilities of
the standard normal bootstrap confidence interval and the basic bootstrap confidence
interval. Sample from a normal population and check the empirical coverage rates for the
sample ... | Coverage probabilities of the basic bootstrap confidence Interval | CC BY-SA 3.0 | null | 2011-04-02T20:42:02.777 | 2015-12-15T23:32:22.733 | 2015-12-15T23:32:22.733 | 4253 | 1894 | [
"r",
"confidence-interval",
"self-study",
"bootstrap",
"monte-carlo"
] |
9105 | 1 | null | null | 16 | 1288 | for the following 3 values 222,1122,45444
[WolframAlpha gives](http://www.wolframalpha.com/input/?i=skew+222%2C1122%2C45444+) 0.706
Excel, using `=SKEW(222,1122,45444)` gives 1.729
What explains the difference?
| Why do Excel and WolframAlpha give different values for skewness | CC BY-SA 2.5 | null | 2011-04-02T21:57:45.523 | 2011-11-20T04:41:40.967 | 2011-11-20T04:41:40.967 | 1381 | 276 | [
"excel",
"software",
"descriptive-statistics",
"mathematica"
] |
9106 | 2 | null | 9105 | 19 | null | They are using different methods to compute the skew. Searching in the help pages for `skewness()` within the R package `e1071` yields:
```
Joanes and Gill (1998) discuss three methods for estimating skewness:
Type 1:
g_1 = m_3 / m_2^(3/2). This is the typical definition used in many older textbooks.
Type 2:
G_1 = g_1... | null | CC BY-SA 2.5 | null | 2011-04-02T22:07:09.333 | 2011-04-02T22:07:09.333 | null | null | 696 | null |
9107 | 1 | 9108 | null | 4 | 281 | I would like to plot the attached [income distribution dataset](https://i.stack.imgur.com/acNhK.jpg) (rendered as an image) as an area chart.
As you can see, personal income is divided into 26 intervals of varying width. I also have the average and mean income in the intervals.
To convey a truthful area graphic of this... | Categorical or continuous scale for area chart? | CC BY-SA 2.5 | null | 2011-04-02T22:08:57.993 | 2011-04-03T12:15:46.313 | 2011-04-03T09:10:18.443 | 930 | 4003 | [
"distributions",
"data-visualization",
"data-transformation",
"categorical-data"
] |
9108 | 2 | null | 9107 | 4 | null | This is not exactly what you asked, but still may be helpful.
You may rely on the `plot.histogram` command in R. The usual use is to run a `hist` command, which prepares an object of class `histogram` and passes it to `plot.histogram`. You may prepare a customized `histogram` object yourself and plot it with `plot.hist... | null | CC BY-SA 2.5 | null | 2011-04-02T22:43:06.963 | 2011-04-03T12:15:46.313 | 2011-04-03T12:15:46.313 | 3911 | 3911 | null |
9109 | 1 | null | null | 2 | 1700 | I have finally been able to wrap my head around the mechanics of how to initialize and train a multivariate Gaussian mixture model using expectation maximization algorithm. So I wonder how difficult this GMM and EM task is in comparison to all other common algorithms and models in machine learning. I appreciate any fee... | How difficult is it to train a gaussian mixture model compared to other models? | CC BY-SA 2.5 | null | 2011-04-02T22:51:25.520 | 2011-04-04T06:35:27.023 | 2011-04-03T09:19:38.667 | 930 | 2729 | [
"mixed-model",
"normal-distribution",
"maximum-likelihood"
] |
9110 | 2 | null | 9085 | 2 | null | Anyone who answers R, or any of it's "GUIs" didn't read the question.
There is a program specifically designed for this and it's called JMP. Yes, it's expensive, though it has a free trial, and is incredibly cheap for students or college staff (like $50 cheap).
There is also RapidMiner, which is a workflow-based GUI ... | null | CC BY-SA 2.5 | null | 2011-04-02T22:53:03.193 | 2011-04-02T22:53:03.193 | null | null | 74 | null |
9111 | 1 | 9112 | null | 9 | 27399 | I am assuming R has this built-in. How do I reference it?
| In R how do I reference\lookup in the cdf of standard normal distribution table? | CC BY-SA 2.5 | null | 2011-04-02T19:10:34.457 | 2011-04-08T20:42:13.073 | 2011-04-08T20:42:13.073 | 919 | 4008 | [
"r",
"normal-distribution"
] |
9112 | 2 | null | 9111 | 13 | null | The functions you are looking for are either `dnorm`, `pnorm` or `qnorm`, depending on exactly what you are looking for.
`dnorm(x)` gives the density function at `x`.
`pnorm(x)` gives the probability that a random value is less than `x`.
`qnorm(p)` is the inverse of `pnorm`, giving the value of `x` for which getting a ... | null | CC BY-SA 2.5 | null | 2011-04-02T20:21:03.230 | 2011-04-03T01:43:38.397 | 2011-04-03T01:43:38.397 | 72 | 72 | null |
9114 | 2 | null | 9109 | 3 | null | [this paper](http://www.uv.es/~bernardo/Kernel.pdf) is a small gem about fitting mixtures of gaussians to the data using Bayesian methods.
Big plus: no maximising algorithms required, so should be quick compared to EM algorithm. Also paper has testing data, so you can test each method against this one if you want. An... | null | CC BY-SA 2.5 | null | 2011-04-03T01:42:41.927 | 2011-04-03T01:42:41.927 | null | null | 2392 | null |
9115 | 2 | null | 9107 | 3 | null | A histogram with a continuous scale as described by GaBorgulya is clearly the way to go. When the blocks are wider, you need to adjust the density appropriately: the block 380-399 with 42246 people should be about 1.6 times the density of the block 400-499 with 132485 people.
Except for the extremes of 0 and 1000+... | null | CC BY-SA 2.5 | null | 2011-04-03T01:51:49.713 | 2011-04-03T01:51:49.713 | null | null | 2958 | null |
9116 | 1 | 9118 | null | 0 | 4822 | First off, be warned: I am a complete stats novice. I'd like to learn, but at the moment I have an intense business problem to solve that I think (hope?) is straightforward enough that it could be answered easily. I will try to explain as simply as I can so I don't muck it up.
In short, i'm trying to find the right way... | How to properly weight contributing percentages? | CC BY-SA 2.5 | null | 2011-04-03T03:04:46.487 | 2011-04-03T03:18:51.280 | 2011-04-03T03:18:29.550 | 4004 | 4004 | [
"pie-chart"
] |
9117 | 2 | null | 9085 | 7 | null | I can recommend Tableau as a good tool for data exploration and visualization, simply because of the different ways that you can explore and view the data, simply by dragging and dropping. The graphs are fairly sharp and you can easily output to PDF for presentation purposes. If you want you can extend it with some "p... | null | CC BY-SA 2.5 | null | 2011-04-03T03:07:13.240 | 2011-04-03T03:07:13.240 | null | null | 3489 | null |
9118 | 2 | null | 9116 | 1 | null | It might be easier than I thought -- please let me know if this is correct or on the right path.
Let's assume the following data:
- Company A: 70 percentage of capture across 10 contracts
- Company B: 60 percentage of capture across 100 contracts
- Company C: 50 percentage of capture across 40 contracts
First, I f... | null | CC BY-SA 2.5 | null | 2011-04-03T03:18:51.280 | 2011-04-03T03:18:51.280 | null | null | 4004 | null |
9119 | 2 | null | 8567 | 3 | null | The dlmMLE function in the dlm package will compute the likelihood. Then
AIC = -2 log(likelihood) + 2p
where p is the number of parameters estimated. You might like to read the [vignette for the dlm package](http://cran.r-project.org/web/packages/dlm/vignettes/dlm.pdf) which contains a lot of helpful information and ... | null | CC BY-SA 2.5 | null | 2011-04-03T04:34:24.763 | 2011-04-03T04:34:24.763 | null | null | 159 | null |
9120 | 2 | null | 2504 | 6 | null | In order to test such a vague hypothesis, you need to average out over all densities with finite variance, and all densities with infinite variance. This is likely to be impossible, you basically need to be more specific. One more specific version of this and have two hypothesis for a sample $D\equiv Y_{1},Y_{2},\dot... | null | CC BY-SA 2.5 | null | 2011-04-03T05:03:28.207 | 2011-04-03T22:43:02.550 | 2011-04-03T22:43:02.550 | 2392 | 2392 | null |
9121 | 1 | 11424 | null | 3 | 1675 | I've seen couple of articles reporting individuals standard deviations and/or standard errors for groups even after implementing ANOVA. My understanding is that groups SE's should be based on experimental error mean square. Any comment?
| Individuals standard deviations and/or standard errors for groups after implementing ANOVA? | CC BY-SA 2.5 | null | 2011-04-03T06:28:11.013 | 2011-06-01T10:11:22.020 | 2011-06-01T10:11:22.020 | 930 | 3903 | [
"anova",
"standard-deviation",
"experiment-design",
"standard-error"
] |
9124 | 1 | null | null | 2 | 4679 | In S-plus estimates of percentiles for a survival function can be obtained using the `qkaplanMeier` function (on the results of a call to kaplanMeier) like that:
```
kfit <-kaplanMeier(censor(TIME,STATUS)~1)
qkaplanMeier(kfit, c(.25, .5, .75))
```
How can I do this in R?. Those functions do not exist anymore. What if ... | Estimates and C.I. of percentiles for a survival function | CC BY-SA 2.5 | null | 2011-04-03T11:06:21.313 | 2023-04-29T06:23:20.560 | 2011-04-03T11:18:23.073 | 339 | 339 | [
"r",
"confidence-interval",
"survival"
] |
9125 | 2 | null | 9124 | 3 | null | The [CRAN task view on survival analysis](http://cran.r-project.org/web/views/Survival.html) says:
Kaplan-Meier:
The
`survfit`
function from the
[survival](http://cran.r-project.org/web/packages/survival/index.html)
package
computes the Kaplan-Meier estimator for truncated and/or censored data.
[rms](http://cran.r-proj... | null | CC BY-SA 4.0 | null | 2011-04-03T11:40:56.150 | 2023-04-29T06:23:20.560 | 2023-04-29T06:23:20.560 | 362671 | 2958 | null |
9126 | 2 | null | 9124 | 2 | null | The `bootkm()` function in [Hmisc](http://cran.r-project.org/web/packages/Hmisc/index.html) provides bootstraped estimate of the probability of survival, as well as the estimate of the quantile of the survival distribution (through either `describe` or `quantile` applied onto the result of `bootkm`).
| null | CC BY-SA 2.5 | null | 2011-04-03T12:44:55.260 | 2011-04-03T12:44:55.260 | null | null | 930 | null |
9127 | 1 | null | null | 3 | 2684 | Where can I obtain all hourly weather data available?
Notes:
- Ideally, this would include the history of all data currently published at:
here
here
here
- I've obtained 10 years of METAR data from wunderground.com like this:
curl -H 'Cookie: Prefs=|SHOWMETAR:1|;' -o data.txt 'http://www.wunderground.com/histor... | How to fetch all historically available hourly weather data? | CC BY-SA 4.0 | null | 2011-04-03T14:27:58.373 | 2022-11-22T02:20:05.457 | 2022-11-22T02:20:05.457 | 362671 | null | [
"time-series",
"dataset"
] |
9128 | 2 | null | 9085 | 4 | null | As John said, data exploration doesn't require much programming in R. Here's a list of data exploration commands you can give people. (I just came up with this; you can surely expand it.)
Export the data from whatever package it's in. (Exporting numerical data without quotation marks is convenient.) Then read the data ... | null | CC BY-SA 2.5 | null | 2011-04-03T14:49:47.713 | 2011-04-03T14:55:02.353 | 2011-04-03T14:55:02.353 | 3874 | 3874 | null |
9129 | 1 | null | null | 5 | 1303 | I am currently looking for some Information Retrieval techniques.
I have a SQL database table containing strings. It has 1000 records, each being a random sentence I picked from random web sites. I need to get the term frequency and represent each string into a vector. I also need to cluster the records, e.g. using k-m... | How to compute term frequency and find clusters in a dataset composed of strings? | CC BY-SA 2.5 | null | 2011-04-03T15:33:54.147 | 2011-09-02T12:37:19.820 | 2011-04-03T16:26:29.463 | 930 | 4020 | [
"clustering",
"information-retrieval"
] |
9130 | 2 | null | 9127 | 3 | null | This may be simplistic, but if you have a consistent directory structure on the NOAA site (they usually are), you can recursive wget the entire thing, then sort through it at your leisure.
```
wget -r http://weather.noaa.gov/pub/SL.us008001/DF.an/
```
This will grab everything recursively from that URL and deeper. It... | null | CC BY-SA 2.5 | null | 2011-04-03T15:41:42.047 | 2011-04-03T15:41:42.047 | null | null | 781 | null |
9131 | 1 | 9144 | null | 22 | 10010 | Let's take the following example:
```
set.seed(342)
x1 <- runif(100)
x2 <- runif(100)
y <- x1+x2 + 2*x1*x2 + rnorm(100)
fit <- lm(y~x1*x2)
```
This creates a model of y based on x1 and x2, using a OLS regression. If we wish to predict y for a given x_vec we could simply use the formula we get from the `summary(fit)`.... | Obtaining a formula for prediction limits in a linear model (i.e.: prediction intervals) | CC BY-SA 4.0 | null | 2011-04-03T18:24:49.593 | 2021-09-27T14:11:17.713 | 2019-04-15T11:33:54.723 | 253 | 253 | [
"r",
"regression",
"predictive-models",
"prediction-interval"
] |
9132 | 1 | 9133 | null | 2 | 2798 | We are trying to predict values of variable A having N other variables. What we do, we calculate Pearson correlation between A and each of the other N variables, for last M values, using fixed M. We use variable with largest correlation coefficient as predictor.
This scheme works fine when analyzing N variables during ... | Alternatives to Pearson correlation | CC BY-SA 2.5 | null | 2011-04-03T20:02:37.747 | 2011-04-03T20:47:44.887 | 2011-04-03T20:39:22.490 | null | 4010 | [
"correlation"
] |
9133 | 2 | null | 9132 | 3 | null | This is generally hard to tell without knowing what the problem exactly is, but I would advise you to try some machine learning methods. For start you may try random forest, which is almost trivial to apply and quite probably will achieve better accuracy than just using one, best-correlated variable.
Also, it will prod... | null | CC BY-SA 2.5 | null | 2011-04-03T20:47:44.887 | 2011-04-03T20:47:44.887 | null | null | null | null |
9134 | 2 | null | 9131 | 9 | null | Are you by chance after the different types of prediction intervals? The `predict.lm` manual page has
```
## S3 method for class 'lm'
predict(object, newdata, se.fit = FALSE, scale = NULL, df = Inf,
interval = c("none", "confidence", "prediction"),
level = 0.95, type = c("response", "terms"),
... | null | CC BY-SA 2.5 | null | 2011-04-03T21:04:33.583 | 2011-04-03T21:04:33.583 | null | null | 334 | null |
9135 | 1 | null | null | 4 | 205 | Given two point sets, $B$ consisting of blue points, and $R$ of red points, on the plane.
The problem is to formulate a theoretical model to compare the average runtime of the Computational Geometric (CG) Algorithm and the [ Perceptron Learning (PL) ](http://en.wikipedia.org/wiki/Perceptron#Learning_algorithm) for shat... | Computational geometry vs perceptron for shattering | CC BY-SA 2.5 | null | 2011-04-03T21:12:07.560 | 2021-05-22T00:59:36.957 | 2021-05-22T00:59:36.957 | 11887 | 4011 | [
"neural-networks",
"algorithms",
"geometry"
] |
9136 | 2 | null | 8911 | 0 | null | Try [http://tukhi.com](http://tukhi.com). It is not clear whether or not they have a Mac OS X Excel version, but they have contact info on that site. It is pretty amazing. Heh, you could always run Window in a VM.
| null | CC BY-SA 2.5 | null | 2011-04-03T21:31:49.947 | 2011-04-03T21:31:49.947 | null | null | null | null |
9137 | 1 | 9139 | null | 8 | 2923 | Say there are 3 companies A, B and C. Each company has a quality rating from 0 to 100 and a price in USD.
```
Company Quality Price
A 80 7.9
B 70 8.0
C 75 8.1
```
How do I determine the best quality-price trade-off? What kind of analysis should I use?
| Quality-price trade-off | CC BY-SA 2.5 | null | 2011-04-03T22:17:08.980 | 2013-06-28T13:24:24.787 | 2013-06-28T13:24:24.787 | 919 | 4013 | [
"valuation"
] |
9139 | 2 | null | 9137 | 11 | null | The [Keeney-Raiffa approach to Multi-attribute valuation theory](http://rads.stackoverflow.com/amzn/click/0521438837) is well-grounded practically and theoretically, has been successfully applied to many problems, and--when applied to problems with just two attributes--is particularly simple. It proceeds by systematic... | null | CC BY-SA 2.5 | null | 2011-04-03T23:04:03.450 | 2011-04-03T23:04:03.450 | null | null | 919 | null |
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