Id stringlengths 1 6 | PostTypeId stringclasses 7
values | AcceptedAnswerId stringlengths 1 6 ⌀ | ParentId stringlengths 1 6 ⌀ | Score stringlengths 1 4 | ViewCount stringlengths 1 7 ⌀ | Body stringlengths 0 38.7k | Title stringlengths 15 150 ⌀ | ContentLicense stringclasses 3
values | FavoriteCount stringclasses 3
values | CreationDate stringlengths 23 23 | LastActivityDate stringlengths 23 23 | LastEditDate stringlengths 23 23 ⌀ | LastEditorUserId stringlengths 1 6 ⌀ | OwnerUserId stringlengths 1 6 ⌀ | Tags list |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
2941 | 2 | null | 2938 | 1 | null | I would say that the more or less canonical way to compare two distributions would be a chi-squared test. The statistic is not normalized, though, and it depends on how you choose the bins. The last point can of course be seen as a feature, not a bug: choosing bins appropriately allows you to look more closely for simi... | null | CC BY-SA 2.5 | null | 2010-09-21T07:46:52.397 | 2010-09-21T07:46:52.397 | null | null | 1352 | null |
2942 | 2 | null | 2917 | 7 | null | I can only agree with John. Furthermore, perhaps [this](http://eskes.psychiatry.dal.ca/Files/2003_-_Saville_-_Basic_statistics_and_the_inconsistency_of_m.pdf) paper by David Saville helps you with some formula to recalculate variability measures from LSDs et al.:
Saville D.J. (2003). [Basic statistics and the inconsist... | null | CC BY-SA 2.5 | null | 2010-09-21T08:55:35.920 | 2010-09-23T16:56:50.490 | 2010-09-23T16:56:50.490 | 442 | 442 | null |
2944 | 2 | null | 2860 | 3 | null | Following the comments exchange with Ebony (see Whuber's answer). I gather that in Ebony's application, $p$ is much larger than $n$ which is itself very large. In this case the complexity of computing the eigen decomposition is in the order of $O(n^3)$. Two solutions spring to mind:
- partial decomposition: assuming $... | null | CC BY-SA 2.5 | null | 2010-09-21T11:45:56.080 | 2010-09-22T00:35:28.130 | 2010-09-22T00:35:28.130 | 603 | 603 | null |
2945 | 2 | null | 2828 | 1 | null | From Yaroslav's comments to Henrik's answer:
but cross-validation seems to just postpone the task of assessing complexity. If you use data to pick your parameters and your model as in cross-validation, the relevant question becomes how estimate the amount of data needed for this "meta"-fitter to perform well
I wonder ... | null | CC BY-SA 2.5 | null | 2010-09-21T12:08:33.850 | 2010-09-21T12:35:49.923 | 2010-09-21T12:35:49.923 | 603 | 603 | null |
2946 | 2 | null | 2932 | 4 | null | Interesting reference. Its value for me lies in questioning the ability of measure theoretic probability to capture an "intuition" about probability (whatever that might mean) and going on to propose an intriguing distinction; namely, between a set of measure zero having a measure zero neighborhood and a set of measur... | null | CC BY-SA 3.0 | null | 2010-09-21T14:28:05.750 | 2012-01-08T17:18:07.957 | 2012-01-08T17:18:07.957 | 919 | 919 | null |
2947 | 2 | null | 2938 | 2 | null | I [recently](http://www.thinkingaboutthinking.org/wp-content/uploads/2010/05/Lawrence_BRM_in_press.pdf) used the correlation between the empirical CDF and the fitted CDF to quantify goodness-of-fit, and I wonder if this approach might also be useful in the current case, which as I understand it involves comparing two e... | null | CC BY-SA 2.5 | null | 2010-09-21T15:15:12.333 | 2010-09-21T15:15:12.333 | null | null | 364 | null |
2948 | 1 | 2951 | null | 47 | 28131 | I'm wondering if someone could suggest what are good starting points when it comes to performing community detection/graph partitioning/clustering on a graph that has weighted, undirected edges. The graph in question has approximately 3 million edges and each edge expresses the degree of similarity between the two vert... | How to do community detection in a weighted social network/graph? | CC BY-SA 3.0 | null | 2010-09-21T15:50:45.697 | 2018-01-15T12:14:55.440 | 2016-05-18T16:29:25.253 | 7290 | 1007 | [
"clustering",
"data-visualization",
"networks",
"partitioning",
"modularity"
] |
2950 | 1 | null | null | 3 | 3178 | I am trying to determine significant differences between groups of data using a k-nearest neighbor randomization test in R. This test basically looks for similarity amongst homogenous groups and separates them out using random clustering. In the literature, this test is called a "K-nearest neighbor (kNN) randomization ... | kNN randomization test in R? | CC BY-SA 2.5 | null | 2010-09-21T17:19:53.943 | 2012-05-16T06:19:59.067 | 2011-01-18T00:37:31.553 | null | null | [
"r",
"k-nearest-neighbour"
] |
2951 | 2 | null | 2948 | 23 | null | igraph implementation of Newman's modularity clustering (fastgreedy function) can be used with weighted edges as well. Just add weight attribute to the edges and analyse as usual. In my experience, it run even faster with weights as there are less ties.
| null | CC BY-SA 2.5 | null | 2010-09-21T17:20:48.360 | 2010-09-21T17:20:48.360 | null | null | 1387 | null |
2952 | 2 | null | 2948 | 10 | null | I know that [Gephi](http://gephi.org/) can process undirected weighted graph, but I seem to remember it has to be stored in [GDF](http://gephi.org/users/supported-graph-formats/gdf-format/), which is pretty close to CSV, or Ucinet [DL](http://gephi.org/users/supported-graph-formats/ucinet-dl-format/). Be aware that it'... | null | CC BY-SA 2.5 | null | 2010-09-21T17:45:48.903 | 2010-09-21T21:07:25.300 | 2010-09-21T21:07:25.300 | 930 | 930 | null |
2953 | 2 | null | 2950 | 3 | null | kmeans() (you can find it by typing `??` followed by the name of what you want as in `??kmean`). In general, a good trick is to first look it up on a R specific search [engine](http://rseek.org/)
| null | CC BY-SA 2.5 | null | 2010-09-21T18:13:57.677 | 2010-09-21T18:44:13.107 | 2010-09-21T18:44:13.107 | 603 | 603 | null |
2954 | 2 | null | 2950 | 3 | null | You appear to have confused "cluster analysis" with "classification". The former is where you don't know the groupings and wish to determine them from the training data to hand. Classification is where you know the groups and want to predict them.
There are a few packages in R that do this. For example, look at the res... | null | CC BY-SA 2.5 | null | 2010-09-21T21:34:04.403 | 2010-09-21T22:20:21.663 | 2010-09-21T22:20:21.663 | 1390 | 1390 | null |
2955 | 2 | null | 2950 | 3 | null | I am not sure to understand your question since you talk about k-means, which is basically an unsupervised method (i.e. where classes are not known a priori), while at the same time you are saying that you already identified groups of individuals. So I would suggest to look at classification methods, or other supervise... | null | CC BY-SA 2.5 | null | 2010-09-21T21:45:12.007 | 2010-09-21T22:05:16.047 | 2010-09-21T22:05:16.047 | 930 | 930 | null |
2956 | 1 | null | null | 18 | 616 | I'm the author of the [ez package](http://cran.r-project.org/package=ez) for R, and I'm working on an update to include automatic computation of likelihood ratios (LRs) in the output of ANOVAs. The idea is to provide a LR for each effect that is analogous to the test of that effect that the ANOVA achieves. For example,... | Have I computed these likelihood ratios correctly? | CC BY-SA 2.5 | null | 2010-09-21T23:40:39.317 | 2018-08-27T16:09:42.907 | 2018-08-27T16:09:42.907 | 11887 | 364 | [
"r",
"anova",
"likelihood-ratio"
] |
2957 | 1 | 7716 | null | 20 | 3140 | The Gauss-Markov theorem tells us that the OLS estimator is the best linear unbiased estimator for the linear regression model.
But suppose I don't care about linearity and unbiasedness. Then is there some other (possible nonlinear/biased) estimator for the linear regression model which is the most efficient under the... | OLS is BLUE. But what if I don't care about unbiasedness and linearity? | CC BY-SA 2.5 | null | 2010-09-22T01:15:49.873 | 2022-09-17T16:31:59.187 | 2010-10-19T07:20:15.037 | 449 | 1393 | [
"regression",
"unbiased-estimator"
] |
2958 | 1 | null | null | 2 | 776 | I know the value for the 16% quartile, so I know the additive deviation for the given distribution. How do I find the deviation of the log of the given distribution on a multiplicative scale?
| How do you calculate the standard deviation on a multiplicative scale for a distribution that has been transformed logarithmically? | CC BY-SA 2.5 | null | 2010-09-22T01:21:15.247 | 2010-09-22T08:22:25.940 | null | null | null | [
"distributions",
"standard-deviation",
"logarithm"
] |
2959 | 2 | null | 2957 | 9 | null | I don't know if you are OK with the Bayes Estimate? If yes, then depending on the Loss function you can obtain different Bayes Estimates. A theorem by Blackwell states that Bayes Estimates are never unbiased. A decision theoretic argument states that every admissible rule ((i.e. or every other rule against which it is ... | null | CC BY-SA 2.5 | null | 2010-09-22T01:47:20.280 | 2011-02-28T16:24:13.567 | 2011-02-28T16:24:13.567 | 1307 | 1307 | null |
2960 | 2 | null | 2925 | 0 | null | an easy way to generate symmetric bernoulli trials is to flip a coin twice. if the first toss is H and the second is T, say X = 1. if it's the other way round, say X = 0. if the two tosses match [2H or 2T], discard the outcome and continue. no matter what the bias of the coin, X will be symmetric bernoulli.
| null | CC BY-SA 2.5 | null | 2010-09-22T03:42:41.153 | 2010-09-22T03:42:41.153 | null | null | 1112 | null |
2961 | 2 | null | 2957 | 5 | null | There is a nice review paper by [Kay and Eldar](http://webee.technion.ac.il/Sites/People/YoninaEldar/Download/67j-04490210.pdf) on biased estimation for the purpose of finding estimators with minimum mean square error.
| null | CC BY-SA 2.5 | null | 2010-09-22T04:00:30.520 | 2010-09-22T04:00:30.520 | null | null | 352 | null |
2962 | 1 | 125649 | null | 11 | 12982 | The statistics book I am reading recommends omega squared to measure the effects of my experiments. I have already proven using a split plot design (mix of within-subjects and between-subjects design) that my within-subjects factors are statistically significant with p<0.001 and F=17.
Now I'm looking to see how big is ... | Omega squared for measure of effect in R? | CC BY-SA 2.5 | null | 2010-09-22T04:38:22.253 | 2019-01-27T17:08:35.430 | 2011-01-14T19:31:12.340 | 449 | 1320 | [
"r",
"anova",
"effect-size",
"split-plot"
] |
2963 | 2 | null | 2925 | 7 | null | Just as another source of verifiable randomness: random.org generates random numbers from atmospheric noise. They publish [a daily file (most days) of random bits](http://www.random.org/files/); the first digit of each day's file might prove suitably verifiable to your parties.
---
Update 2013-11-12: Access to thes... | null | CC BY-SA 3.0 | null | 2010-09-22T05:25:07.513 | 2013-11-12T19:18:29.360 | 2013-11-12T19:18:29.360 | 71 | 71 | null |
2965 | 2 | null | 2958 | 2 | null | I assume you are referring to something like the estimated coefficients in a logistic regression. These are the log-odds. The estimates usually have a standard error and symmetrical confidence interval.
For example lets say an estimated log odds is 2 with an SE of 0.5 and 95% CI of 1.02 to 2.98. The odds ratio you cal... | null | CC BY-SA 2.5 | null | 2010-09-22T08:22:25.940 | 2010-09-22T08:22:25.940 | null | null | 521 | null |
2966 | 1 | 2967 | null | 9 | 214 | Context
- you have 200 observations of an individual's running time for the 100 metres measured once a day for 200 days.
- Assume the individual was not a runner before commencement of practice
- Based on the observed data and the 199 other observations, you want to estimate the latent time it would take the individ... | Estimating latent performance potential based on a sequence of observations | CC BY-SA 2.5 | null | 2010-09-22T10:32:51.633 | 2021-03-21T17:39:32.187 | 2015-08-31T19:19:04.880 | 11887 | 183 | [
"time-series",
"bayesian",
"latent-variable",
"isotonic"
] |
2967 | 2 | null | 2966 | 7 | null | You need to perform an isotonic (i.e. monotonic non decreasing) nonparametric regression (see page 6 of [this](https://www.semanticscholar.org/paper/%E2%80%98monoProc%E2%80%99-Version-1.0-5-Strictly-monotone-and-in-R-Scheder/196e7585c8a525c6feaac27147b29bebc0f9b43b) document for an example), then use $\hat{E}(y|x)+ \de... | null | CC BY-SA 4.0 | null | 2010-09-22T11:01:39.503 | 2021-03-21T17:39:32.187 | 2021-03-21T17:39:32.187 | 603 | 603 | null |
2968 | 2 | null | 2925 | 6 | null | Many countries have state lottery which is regularly audited, and whose results are announced online: e.g. [UK national lottery](https://www.national-lottery.co.uk/player/p/results/lotto.ftl). You just need to construct an appropriate function which maps this output space to your desired output.
A continuous distribut... | null | CC BY-SA 2.5 | null | 2010-09-22T11:53:01.753 | 2010-09-22T11:53:01.753 | null | null | 495 | null |
2969 | 2 | null | 2962 | 1 | null | I'd suggest that generalized eta square is considered ([ref](http://www.ncbi.nlm.nih.gov/pubmed/14664681), [ref](http://brm.psychonomic-journals.org/content/37/3/379.short)) a more appropriate measure of effect size. It is included in the ANOVA output in the [ez package](http://cran.r-project.org/package=ez) for R.
| null | CC BY-SA 2.5 | null | 2010-09-22T12:32:55.730 | 2010-09-22T12:32:55.730 | null | null | 364 | null |
2970 | 2 | null | 2966 | 2 | null | Just a guess.
First I would explore transformations of the data, such as converting time to speed or acceleration. Then I would consider the log of that, since it obviously won't be negative.
Then, since you are interested in the asymptote, I would try fitting (by least squares) a simple exponential to the transformed ... | null | CC BY-SA 2.5 | null | 2010-09-22T12:36:38.767 | 2010-09-22T13:28:08.330 | 2010-09-22T13:28:08.330 | 1270 | 1270 | null |
2971 | 1 | 2985 | null | 17 | 1551 | If you fit a non linear function to a set of points (assuming there is only one ordinate for each abscissa) the result can either be:
- a very complex function with small residuals
- a very simple function with large residuals
Cross validation is commonly used to find the "best" compromise between these two extrem... | What is the definition of "best" as used in the term "best fit" and cross validation? | CC BY-SA 2.5 | null | 2010-09-22T14:11:12.697 | 2013-02-11T12:23:57.870 | 2010-09-23T10:54:12.720 | null | 1134 | [
"model-selection",
"cross-validation"
] |
2972 | 1 | 2989 | null | 7 | 1503 | A little while back, J.M. [suggested](https://stats.stackexchange.com/questions/2746/how-to-efficiently-generate-positive-semi-definite-correlation-matrices/2786#2786) using the [Stewart](http://www.jstor.org/stable/2156882) algorithm for generating $n$ by $n$ pseudo random orthogonal matrices in $O(n^2)$ time. He furt... | Pseudo-random orthogonal matrix generation | CC BY-SA 2.5 | null | 2010-09-22T14:25:01.887 | 2010-09-23T10:45:55.277 | 2017-04-13T12:44:26.710 | -1 | 603 | [
"random-generation",
"matrix"
] |
2973 | 2 | null | 665 | 3 | null | Statistics is the pursuit of truth in the face of uncertainty. Probability is the tool that allows us to quantify uncertainty.
(I have provided another, longer, answer that assumed that what was being asked was something along the lines of "how would you explain it to your grandmother?")
| null | CC BY-SA 2.5 | null | 2010-09-22T14:31:43.833 | 2010-09-22T14:31:43.833 | null | null | 666 | null |
2975 | 1 | 2978 | null | 5 | 2290 | How do I detrend or normalize multiple series of data so that I can inter-compare between the series?
---
Specifics below may not be appropriate for this forum. Please let me know and I can remove or re-phrase, but I think it might be helpful to fully understand the generic question above.
I have a data set that I ... | Normalizing or detrending groups of samples | CC BY-SA 2.5 | null | 2010-09-22T16:35:45.250 | 2010-09-26T07:45:00.947 | 2017-04-13T12:33:47.693 | -1 | 957 | [
"time-series",
"data-visualization",
"variance",
"normalization"
] |
2976 | 1 | 2998 | null | 31 | 31330 | Questions:
- I have a large correlation matrix. Instead of clustering individual correlations, I want to cluster variables based on their correlations to each other, ie if variable A and variable B have similar correlations to variables C to Z, then A and B should be part of the same cluster. A good real life example ... | Clustering variables based on correlations between them | CC BY-SA 2.5 | null | 2010-09-22T17:01:37.580 | 2017-04-06T14:04:02.507 | 2015-11-24T10:38:37.860 | 28666 | 1250 | [
"correlation",
"clustering",
"correlation-matrix"
] |
2977 | 2 | null | 2971 | 9 | null | I will offer a brief intuitive answer (at a fairly abstract level) till a better answer is offered by someone else:
First, note that complex functions/models achieve better fit (i.e., have lower residuals) as they exploit some local features (think noise) of the dataset that are not present globally (think systematic ... | null | CC BY-SA 2.5 | null | 2010-09-22T17:04:06.030 | 2010-09-22T17:19:31.487 | 2010-09-22T17:19:31.487 | null | null | null |
2978 | 2 | null | 2975 | 5 | null | Multi-level modelling where your data are grouped by flight as a random variable sounds like a good analysis method for this problem. In R the code might be
```
library(lme4) #load the package)
lmer(temp ~ region + (1|flight))
```
This is doable in a variety of statistics packages. If region is simply in region or o... | null | CC BY-SA 2.5 | null | 2010-09-22T17:21:32.487 | 2010-09-26T07:45:00.947 | 2010-09-26T07:45:00.947 | 601 | 601 | null |
2979 | 2 | null | 2904 | 0 | null | You could avoid the problem altogether by simply estimating
W = alphaH * H + alphaM * M + alphaL * L + X * beta
using 2sls. The fact that H,M, and L are discrete doesn't violate any of the assumptions of 2sls. Of course, using maximum likelihood will produce more efficient estimates, but it relies on more assumptions. ... | null | CC BY-SA 2.5 | null | 2010-09-22T17:41:07.970 | 2010-09-22T17:41:07.970 | null | null | 1229 | null |
2980 | 2 | null | 2925 | 1 | null | I didn't quite understand what you meant by "on the basis of an external event." But you can certainly flip a fair coin in a manner that a remote user can cryptographically verify.
Consider this algorithm:
- Bob picks a uniformly random boolean value, TRUE or FALSE. He also chooses a large random number. He sends Alic... | null | CC BY-SA 2.5 | null | 2010-09-22T18:26:26.657 | 2010-09-23T02:54:38.517 | 2010-09-23T02:54:38.517 | 1122 | 1122 | null |
2981 | 1 | 2986 | null | 7 | 3553 | With the help of several people in this community I have been wetting my feet in clustering some social network data using [igraph's implementation of modularity-based clustering](http://igraph.sourceforge.net/doc/R/fastgreedy.community.html).
I am having some trouble interpreting the output of this routine and how to ... | Interpreting output of igraph's fastgreedy.community clustering method | CC BY-SA 3.0 | null | 2010-09-22T18:49:28.433 | 2016-05-20T16:38:16.397 | 2020-06-11T14:32:37.003 | -1 | 1007 | [
"clustering",
"networks",
"partitioning",
"igraph",
"modularity"
] |
2982 | 1 | 3005 | null | 12 | 3582 | I am trying to summarize what I understood so far in penalized multivariate analysis with high-dimensional data sets, and I still struggle through getting a proper definition of soft-thresholding vs. Lasso (or $L_1$) penalization.
More precisely, I used sparse PLS regression to analyze 2-block data structure including ... | Soft-thresholding vs. Lasso penalization | CC BY-SA 2.5 | null | 2010-09-22T20:53:20.303 | 2010-12-21T15:30:40.507 | 2010-12-21T15:30:40.507 | 930 | 930 | [
"multivariate-analysis",
"lasso",
"feature-selection",
"genetics"
] |
2983 | 2 | null | 2971 | 1 | null | The error function is the error of your model (function) on the training data. The complexity is some norm (e.g., squared l2 norm) of the function you are trying to learn. Minimizing the complexity term essentially favors smooth functions, which do well not just on the training data but also on the test data. If you re... | null | CC BY-SA 2.5 | null | 2010-09-22T20:59:33.923 | 2010-09-22T21:05:22.323 | 2010-09-22T21:05:22.323 | 881 | 881 | null |
2984 | 2 | null | 2971 | 1 | null | From an optimization point of view, the problem (with $(p,q)\geq 1,\;\lambda>0$),
$(1)\;\underset{\beta|\lambda,x,y}{Arg\min.}||y-m(x,\beta)||_p+\lambda||\beta||_q$
is equivalent to
$(2)\;\underset{\beta|\lambda,x,y}{Arg\min.}||y-m(x,\beta)||_p$
$s.t.$ $||\beta||_q\leq\lambda$
Which simply incorporates unto the obj... | null | CC BY-SA 2.5 | null | 2010-09-22T21:47:06.203 | 2010-09-23T00:56:12.477 | 2010-09-23T00:56:12.477 | 603 | 603 | null |
2985 | 2 | null | 2971 | 7 | null | I think this is an excellent question. I am going to paraphase it just to be sure I have got it right:
>
It would seem that there are lots of
ways to choose the complexity penalty
function $c$ and error penalty
function $e$. Which choice is `best'.
What should best even mean?
I think the answer (if there is... | null | CC BY-SA 2.5 | null | 2010-09-22T22:11:42.100 | 2010-09-22T23:28:21.123 | 2010-09-22T23:28:21.123 | 352 | 352 | null |
2986 | 2 | null | 2981 | 3 | null | The function which is used for this purpose:
community.to.membership(graph, merges, steps, membership=TRUE, csize=TRUE)
this can be used to extract membership based on the fastgreedy.community function results.
You have to provide number of steps - how many merges should be performed. The optimal number of steps(merg... | null | CC BY-SA 2.5 | null | 2010-09-22T23:05:26.657 | 2010-09-22T23:05:26.657 | null | null | 1396 | null |
2987 | 2 | null | 7 | 2 | null | This is probably the most complete list you'll find: [Some Datasets Available on the Web](http://www.datawrangling.com/some-datasets-available-on-the-web)
| null | CC BY-SA 2.5 | null | 2010-09-22T23:57:02.360 | 2010-09-22T23:57:02.360 | null | null | 635 | null |
2988 | 1 | null | null | 12 | 8192 | How does one calculate the sample size needed for a study in which a cohort of subjects will have a single continuous variable measured at the time of a surgery and then two years later they will be classified as functional outcome or impaired outcome.
We would like to see if that measurement could have predicted the b... | Sample size calculation for univariate logistic regression | CC BY-SA 2.5 | null | 2010-09-23T00:12:51.567 | 2010-12-03T11:40:53.517 | null | null | 104 | [
"logistic",
"sample-size"
] |
2989 | 2 | null | 2972 | 9 | null | It's in the Test Matrix Toolbox, not the Matrix Computation Toolbox. The M-file `qmult.m` (premultiplication by a Haar-distributed pseudorandom orthogonal matrix) can be found [here](http://www.netlib.org/toms/694) or [here](http://people.sc.fsu.edu/~jburkardt/m_src/test_matrix/qmult.m).
| null | CC BY-SA 2.5 | null | 2010-09-23T00:22:16.190 | 2010-09-23T00:22:16.190 | null | null | 830 | null |
2998 | 2 | null | 2976 | 16 | null | Here's a simple example in R using the `bfi` dataset: bfi is a dataset of 25 personality test items organised around 5 factors.
```
library(psych)
data(bfi)
x <- bfi
```
A hiearchical cluster analysis using the euclidan distance between variables based on the absolute correlation between variables can be obtained lik... | null | CC BY-SA 2.5 | null | 2010-09-23T01:38:24.977 | 2010-10-29T11:20:13.590 | 2010-10-29T11:20:13.590 | 183 | 183 | null |
2999 | 2 | null | 2988 | 7 | null | Sample size calculations for logistic regression are complex. I wont attempt to summarise it here. Reasonably accessible solutions to this problem are found in:
[Hsieh FY. Sample size tables for logistic regression. Statistics in Medicine. 1989 Jul;8(7):795-802.](http://onlinelibrary.wiley.com/doi/10.1002/sim.478008070... | null | CC BY-SA 2.5 | null | 2010-09-23T01:56:37.513 | 2010-09-23T01:56:37.513 | null | null | 521 | null |
3000 | 2 | null | 2988 | 3 | null | a simple question about sample size is: how large a sample is needed to get a 95% confidence interval no longer than 2d for the [unknown] mean of the data distribution. another variant is: how large a sample is needed to have power 0.9 at $\theta = 1$ when testing H$_0: \theta = 0$. you don't seem to specify any criter... | null | CC BY-SA 2.5 | null | 2010-09-23T02:13:36.460 | 2010-09-26T01:42:35.190 | 2010-09-26T01:42:35.190 | 1112 | 1112 | null |
3001 | 1 | 3002 | null | 7 | 339 | So let's say you have a distribution where X is the 16% quantile. Then you take the log of all the values of the distribution. Would log(X) still be the 16% quantile in the log distribution?
| Log graph question | CC BY-SA 2.5 | null | 2010-09-23T03:57:03.107 | 2010-11-02T13:51:35.310 | 2010-11-02T13:51:35.310 | 8 | 1395 | [
"standard-deviation",
"logarithm",
"quantiles",
"self-study"
] |
3002 | 2 | null | 3001 | 10 | null | Yes. Quantiles can be transformed under any monotonically increasing transformation.
To see this, suppose $Y$ is the random variable and $q_{0.16}$ is the 16% quantile. Then
$$
\text{Pr}(Y\le q_{0.16}) = \text{Pr}(\log(Y)\le\log(q_{0.16})) = 0.16.
$$
Generally, if $f$ is monotonic and increasing then
$$
\text{Pr}(Y\le... | null | CC BY-SA 2.5 | null | 2010-09-23T04:30:57.217 | 2010-09-24T03:06:00.800 | 2010-09-24T03:06:00.800 | 159 | 159 | null |
3003 | 2 | null | 7 | 2 | null | Peter Skomoroch maintains a list of datasets at [http://www.datawrangling.com/some-datasets-available-on-the-web](http://www.datawrangling.com/some-datasets-available-on-the-web). Many of the links provided as to places that list datasets.
| null | CC BY-SA 2.5 | null | 2010-09-23T06:10:48.023 | 2010-09-23T06:10:48.023 | null | null | 1392 | null |
3004 | 2 | null | 2971 | 3 | null | A lot of people have excellent answers, here is my $0.02.
There are two ways to look at "best model", or "model selection", speaking statistically:
1 An explanation that is as simple as
possible, but no simpler (Attrib.
Einstein)
```
- This is also called Occam's Razor, as explanation applies here.
- Have a conc... | null | CC BY-SA 2.5 | null | 2010-09-23T07:39:23.560 | 2010-09-23T07:39:23.560 | null | null | 1307 | null |
3005 | 2 | null | 2982 | 2 | null | What i'll say holds for regression, but should be true for PLS also. So it's not a bijection because depeding on how much you enforce the constrained in the $l1$, you will have a variety of 'answers' while the second solution admits only $p$ possible answers (where $p$ is the number of variables) <-> there are more sol... | null | CC BY-SA 2.5 | null | 2010-09-23T07:51:46.680 | 2010-09-23T23:33:14.197 | 2010-09-23T23:33:14.197 | 603 | 603 | null |
3006 | 1 | 3010 | null | 9 | 1290 | An anonymous reader posted [the following question on my blog](http://jeromyanglim.blogspot.com/2009/11/tips-for-writing-up-research-in.html?showComment=1285227037363#c3803430820186070755).
Context:
The reader wanted to run a factor analysis on scales from a questionnaire - but the data was from paired husbands and wi... | Factor analysis of dyadic data | CC BY-SA 2.5 | null | 2010-09-23T07:58:18.980 | 2020-07-13T03:26:44.700 | 2020-07-13T03:26:44.700 | 11887 | 183 | [
"independence",
"factor-analysis",
"dyadic-data"
] |
3008 | 2 | null | 2988 | 7 | null | I usually find it easier and faster to run a simulation. Papers take a long time to read, to understand and finally come to the conclusion that they don't apply in the special case one is interested in.
Therefore, I would just pick a number of subjects, simulate the covariate you are interested in (distributed as you b... | null | CC BY-SA 2.5 | null | 2010-09-23T08:16:37.353 | 2010-10-02T20:05:22.870 | 2010-10-02T20:05:22.870 | 1352 | 1352 | null |
3009 | 2 | null | 3006 | 6 | null | Yes, s/he can run a factor analysis on dyadic data. I would start with Kenny et al.'s (2006) "[Dyadic Data Analysis](http://davidakenny.net/kkc/kkc.htm)". It is a great and extremly helpful book!
Another option is "Modeling Dyadic and Interdependent Data in the Developmental and Behavioral Sciences" (Card et al. 2008)... | null | CC BY-SA 2.5 | null | 2010-09-23T09:07:55.583 | 2010-09-23T09:17:29.330 | 2010-09-23T09:17:29.330 | 307 | 307 | null |
3010 | 2 | null | 3006 | 7 | null | Structural equation models are better suited for this kind of data, e.g. by introducing an extra factor for couple which allows to account for the dependence structure (paired responses). David A. Kenny reviewed the main points for [analysis dyadic data](http://davidakenny.net/dyad.htm); although it doesn't focus on qu... | null | CC BY-SA 2.5 | null | 2010-09-23T09:34:29.073 | 2010-09-23T09:42:17.033 | 2010-09-23T09:42:17.033 | 930 | 930 | null |
3011 | 2 | null | 2971 | 5 | null | In a general machine-learning view the answer is fairly simple: we want to build model that will have the highest accuracy when predicting new data (unseen during training). Because we cannot directly test this (we don't have data from the future) we do Monte Carlo simulation of such a test -- and this is basically the... | null | CC BY-SA 2.5 | null | 2010-09-23T10:31:04.193 | 2010-09-23T10:31:04.193 | null | null | null | null |
3012 | 5 | null | null | 0 | null | null | CC BY-SA 2.5 | null | 2010-09-23T10:59:13.660 | 2010-09-23T10:59:13.660 | 2010-09-23T10:59:13.660 | null | null | null | |
3013 | 4 | null | null | 0 | null | Model selection is a problem of judging which model from some set performs best. Popular methods include $R^2$, AIC and BIC criteria, test sets, and cross-validation. To some extent, feature selection is a subproblem of model selection. | null | CC BY-SA 3.0 | null | 2010-09-23T10:59:13.660 | 2011-06-15T03:58:08.313 | 2011-06-15T03:58:08.313 | 919 | null | null |
3014 | 5 | null | null | 0 | null | Tag Usage
Clustered-standard-errors and/or cluster-samples should be tagged as such; do not use the "clustering" tag for them. Both these methodologies take clusters as given, rather than discovered.
Overview
Clustering, or cluster analysis, is a statistical technique of uncovering groups of units in multivariate data.... | null | CC BY-SA 4.0 | null | 2010-09-23T11:06:19.013 | 2020-12-08T14:26:18.850 | 2020-12-08T14:26:18.850 | 11887 | null | null |
3015 | 4 | null | null | 0 | null | Cluster analysis is the task of partitioning data into subsets of objects according to their mutual "similarity," without using preexisting knowledge such as class labels. [Clustered-standard-errors and/or cluster-samples should be tagged as such; do NOT use the "clustering" tag for them.] | null | CC BY-SA 3.0 | null | 2010-09-23T11:06:19.013 | 2016-03-09T10:33:39.543 | 2016-03-09T10:33:39.543 | 3277 | null | null |
3016 | 5 | null | null | 0 | null | Overview
Time series are data observed over time (either in continuous time or at discrete time periods).
Time series analysis includes trend identification, temporal pattern recognition, spectral analysis, and forecasting future values based on the past.
The salient characteristic of methods of time series analysis (a... | null | CC BY-SA 4.0 | null | 2010-09-23T11:14:30.340 | 2020-11-02T13:57:35.530 | 2020-11-02T13:57:35.530 | 53690 | null | null |
3017 | 4 | null | null | 0 | null | Time series are data observed over time (either in continuous time or at discrete time periods). | null | CC BY-SA 2.5 | null | 2010-09-23T11:14:30.340 | 2011-03-11T19:39:55.637 | 2011-03-11T19:39:55.637 | 919 | null | null |
3018 | 5 | null | null | 0 | null | [Hypothesis testing](https://en.wikipedia.org/wiki/Statistical_hypothesis_testing) assesses whether data are inconsistent with a given hypothesis rather than being an effect of random fluctuations.
| null | CC BY-SA 3.0 | null | 2010-09-23T11:28:36.817 | 2017-09-27T18:47:47.197 | 2017-09-27T18:47:47.197 | 7290 | null | null |
3019 | 4 | null | null | 0 | null | Hypothesis testing assesses whether data are inconsistent with a given hypothesis rather than being an effect of random fluctuations. | null | CC BY-SA 3.0 | null | 2010-09-23T11:28:36.817 | 2017-09-27T18:47:47.197 | 2017-09-27T18:47:47.197 | 7290 | null | null |
3022 | 2 | null | 3001 | 1 | null | Yes.
When you say that "X is the 16% quantile", what it means is that 16% of the sample have a lower value than X. The log of any number smaller than X is smaller than log(X) and the log of any number greater than X is greater than log(X), so the ordering is not changed.
| null | CC BY-SA 2.5 | null | 2010-09-23T18:45:20.550 | 2010-09-23T18:45:20.550 | null | null | 666 | null |
3023 | 2 | null | 2686 | 1 | null | A simple solution that does not require the acquisition of specialized knowledge is to use [control charts](http://en.wikipedia.org/wiki/Control_chart). They're ridiculously easy to create and make it easy to tell special cause variation (such as when you are out of town) from common cause variation (such as when you h... | null | CC BY-SA 2.5 | null | 2010-09-23T19:16:36.727 | 2010-09-23T19:16:36.727 | null | null | 666 | null |
3024 | 1 | 3027 | null | 36 | 60577 | I understand that for certain datasets such as voting it performs better. Why is Poisson regression used over ordinary linear regression or logistic regression? What is the mathematical motivation for it?
| Why is Poisson regression used for count data? | CC BY-SA 2.5 | null | 2010-09-23T19:38:40.190 | 2022-08-22T00:39:09.940 | 2013-10-04T02:20:09.050 | 7290 | 1392 | [
"count-data",
"poisson-regression"
] |
3025 | 2 | null | 2971 | 4 | null | Great discussion here, but I think of cross-validation in a different way from the answers thus far (mbq and I are on the same page I think). So, I'll put in my two cents at the risk of muddying the waters...
Cross-validation is a statistical technique for assessing the variability and bias, due to sampling error, in a... | null | CC BY-SA 2.5 | null | 2010-09-23T20:19:30.067 | 2010-09-23T20:19:30.067 | null | null | 1080 | null |
3026 | 2 | null | 3024 | 2 | null | My understanding is primarily because counts are always positive and discrete, the Poisson can summarize such data with one parameter. The main catch being that the variance equals the mean.
| null | CC BY-SA 2.5 | null | 2010-09-23T20:28:48.930 | 2010-09-23T20:28:48.930 | null | null | null | null |
3027 | 2 | null | 3024 | 59 | null | [Poisson distributed](http://en.wikipedia.org/wiki/Poisson_distribution) data is intrinsically integer-valued, which makes sense for count data. Ordinary Least Squares (OLS, which you call "linear regression") assumes that true values are [normally distributed](http://en.wikipedia.org/wiki/Normal_distribution) around t... | null | CC BY-SA 2.5 | null | 2010-09-23T20:42:45.613 | 2010-09-23T20:42:45.613 | null | null | 1352 | null |
3028 | 2 | null | 3024 | 26 | null | Essentially, it's because linear and logistic regression make the wrong kinds of assumptions about what count outcomes look like. Imagine your model as a very stupid robot that will relentlessly follow your orders, no matter how nonsensical those orders are; it completely lacks the ability to evaluate what you tell it... | null | CC BY-SA 2.5 | null | 2010-09-23T20:52:15.760 | 2010-09-23T20:52:15.760 | null | null | 71 | null |
3029 | 2 | null | 3024 | 3 | null | Others have basically said the same thing I'm going to but I thought I'd add my take on it. It depends on what you're doing exactly but a lot of times we like to conceptualize the problem/data at hand. This is a slightly different approach compared to just building a model that predicts pretty well. If we are trying... | null | CC BY-SA 2.5 | null | 2010-09-23T23:10:50.283 | 2010-09-23T23:10:50.283 | null | null | 1028 | null |
3030 | 2 | null | 3024 | 5 | null | Mathematically if you start with the simple assumption that the probability of an event occurring in a defined interval $T = 1$ is $\lambda$ you can show the expected number of events in the interval $T = t$ is is $\lambda.t$, the variance is also $\lambda.t$ and the [probability distribution ](http://www.umass.edu/wsp... | null | CC BY-SA 2.5 | null | 2010-09-23T23:28:22.660 | 2010-09-23T23:28:22.660 | null | null | 521 | null |
3031 | 1 | null | null | 3 | 1911 | I just loaded a csv file in R. When I ran the `summary` command for one of the columns, I got the following:
```
> Error: unexpected symbol in "summary k_low"
```
I'm pretty sure I know what the 'unexpected symbol' is: for observations in which we had no reliable data for the variable `k_low`, we simply entered a per... | How to escape symbolic value in R | CC BY-SA 2.5 | null | 2010-09-24T02:56:56.010 | 2010-09-24T13:58:47.937 | 2010-09-24T13:58:47.937 | 930 | 1410 | [
"r"
] |
3032 | 2 | null | 3031 | 6 | null | It looks like you just left out the parentheses. Try
```
summary(k_low)
```
| null | CC BY-SA 2.5 | null | 2010-09-24T03:03:32.427 | 2010-09-24T03:03:32.427 | null | null | 159 | null |
3034 | 2 | null | 3031 | 4 | null | Looks like Rob got it right but I'll illustrate how to fix the period problem.
```
> testdata <- c(1, 2, 3, ".")
> testdata
[1] "1" "2" "3" "."
> summary(testdata)
Length Class Mode
4 character character
> #That's not what we want....
> cleandata <- as.numeric(testdata)
Warning message:
NAs introd... | null | CC BY-SA 2.5 | null | 2010-09-24T03:13:31.783 | 2010-09-24T03:13:31.783 | null | null | 1028 | null |
3035 | 2 | null | 423 | 142 | null | I just came across this and loved it:
([http://xkcd.com/795/](http://xkcd.com/795/)).
| null | CC BY-SA 3.0 | null | 2010-09-24T04:09:04.750 | 2012-12-15T20:29:26.853 | 2012-12-15T20:29:26.853 | 919 | 253 | null |
3036 | 2 | null | 423 | 102 | null | this too:
| null | CC BY-SA 2.5 | null | 2010-09-24T04:13:25.630 | 2010-09-24T04:13:25.630 | null | null | 253 | null |
3037 | 2 | null | 3031 | 1 | null | Another point is that when you read in the data, it sound as if your "." is really a missing value.
So what you might wish to do when reading the data, is something like this:
k_low <- read.date(..., na.strings = ".")
| null | CC BY-SA 2.5 | null | 2010-09-24T04:16:19.980 | 2010-09-24T04:16:19.980 | null | null | 253 | null |
3038 | 1 | null | null | 41 | 23896 | Imagine you have a study with two groups (e.g., males and females) looking at a numeric dependent variable (e.g., intelligence test scores) and you have the hypothesis that there are no group differences.
Question:
- What is a good way to test whether there are no group differences?
- How would you determine the samp... | How to test hypothesis of no group differences? | CC BY-SA 3.0 | null | 2010-09-24T05:24:00.240 | 2016-10-21T14:44:52.830 | 2016-10-21T14:44:52.830 | 35989 | 183 | [
"hypothesis-testing",
"t-test",
"equivalence",
"tost"
] |
3039 | 2 | null | 3031 | 2 | null | When using [read.table, read.csv or read.delim](http://127.0.0.1:20298/library/utils/html/read.table.html) use:
```
read.table(file,..., na.strings = ".", ...)
```
na.strings - a character vector of strings which are to be interpreted as NA values.
Blank fields are also considered to b... | null | CC BY-SA 2.5 | null | 2010-09-24T05:37:32.793 | 2010-09-24T05:37:32.793 | null | null | 521 | null |
3040 | 2 | null | 3038 | 21 | null | I think you are asking about [testing for equivalence](http://web.archive.org/web/20120119090119/http://www.graphpad.com/library/BiostatsSpecial/article_182.htm). Essentially you need to decide how large a difference is acceptable for you to still conclude that the two groups are effectively equivalent. That decision d... | null | CC BY-SA 3.0 | null | 2010-09-24T05:50:39.257 | 2013-04-04T16:34:49.933 | 2017-04-13T12:44:41.967 | -1 | 521 | null |
3044 | 2 | null | 3038 | 13 | null | Following Thylacoleo's answer, I did a little research.
The [equivalence](http://cran.r-project.org/web/packages/equivalence/equivalence.pdf) package in R has the `tost()` function.
See Robinson and Frose (2004) "[Model validation using equivalence tests](http://research.eeescience.utoledo.edu/lees/papers_PDF/Robinson_... | null | CC BY-SA 2.5 | null | 2010-09-24T08:25:03.020 | 2010-09-24T08:25:03.020 | null | null | 183 | null |
3045 | 2 | null | 2746 | 0 | null | A cheap and cheerful approach I've used for testing is to generate m N(0,1) n-vectors V[k] and then use P = d*I + Sum{ V[k]*V[k]'} as an nxn psd matrix. With m < n this will be singular for d=0, and for small d will have high condition number.
| null | CC BY-SA 2.5 | null | 2010-09-24T09:19:30.617 | 2010-09-24T09:19:30.617 | null | null | null | null |
3046 | 2 | null | 3038 | 5 | null | In the medical sciences, it is preferable to use a confidence interval approach as opposed to two one-sided tests (tost). I also recommend graphing the point estimates, CIs, and a priori-determined equivalence margins to make things very clear.
Your question would likely be addressed by such an approach.
The CONSORT gu... | null | CC BY-SA 2.5 | null | 2010-09-24T09:43:08.590 | 2010-09-24T09:43:08.590 | null | null | 561 | null |
3047 | 2 | null | 3038 | 16 | null | Besides the already mentioned possibility of some kind of equivalence test, of which most of them, to the best of my knowledge, are mostly routed in the good old frequentist tradition, there is the possibility of conducting tests which really provide a quantification of evidence in favor of a null-hyptheses, namely bay... | null | CC BY-SA 3.0 | null | 2010-09-24T11:39:12.360 | 2013-06-05T18:31:14.217 | 2017-04-13T12:44:37.420 | -1 | 442 | null |
3048 | 1 | 3066 | null | 29 | 12538 | I have a matrix where a(i,j) tells me how many times individual i viewed page j. There are 27K individuals and 95K pages. I would like to have a handful of "dimensions" or "aspects" in the space of pages which would correspond to sets of pages which are often viewed together. My ultimate goal is to then be able to comp... | How to do dimensionality reduction in R | CC BY-SA 2.5 | null | 2010-09-24T11:44:24.637 | 2016-11-21T11:44:06.490 | 2010-09-24T13:19:19.210 | 930 | 1007 | [
"r",
"clustering",
"dimensionality-reduction"
] |
3049 | 2 | null | 3038 | 8 | null | There are a few papers I know of that could be helpful to you:
Tryon, W. W. (2001). [Evaluating statistical difference, equivalence, and indeterminacy using inferential confidence intervals: An integrated alternative method of conducting null hypothesis statistical tests.](http://dx.doi.org/10.1037//1082-989X.6.4.371) ... | null | CC BY-SA 2.5 | null | 2010-09-24T11:50:32.963 | 2010-09-24T11:57:15.200 | 2010-09-24T11:57:15.200 | 183 | 442 | null |
3050 | 2 | null | 3048 | 4 | null | It is certainly a clustering problem. Check out Rs `cluster` package to get an overview of algorithm options (`pam` and `agnes` are the best options to start; they represent two main streams in clustering -- [centroids](http://en.wikipedia.org/wiki/K-means_clustering) and [hierarchical](http://en.wikipedia.org/wiki/Hie... | null | CC BY-SA 3.0 | null | 2010-09-24T12:37:53.703 | 2016-11-21T11:44:06.490 | 2016-11-21T11:44:06.490 | 20833 | null | null |
3051 | 1 | 3054 | null | 25 | 54995 | I have a vector of values that I would like to report the average in windows along a smaller slide.
For example, for a vector of the following values:
```
4, 5, 7, 3, 9, 8
```
A window size of 3 and a slide of 2 would do the following:
```
(4+5+7)/3 = 5.33
(7+3+9)/3 = 6.33
(9+8)/3 = 5.67
```
And return a vector of th... | Mean of a sliding window in R | CC BY-SA 2.5 | null | 2010-09-24T14:41:31.997 | 2021-01-04T13:01:05.197 | 2010-09-24T17:17:56.337 | null | 1024 | [
"r"
] |
3052 | 1 | 3065 | null | 11 | 5479 | I'm examining some genomic coverage data which is basically a long list (a few million values) of integers, each saying how well (or "deep") this position in the genome is covered.
I would like to look for "valleys" in this data, that is, regions which are significantly "lower" than their surrounding environment.
Note... | How to look for valleys in a graph? | CC BY-SA 2.5 | null | 2010-09-24T15:09:52.740 | 2010-09-26T20:15:09.507 | 2010-09-24T18:02:47.390 | 634 | 634 | [
"r",
"distributions",
"statistical-significance",
"data-visualization"
] |
3053 | 2 | null | 3051 | 5 | null | This simple line of code does the thing:
```
((c(x,0,0) + c(0,x,0) + c(0,0,x))/3)[3:(length(x)-1)]
```
if `x` is the vector in question.
| null | CC BY-SA 2.5 | null | 2010-09-24T15:27:13.803 | 2010-09-24T17:16:59.560 | 2010-09-24T17:16:59.560 | null | 1414 | null |
3054 | 2 | null | 3051 | 30 | null | Function `rollapply` in package zoo gets you close:
```
> require(zoo)
> TS <- zoo(c(4, 5, 7, 3, 9, 8))
> rollapply(TS, width = 3, by = 2, FUN = mean, align = "left")
1 3
5.333333 6.333333
```
It just won't compute the last value for you as it doesn't contain 3 observations. Maybe this will be sufficien... | null | CC BY-SA 2.5 | null | 2010-09-24T15:36:42.200 | 2010-09-25T09:03:09.933 | 2010-09-25T09:03:09.933 | 1390 | 1390 | null |
3055 | 2 | null | 3052 | 2 | null | There are many options for this, but one good one: you can use the `msExtrema` function in the [msProcess package](http://cran.r-project.org/web/packages/msProcess/index.html).
Edit:
In financial performance analysis, this kind of analysis is often performed using a "drawdown" concept. The `PerformanceAnalytics` packa... | null | CC BY-SA 2.5 | null | 2010-09-24T15:43:45.877 | 2010-09-24T16:36:44.277 | 2010-09-24T16:36:44.277 | 5 | 5 | null |
3056 | 2 | null | 2909 | 2 | null | You might start by looking at the [drawdown distribution functions in fBasics](http://help.rmetrics.org/fBasics/html/stats-maxdd.html). So you could easily simulate the brownian motion with drift and apply these functions as a start.
| null | CC BY-SA 2.5 | null | 2010-09-24T16:03:15.173 | 2010-10-28T16:01:23.113 | 2010-10-28T16:01:23.113 | 5 | 5 | null |
3057 | 2 | null | 3052 | 4 | null | I'm completely ignorant of these data, but assuming the data are ordered (not in time, but by position?) it makes sense to make use of time series methods. There are lots of methods for identifying temporal clusters in data. Generally they are used to find high values but can be used for low values grouped together. I'... | null | CC BY-SA 2.5 | null | 2010-09-24T16:32:47.110 | 2010-09-24T16:32:47.110 | null | null | null | null |
3058 | 2 | null | 3051 | 1 | null | This will get you the window means and the index of the first value of the window:
```
#The data
x <- c(4, 5, 7, 3, 9, 8)
#Set window size and slide
win.size <- 3
slide <- 2
#Set up the table of results
results <- data.frame(index = numeric(), win.mean = numeric())
#i indexes the first value of the window (the sill?... | null | CC BY-SA 2.5 | null | 2010-09-24T16:40:33.070 | 2010-09-24T16:40:33.070 | null | null | 71 | null |
3059 | 1 | null | null | 7 | 480 | I'm investigating the effect of a continuous variable A on a measurement variable M stratified by another factor variable C in an observational dataset.
Due to heteroscedasticity I decided to use a bootstrapped regression analysis. However looking at the data, the background set of variables are not evenly distributed... | Bootstrapped regression with total data or bootstrap with matched data? | CC BY-SA 2.5 | null | 2010-09-24T17:04:59.187 | 2010-09-27T00:25:43.633 | 2010-09-27T00:25:43.633 | 159 | 1291 | [
"bootstrap"
] |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.