Home > Error In > Error In .data.frame X .internalsamplelengthx Size Replace# Error In .data.frame X .internalsamplelengthx Size Replace

## K Means Cannot Take A Sample Larger Than The Population When 'replace = False'

## R Error In Sample.int(m, K) : Cannot Take A Sample Larger Than The Population When 'replace = False'

## Hi, I find convenient to use a custom function for this: sample.df <- function (df, N = 1000, ...) { df[sample(nrow(df), N, ...), ] } sample.df(daf1,1000) Hope this helps,

of 8 variables: $ **V1: num 0** 0 0 0 0 0 0 0 0 0 ... ... Is your gly a vector? -Peter Ehlers Hayes, Rachel M wrote: > Hi All, > > > > I want to take a simple random sample from a large V-brake arm not returning to "open" position more hot questions question feed default about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Hi All, I want to take a simple random sample from a large dataset, gly, but I'm getting an error message. news

What would be a good approach to make sure my advisor goes through all the report? In the case of sample(), you will find that it requires a *vector* input. Free forum by Nabble Edit this page R › R help Search everywhere only in this topic Advanced Search help sample from large dataset - misleading error? ‹ Previous Topic Next Register here! you can try this out

What the heck happened?? David Winsemius Threaded Open this post in threaded view ♦ ♦ | Report Content as Inappropriate ♦ ♦ Re: help sample from large dataset - misleading error? Related 7How do I sample without replacement using a sampling-with-replacement function?1How to resample matrices to test for the robustness of their correlation?10How to resample in R without repeating permutations?3Resampling and Huber-White3Resampling Free forum by Nabble Edit this page [R] Using sample() with a data frame ?

- x0 = c(-1.9,-1.9) gd = grad.descent(simpleFun,x0) gd$x ## [1] -5.437919e-07 -1.490486e-02 gd$k ## [1] 144 Note: the minimum here is achieved at (0,0), so this is right Let’s look at the
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- ArghhhHow do I go about sampeling from a data frame?:-( Martin______________________________________________R-help at r-project.org mailing listhttps://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guidehttp://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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- asked 4 years ago viewed 1957 times active 4 years ago 13 votes · comment · stats Linked 1 How to resample matrices to test for the robustness of their correlation?

Using parameter expansion to generate arguments list for `mkdir -p` Polyline split at node in QGIS Is masking before unsigned left shift in C/C++ too paranoid? How to create a plot with inclined axes? Uh oh! I use: > >> daf2 <- **sample(daf1, 1000, replace** = FALSE, prob = NULL) > Error in `[.data.frame`(x, .Internal(sample(length(x), size, > replace, : > cannot take a sample larger than the

dim(gly) [1] 112371 37 > s1 <- sample(gly,100) Error in `[.data.frame`(x, .Internal(sample(length(x), size, replace, : cannot take a sample larger than the population when 'replace = FALSE' R Error In Sample.int(m, K) : Cannot Take A Sample Larger Than The Population When 'replace = False' Try our newsletter Sign up for our newsletter and get our top new questions delivered to your inbox (see an example). This is a vector. if (all(abs(grad.cur) < stopping.deriv)) { k = k-1; break } # Move in the opposite direction of the grad xmat[,k] = xmat[,k-1] - step.size * grad.cur } xmat = xmat[,1:k] #

However, for backward compatibility, you can let the sample argument x be an integer giving the sample size. I'd forgotten how fun 'The R Inferno' is (and educational.) –jbowman Feb 20 '12 at 19:05 add a comment| 2 Answers 2 active oldest votes up vote 3 down vote You If so how? Any help? > > > > dim(gly) > > [1] 112371 37 > >> s1 <- sample(gly,100) > > Error in `[.data.frame`(x, .Internal(sample(length(x), size, replace, : > >

This might do the trick: ddply(data1,'a',function(x) x[sample.int(NROW(x),20,replace=TRUE),]) share|improve this answer answered Mar 28 '12 at 18:00 Tommy 23.7k36067 +1 for the function sample.int @Tommy –Stat-R Mar 28 '12 at For a data frame, length returns the number of columns: > foo <- as.data.frame(matrix(0,100,8)) > str(foo) 'data.frame': 100 obs. K Means Cannot Take A Sample Larger Than The Population When 'replace = False' Editing will bump your question back up to the top of the queue & let people know that the answers you've gotten so far aren't what you needed, so it accomplishes Any help? > > > > dim(gly) > > [1] 112371 37 > > > s1 <- sample(gly,100) > > Error in `[.data.frame`(x, .Internal(sample(length(x), size, replace, : > >

I went on and wrote this code: head(tjornres) essayres = tjornres # copy of the data R = 999 # the number of replicates cor.values = numeric(R) # store the data navigate to this website Value sample.int returns a sample from 1:n. What you can do is sample from the vector 1:nrow(essayres) and use the resulting sample to specify the rows of essayres in the relevant group, as in the answer that appeared Messages sorted by: [ date ] [ thread ] [ subject ] [ author ] More information about the R-help mailing list Grokbase › Groups › R › r-help › August

up vote 3 down vote It would seem that if you want to sample a category that has less than 20 rows, you'd need replace=TRUE... Have R calculate the derivatives symbolically: (d1 = D(expression((1/2*x^2-1/4*y^2+3)*cos(2*x+1-exp(y))), "x")) ## 1/2 * (2 * x) * cos(2 * x + 1 - exp(y)) - (1/2 * x^2 - 1/4 * If n>1, this does not give overall selection probabilities proportional to prob; the actual selection probabilities are between those implied by prob and equal probabilities. More about the author Browse other questions tagged r plyr sampling or ask your own question.

If prob is supplied and replace=FALSE, then values are drawn sequentially with probabilities proportional to prob, excluding elements already drawn. baptiste auguie Threaded Open this post in threaded view ♦ ♦ | Report Content as Inappropriate ♦ ♦ Re: Using sample() with a data frame ? School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag Baptiste auguie at Aug 25, 2008 at 11:51 am ⇧ Hi,I find convenient

If TRUE, sampling is done with replacement; otherwise sampling is without replacement. Jorge I Velez Threaded Open this post in threaded view ♦ ♦ | Report Content as Inappropriate ♦ ♦ Re: help sample from large dataset - misleading error? Jorge I Velez Threaded Open this post in threaded view ♦ ♦ | Report Content as Inappropriate ♦ ♦ Re: help sample from large dataset - misleading error? The idea is simple—just calculate the gradient of the function you’re trying to (say) minimizing, move in the direction of its negative gradient, and repeat With our knowledge of grad(), we

Given your question statement, I strongly suspect you don't really want two groups at all, but rather something like: N <- nrow(essayres) for (i in 1:R) { foo <- essayres[sample(1:N, size=435, Forgot Password? This means that they can be passed to functions as arguments and returned by functions as outputs as well Functions of functions: computationally We often want to do very similar things click site Copyright © 2016 Rezider, Inc.

Arghhh > > How do I go about sampeling from a data frame? > > :-( Martin > > ______________________________________________ > [hidden email] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the Examples sample(state.name, 10) # pick 10 unique states at random sample(1e6, 75) # pick 75 numbers between 1 and one million sample.int(50) # random permutation of numbers 1:50 # Bernoulli(.3) sample You are absolutely right. Success!

Also if I run one resampling, sample(essayres, size=435, replace=F), I get this error message: Error in `[.data.frame`(x, .Internal(sample(length(x), size, replace, :cannot take a sample larger than the population when 'replace =

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