Re: [R] Sample function and prob argument

On 05/06/2019 4:34 a.m., le Gleut, Ronan wrote: Dear R-help mailing list, First of all, many many thanks for your great work on the R project! I have a very small issue regarding the sample function. Depending if we specify values for the prob argument, we don't get the same result fo

[R] Sample function and prob argument

Dear R-help mailing list, First of all, many many thanks for your great work on the R project! I have a very small issue regarding the sample function. Depending if we specify values for the prob argument, we don't get the same result for a random sampling with replacement and with equal pr

Re: [R] Sample function in R

On Sun, Feb 12, 2012 at 01:57:18PM -0500, SUPAKORN LAOHAPITAKVORN wrote: > This is what I got: > > > sessionInfo() > R version 2.14.1 (2011-12-22) > Platform: x86_64-pc-mingw32/x64 (64-bit) > > locale: > [1] LC_COLLATE=English_United States.1252 > [2] LC_CTYPE=English_United States.1252 > [3] LC_

Re: [R] Sample function in R

Supakorn, Try: rm(sample) #then sample(x) Etienne 2012/2/12 SUPAKORN LAOHAPITAKVORN > This is what I got: > > > sessionInfo() > R version 2.14.1 (2011-12-22) > Platform: x86_64-pc-mingw32/x64 (64-bit) > > locale: > [1] LC_COLLATE=English_United States.1252 > [2] LC_CTYPE=English_United States.

Re: [R] Sample function in R

This is what I got: > sessionInfo() R version 2.14.1 (2011-12-22) Platform: x86_64-pc-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=English_United States.1252 [2] LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C [5] LC_TIME=English_United States.1252

Re: [R] Sample function in R

Hi, The following is what I get: > x =1:12 > x [1] 1 2 3 4 5 6 7 8 9 10 11 12 > sample(x) [1] 9 6 12 5 3 4 1 11 8 7 10 2 > sample(x, size = 2) [1] 9 4 What's the output of sessionInfo() and ls() ? Perhaps you have a different "sample" function in your workspace? HTH, Jorge

Re: [R] Sample function in R

sessionInfo()? Can you replicate this behavior in a R --vanilla session? This seems very odd and I presume you've overwritten sample() somewhere in your workspace. Michael On Sun, Feb 12, 2012 at 12:52 PM, SUPAKORN LAOHAPITAKVORN wrote: > Hi, > Can anyone help me with the sample () in R? > > If

[R] Sample function in R

Hi, Can anyone help me with the sample () in R? If I sample from x, I should get one integer. Can anyone tell me what's wrong here? > x =1:12 > sample(x) [1] 6.5 And, I cannot get the sample with size = 2 > sample(x, size = 2) Error in sample(x, size = 2) : unused argument(s) (size = 2) > sample

Re: [R] sample function with different proportions

Well, you can have exactly 70:30%, i.e. 70% 1s and 30% 0s, but in random order. For example: Popn <- c(rep(1,70),rep(0,30)) Samp <- sample(Pop) (see '?sample' for this usage -- the result of sample(x) is a random permutation of the elements of x). In probabilistic terms, this is a "condition

Re: [R] sample function with different proportions

On 05/07/2011 2:25 PM, Joshua Wiley wrote: Hi Ana, Look at the documentation for ?sample, specifically, the "prob" argument. In your case this should work: sample(c(0,1), 100, replace = TRUE, prob = c(.3, .7)) note that you may not have *exactly* 70% 1 and 30%, in any given sample. And if y

Re: [R] sample function with different proportions

Hi Ana, Look at the documentation for ?sample, specifically, the "prob" argument. In your case this should work: sample(c(0,1), 100, replace = TRUE, prob = c(.3, .7)) note that you may not have *exactly* 70% 1 and 30%, in any given sample. HTH, Josh On Tue, Jul 5, 2011 at 11:21 AM, Ana Kolar

[R] sample function with different proportions

Hi there, I guess this is an easy one, but still:  I would like to randomly sample 0s and 1s but in a way that I end up having for example 70% of 1s and the rest of 0s and not 50:50 as this function does: sample(c(0,1), 100, replace = TRUE) Any recommendations? Many thanks! Ana [[al

Re: [R] Sample function in R???

On 2011-02-19 10:32, Hongwei Dong wrote: Hi, R users, I'm wondering if there a way in R I can select cases based on a probability vector. if a case is selected, that case is marked as 1, otherwise, 0. For example: x<-12:18 y<-1:7 sample(x,2,replace=FALSE,y) I got: [1] 15 17 What I want to s

[R] Sample function in R???

Hi, R users, I'm wondering if there a way in R I can select cases based on a probability vector. if a case is selected, that case is marked as 1, otherwise, 0. For example: x<-12:18 y<-1:7 sample(x,2,replace=FALSE,y) I got: [1] 15 17 What I want to see is: [1] 0 0 0 1 0 1 0 Thanks. Gary

Re: [R] Sample Function

What is dataset? What is this supposed to be doing? newsample<-dataset[sample(m,replace=T),] --- On Thu, 7/16/09, MarcioRibeiro wrote: > From: MarcioRibeiro > Subject: [R] Sample Function > To: r-help@r-project.org > Received: Thursday, July 16, 2009, 11:55 AM > > Hi

Re: [R] Sample Function

> -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On > Behalf Of MarcioRibeiro > Sent: Thursday, July 16, 2009 8:56 AM > To: r-help@r-project.org > Subject: [R] Sample Function > > > Hi listers, > Supp

[R] Sample Function

Hi listers, Suppose I have a dataset with n=10 observations and I want to sample with replacement. My new sample is of size m=9. So, I am using the following code... newsample<-dataset[sample(m,replace=T),] The problem is that generates the new sample and the last observation of my data set is nev

[R] Sample Function

Hi listers, I am -- View this message in context: http://www.nabble.com/Sample-Function-tp24519071p24519071.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-hel

[R] sample function

Hi, I use sample function in a for loop to get a new random indexing but I don't know why, I get the same indexes in different iterations of for. I have a script that uses for loop calling crossval function (the same definition as crossval in bootstrap package) defined in another script. Both s