similar to: How sample without replacement on more than one variables?

Hello Everyone,   I'm trying to conduct a couple of power analyses and was hoping someone might be able to help. I want to estimate the sample size that would be necessary to adequately power a couple of non-inferiority tests. The first would be a log-rank test and the second would be a Wilcoxon rank-sum test. I want to be able to determine the sample size that would be necessary to test for a

Hello All, '\\' is printed as '\\', but it is actually only one character. Sometimes, I'd rather print it as a single '\'. Is there a function to do so in R? > nchar('\\') [1] 1 > print('\\') [1] "\\" -- Tom

Dear All, I want to check what packages depends on a given package. For example, glmnet depends on Matrix. I want get the names of all the packages that depend on Matrix on CRAN. Is there a way to do so? Thanks! > library(glmnet) Loading required package: Matrix Loading required package: lattice -- Tom

Hi All, I want to enclose with() in a function mean_on_element. Obviously, it is not working. The problem is how to specify the element name with a function body. Does anybody have any suggestion? Thanks! > data=list(x=1:10) > with(data, mean(x)) [1] 5.5 > > mean_on_element=function(data, elem_name) { + with(data, mean(elem_name)) + } > mean_on_element(data, 'x') [1] NA

Hello All, I'm trying to pass the argument col.names to write.csv using '...'. But I got the following warnings. Maybe it is very simple. But I'm not sure what I am wrong. Could you please help point to me what the problem is? ##################### fun=function(x, ...) { fr=parent.frame() tmp=get(x, envir=fr) write.csv( tmp , file=paste(x, '.csv',

Hi, I've got a quite tricky question. I have a txt file, named 'temp.txt', as the following: snp1 snp2 snp3 AA 00 00 GG GG 00 00 AA 00 I want to read the file into R. 1) when I use 'read.table' without 'header=T' option, > temp <- read.table('temp.txt') # I got > temp V1

Hi All, The last line if the following code returns the error right below this paragraph. Essentially, I use the operator %:% to retrieve a variable in a nested frame. Then I want to use the same operator (with '<-') to change its value. I'm not sure if this is possible in R with %:% operator. Would you please help and let me how to correct the code if it is possible? Error in

How can I see more of the structure than displayed by 'str'? Consider the following: tstDF <- data.frame(a=1, row.names='b') > str(tstDF) 'data.frame': 1 obs. of 1 variable: $ a: num 1 The object 'tstDF' has row.names, but I have to suspect they are there -- AND know a function like 'row.names' or 'dimnames' -- to see

Hello, I have a simple problem (checked the archives and the appropriate help pages, to no avail). I want to creat a vector that is length(2000). The vector is to consist of two strings( "std" and "dev") with 1760 "std"s and 240 "dev"s. Furthermore, for each element of the vector, i want the selection of one of the strings to be approx. random. The

Splus used a similar method for sampling from "bigdata" objects. One problem was that sample() is used both for creating a sample and for scrambling the order of a vector. Scrambling the order of a big vector wastes time. It would be nice to be able to tell sample() that we don't care about the order. Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, Oct 17, 2017 at 10:55

Let us consider the current uniform sampling without replacement algorithm. It resides in function do_sample in https://svn.r-project.org/R/trunk/src/main/random.c Its complexity is obviously O(n), where the sample is selected from 1...n, since the algorithm has to create a vector of length n. So when the sample size is much lesser than n, the algorithm is not effective. Algorithms with

Hi all, I am a fourth year EE bachelors student who is very interested in compilers. I have taken the only compilers course offered in my school and did an independent study with my CS professor. Although I'll begin to work in a couple of months, I definitely want to pursue my interest in compiler design and optimization as a PhD in the future. I am very interested in the z3 SMT solver

Dear all I've got stuck when trying to replace values in a vector by selecting replacements from a replacement table. I'm trying to use only base functions. Here's a dummy example: > (x <- rep(letters,2)) [1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l"

Hi All, I meant to take the min row by row. But the result is apparently not what I want. Changing min to pmin solve the problem. > df=data.frame(X=1:10, Y=1:10) > transform(df, Z=min(X,10-Y)) X Y Z 1 1 1 0 2 2 2 0 3 3 3 0 4 4 4 0 5 5 5 0 6 6 6 0 7 7 7 0 8 8 8 0 9 9 9 0 10 10 10 0 I try to look at the source code to understand what transform() does. I know

> From: "Pavel S. Ruzankin" <ruzankin at math.nsc.ru> > Let us consider the current uniform sampling without replacement > algorithm. It resides in function do_sample in > https://svn.r-project.org/R/trunk/src/main/random.c > Its complexity is obviously O(n), where the sample is selected from > 1...n, since the algorithm has to create a vector of length n. So

Hi, The following two 'df's should be the same, although their constructions are different. But the results of paste() are different. I don't see this is explained in ?paste. Could you help me understand why it is like this? > df=data.frame(X=c(1, 2, 3), Y=c(4, 5, 6)) > df X Y 1 1 4 2 2 5 3 3 6 > paste(df) [1] "c(1, 2, 3)" "c(4, 5, 6)" >

Hello, Normally, if I type a function name, it shows the function definition. When the function is masked by a variable with the same name, it doesn't show the function definition any more. Can anyone please tell me a way how to retrieve the function definition even if it is masked by a variable? > x=1:3 > length(x) [1] 3 > length function (x) .Primitive("length") >

Hello All, If I refer to a variable 'x', 'x' will be searched in the current frame or the parent frame, or the parent of the parent frame, etc., until it is found (or not found at all)? Could you please show me what code in R source that handles this? Is it in the C code or the R code? Thanks, Tom

Hi,I want to replace some elements of a dataframe "a", using 2 index vectors ("x" and "y") and a vector of replacement ("z") of the same length of "x" and "y". I've tried a[cbind(x,y)]<-z, but it doesn't work.How can I do it in a simply way?Thanks,

Gabe As mentioned on Twitter, I think the following behavior should be fixed as part of the upcoming changes: R.version.string ## [1] "R Under development (unstable) (2019-02-25 r76160)" .Machine$double.digits ## [1] 53 set.seed(123) RNGkind() ## [1] "Mersenne-Twister" "Inversion"??????? "Rejection" length(table(runif(1e6))) ## [1] 999863 I don't