similar to: Sample of a subsample

For personal aesthetic reasons, I changed the name "data" to "dat". Your code, with a slight modification: set.seed (1357) ## for reproducibility dat <- data.frame(var1=seq(1:40), var2=seq(40,1)) dat$sampleNo <- 0 idx <- sample(seq(1,nrow(dat)), size=10, replace=F) dat[idx,"sampleNo"] <-1 ## yielding > dat var1 var2 sampleNo 1 1 40

Hi David, I was about to post a reply when Bert responded. His answer is good and his comment to use the name 'dat' rather than 'data' is instructive. I am providing my suggestion as well because I think it may address what was causing you some confusion (mainly to use "which", but also the missing !) idx2 <- sample( which( (!data$var1%%2) & data$sampleNo==0 ),

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Hi, Following is excerpted from dynamic-vector-gep.ll. The resulting "extractelement" seems to always return 0.0f regardless the value idx1 and idx2 is holding. Am I missing something here or there is something fishy take place? Thanks Shuxin 101 ; CHECK: test6 102 ; CHECK: insertelement <4 x float> zeroinitializer, float 1.000000e+00, i32 %idx1 103 ; CHECK:

Hi, I have my data in below format. position var1 var2 2 .1 10 3 .29 89 12 .56 100 425 .34 1234 6546 .12 21 .... ..... ..... .... ..... ......

Hi Shuxin I think i might have written that test. And yeah, no matter what values you get you’ll get a 0.0. Its probably a bad test case, but i can’t remember if it exposed a bug in this form or not. Since writing it Chandler rewrote SROA anyway so the original bug is long gone. Thanks, Pete On Apr 5, 2013, at 11:57 AM, Shuxin Yang <shuxin.llvm at gmail.com> wrote: > Hi, > >

# build a sample data frame illustrating the problem ids<-c(rep(1234,5),rep(5436,3),rep(7864,4)) years<-c(seq(1990,1994,by=1),seq(1991,1993,by=1),seq(1990,1993,by=1)) data<-seq(14,25,by=1) data[6]<-NA DF<-data.frame(Id=ids,Year=years,Data=data) DF Id Year Data 1 1234 1990 14 2 1234 1991 15 3 1234 1992 16 4 1234 1993 17 5 1234 1994 18 6 5436 1991 NA 7

Problem on flexmix when trying to apply signature developed in one model to a new sample. Dear R Users, R Core Team, I have a problem when trying to know the classification of the tested cases using two variables with the function of flexmix: After importing the database and creating a matrix: BM<-cbind(Data$var1,Data$var2) I see that the best model has 2 groups and use: ex2

Hi, I am using R 1.9.1 on on i686 PC with SuSE Linux 9.0. I have a data.frame, e.g.: > myData <- data.frame( var1 = c( 1:4 ), var2 = c (5:8 ) ) If I add a new column by > myData$var3 <- myData[ , "var1" ] + myData[ , "var2" ] everything is fine, but if I omit the commas: > myData$var4 <- myData[ "var1" ] + myData[ "var2" ] the name

Hi all, I have been trying to solve this problem and have had no luck so far. I have numeric vectors VAR1, VAR2, and VAR3 which I am trying to cbind. I also have a character vector "VAR1,VAR2,VAR3". How do I manipulate this character vector such that I can input a transformed version of the character vector into cbind and have it recognize that I'm trying to refer to my numeric

Dear List, I have Multi-level Data i= Indivitual Level g= Group Level var1= First Variable of interest var2= Second Variable of interest and I want to count the frequency of "var1" and "var2" on the group level. I found a way, but there must be a much simpler way. data.ml <- data.frame(i=c(1:8),g=as.factor(c(1,1,1,2,2,3,3,3)),var1=c(3,3,3,4,4,4,4 ,4),

Buenos días. Hoy ando un poco (o bastante) espeso y no doy con la tecla de una cosa que seguro que es muy simple.. Pongo un ejemplo. var1 <- c(rep(0,3),rep(1,2)) var2 <- c(rep(1,2),0,0,1) var3 <- c(rep(1,2),rep(0,3)) var4 <- c(rep(1,2),rep(0,3)) datos <- data.frame(fila=1:5,var1, var2, var3, var4) datos datos fila var1 var2 var3 var4 1 1 0 1 1 1 2 2 0

I would like to know how to turn a variable into a string. I have tried as.symbol and as.name but it doesnt work for what I'd like to do Essentially, I'd like to feed the function below with two variables. This works fine in the bit working out number of elements in each variable. In the print(sprintf("OK with %s and %s\n", var1, var2)) line I would like var1 and var2 to be

I have a data frame whose first colum contains the names of the variables and whose second colum contains the values to assign to them: : kkk <- data.frame(vars=c("var1", "var2", "var3"), vals=c(10, 20, 30), stringsAsFactors=F) If I do : assign(kkk$vars[1], kkk$vals[1]) it works : var1 [1] 10 However, if I try with mapply