similar to: question about if else

Hi everybody. The question: I get two vectors 'iFalseFalse' and 'i2'. I think they should be the same but they are not. Is it because R does not handle complicated logical expressions in such cases or I do something wrong? > z1 = c(NA, "", 3, NA, "", 3) > z2 = c("", "", 3, NA, 3, NA) > cV = (as.character(z1)==as.character(z2))

> version _ platform i386-pc-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major 1 minor 8.1 year 2003 month 11 day 21 language R > aa <- as.POSIXct(c('1/1/1980','1/2/1980','')) > ifelse(is.na(aa),as.POSIXct('1/3/1980'),aa) Error in rep.int(unclass(x), times) : Argument "times" is missing,

Hi, everybody. This was an interesting discussion last time and it helped me a lot. Could you please have a look at some feature and tell me why it was designed this way (my questions are under #########) > x = c(1, 10) > y = c(99, 55) > d <- data.frame(x = x, y = y) > d x y 1 1 99 2 10 55 > add <- data.frame(x = 14, y = 99) > add x y 1 14 99 > d <-

The funciton c() works differently for strings and for factors: For strings: > l = c('a', 'b') > l [1] "a" "b" For factors: > l = c(factor('a'), factor('b')) > l [1] 1 1 What should be the right technique for merging factors? -- Svetlana Eden Biostatistician II School of Medicine

Thank you for your answers, I have another question: the behaviour of setdiff(indicesFalse, indicesNA) does not seem predictable to me. > indices [1] 1 2 3 4 5 6 > compareVector [1] NA TRUE TRUE TRUE FALSE NA > indicesNA = indices[is.na(compareVector)] > indicesNA [1] 1 6 > indicesFalse = indices[compareVector == FALSE] > indicesFalse [1] NA 5 NA >

I have two factors l1, l2, and I'd like to merge them. (Remark: The factors can not be converted to charaters) Function c() does not give me the result I want: > l1 = factor(c('aaaa', 'bbbb')) > l2 = factor(c('ccc', 'dd')) > lMerge = factor(c(l1, l2)) > lMerge [1] 1 2 1 2 Levels: 1 2 > I'd like to merge l1 and l2 and to get lMerge

The Details. In the following version. > version _ platform i386-pc-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major 1 minor 8.1 year 2003 month 11 day 21 language R > lookup.xport ignores some datasets in sas export file. File "emptySasData3.xpt" (available at http://biostat.mc.vanderbilt.edu/tmp/emptySasData3.xpt) is a

Hi there, I don't know why the following codes are return different results. > ifelse(3 > 2, 1:3, length(1:3)) [1] 1 > if (3 > 2) 1:3 else length(1:3) [1] 1 2 3 Any hints? Best, Jinsong

Hi R-helpers! I have the following dataframe: firm<-c(rep(1:3,4)) year<-c(rep(2001:2003,4)) X1<-rep(c(10,NA),6) X2<-rep(c(5,NA,2),4) data<-data.frame(firm, year,X1,X2) data So I want to obtain the same dataframe with a variable X3 that is: X1, if X2=NA X2, if X1=NA X1+X2 if X1 and X2 are not NA So my final data is X3<-c(15,NA,12,5,10,2,15,NA,12,5,10,2)

Luca, If speed is important, you might improve performance by making d0 into a true matrix, rather than a data frame (assuming d0 is indeed a data frame at this point). Although data frames may look like matrices, they aren?t, and they have some overhead that matrices don?t. I don?t think you would be able to use the [[nm]] syntax with a matrix, but [ , nm] should work, provided the matrix has

I forgot to explain why my suggestion. The logical condition returns FALSE/TRUE that in R are coded as 0/1. So all you have to do is coerce to integer. This works because the ifelse will return a 1 or a 0 depending on the condition. Meaning exactly the same values. And is more efficient since ifelse creates both vectors, the true part and the false part, and then indexes those vectors in

Thanks Don, for (i in 1:10){ nm <- paste0("V", i) d0[[nm]] <- ifelse( regexpr(d1[i,1], d0$X0) > 0, 1, 0) } is exaclty what I needed. Best regards, Luca 2018-04-25 23:03 GMT+02:00 MacQueen, Don <macqueen1 at llnl.gov>: > Your code doesn't make sense to me in a couple of ways. > > Inside the loop, the first line assigns a value to an

Hi Rui Thank you for your suggestion, I have tested the code suggested by you against that supplied by Don in terms of timing and results are very much aligned: to populate a 5954x899 0/1 matrix on my machine your procedure took 79 secs, while the one with ifelse employed 80 secs, hence unfortunately not really any significant time saved there. Nevertheless thank you for your contribution.

Hi, I came across the below error when I try to do ifelse condition:- Error in "[<-"(`*tmp*`, test, value = rep(yes, length = length(ans))[test]) : incompatible types What I am trying to accomplish is :- for(x in 1:ncol(filterpred)){ sumno<-sum(filterpred[no,x]) sumyes<-sum(filterpred[yes,x]) ifelse(sumno==0 && sumyes !=0,

Thank you for both replies Don & Rui, The very issue here is that there is a search that needs to be done within a text field and I agree with Rui later comment that regexpr might indeed be the time consuming piece of code. I might try to optimise this piece of code later on, but for the time being I am working on the following part of building a neural network to try indeed classifying some

Your code doesn't make sense to me in a couple of ways. Inside the loop, the first line assigns a value to an object named "t". Then, the second line does the same thing, assigns a value to an object named "t". The value of the object named "t" after the second line will be the output of the ifelse() expression, whatever that is. This has the effect of making

Hello, instead of ifelse, the following is exactly the same and much more efficient. d0[[nm]] <- as.integer(regexpr(d1[i,1], d0$X0) > 0) Hope this helps, Rui Barradas On 4/28/2018 8:45 PM, Luca Meyer wrote: > Thanks Don, > > for (i in 1:10){ > nm <- paste0("V", i) > d0[[nm]] <- ifelse( regexpr(d1[i,1], d0$X0) > 0, 1, 0) > }

Hi, I am trying to debug the following code: for (i in 1:10){ t <- paste("d0$V",i,sep="") t <- ifelse(regexpr(d1[i,1],d0$X0)>0,1,0) } and I would like to see what code is actually processing R, how can I do that? More to the point, I am trying to update my variables d0$V1 to d0$V10 according to the presence or absence of some text (contained in the file d1)

Hi, I try to make a simple for loop with a if else statement (First example - Below) and extend it to a more complex loop (Second example). However, my results #First example: x=c(1,2) t=for(i in 1:length(x)){ if (x==1){a=x+1}else if (x==2){a=x} } Returned from R: Warning messages: 1: the condition has length > 1 and only the first element will be used in: if (x == 1) { 2: the condition has

> Thomas Lumley wrote: > > On Tue, 31 May 2005, Duncan Murdoch wrote: > > > > > >>M??kinen Jussi wrote: > >> > >>>Dear All, > >>> > >>>I luckily found the following feature (or problem) when tried to > >>>apply > >>>ifelse-function to an ordered data. > >>> > >>> >