similar to: How to visualise what code is processed within a for loop

The loop never assigns anything to d0, only t. The first line makes t a character string "d0$V1" (or "d0$V2" etc.). The second line assigns either 0 or 1 to t. Looking at this, I don't think you've got into the R psychology (bad news if you want to use R, good news in many other ways). I assume d0 is a list, so could you put the V's into a vector, and then just use

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

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

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

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

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.

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) > }

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

Hello, I have build a syntax to find out if a given substring is included in a larger string that works like this: d1$V1[regexpr("some text = 9",d1$V2)>0] <- 9 and this works all right till "some text" contains standard ASCII set. However, it does not work when accents are included as the following: d1$V1[regexpr("some t?xt = 9",d1$V2)>0] <- 9 I have

Hi, I am a bit rusty with R programming and do not seem to find a solution to add a number of variables to my existing dataframe. Basically I need to add n=dim(d1)[1] variables to my d0 dataframe and I would like them to be named V1, V2, V3, ... , V[dim(d1)[1]) When running the following code: for (t in 1:dim(d1)[1]){ d0$V[t] <- 0 } all I get is a V variable populated with zeros... I am

Hi Luca, How about this? # create some dummy data since I don't have your d0 or d1 > n <- 3 > d0 <- data.frame(a=runif(5),b=runif(5)) # here's the suggested code > d1 <- cbind(d0, matrix(0,nrow(d0),n)) > colnames(d1)[1:n + ncol(d0)] <- paste("V",1:n,sep="") HTH, Eric On Sun, Apr 22, 2018 at 11:13 AM, Luca Meyer <lucam1968 at

I suspect that this is hitting the issue that I mentioned here: https://reviews.llvm.org/D65857#1621335 We may need to do what I suggested there and restrict global tag entropy on non-Android Linux to 7 bits. You can try working around this issue for now by using lld as the linker (-fuse-ld=lld). Peter On Mon, Jun 22, 2020 at 1:37 PM Mitch Phillips via llvm-dev < llvm-dev at

Hi, I am trying to execute a simple hello world program compiled like so: path/to/compiled/clang -o test --target=aarch64-linux-gnu -march=armv8.5-a -fsanitize=hwaddress --sysroot=/usr/aarch64-linux-gnu/ -L/usr/lib/gcc/aarch64-linux-gnu/10.1.0/ -g test.c However, when I look at the disassembly, there is an unknown instruction listed at 0x2d51c: 000000000002d4c0 main: 2d4c0: ff c3 00 d1

Thanks for the confirmation. From the assembly that was sent on the other branch of the thread: > .set .L.str, .L.str.hwasan-3458764513820540928 -3458764513820540928 = 0xd0 << 56 i.e. a "negative" tag. So this appears to be the issue exactly. Peter On Mon, Jun 22, 2020 at 1:55 PM Derrick McKee <derrick.mckee at gmail.com> wrote: > Using lld fixes this issue. >

Hello, I am trying to get a new vector 'x1' based on the not NA-values in column 'a' and 'b'. I found a way but I am sure this is not the best solution. So any ideas on how to "optimize" this would be great! m <- factor(c("a1", "a1", "a2", "b1", "b2", "b3", "d1", "d1"), ordered

Hello R experts, I have go this data frame: 'data.frame': 1 obs. of 20 variables: $ Anno : chr "PREVISIONI VS TARGET" $ OreTot: num 41 $ GioTot: logi NA $ OrGTot: logi NA $ OreCli: num 99 $ GioCli: logi NA $ OrGCli: logi NA $ OreFor: num -27 $ GioFor: logi NA $ OrGFor: logi NA $ OreOrt: num -18 $ GioOrt: logi NA $ OrGOrt: logi NA $ OreSpo: num -6 $ GioSpo: logi

Dear Listers, I try to change the structure of my data. i have an indicator-matrix and want to end up with a factor. i have v1 v2 v3 1 0 0 0 1 0 0 0 1 and want v1 v2 v3 v4 1 0 0 1 0 1 0 2 0 0 1 3 amongst other things i tried the following d <- data.frame(d1=c(1,0,0), d2=c(0,1,0), d3=c(0,0,1)) d$nr <- NA sapply(1:3, function(x)

Hello, I am trying to run the following syntax for all cases within the dataframe "data" d1 <- data[1,c("material")] fileConn<-file("TESTI/d1.txt") writeLines(d1, fileConn) close(fileConn) I am trying to use the for function: for (i in 1:nrow(data)){ d[i] <- data[i,c("material")] fileConn<-file("TESTI/d[i].txt")

Greetings, I am resending this as it doesn't appear in the archives and no one responded...maybe it got routed to /dev/null or something :-) I have asked this question the Scientific Linux mailing list (a few months ago) and got the suggestion I talk to the kernel guys. I pinged a kernel guy I know, and his suggestion was to ask the Nouveau list. So here I am. :-) I have had my work laptop

Hi List, I have an ASUS G73S laptop with our favorite OS installed and all up-to-date. Sound has always "just worked" When I use the Preferences>Sound app I can setup the hardware, input, output and was overall very impressed. One can check each speaker and the laptop has a front left and right and also has a bass speaker which seems to be connected to the left rear. The various