similar to: Reading .gsheet within R

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

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*** APOLOGIZES FOR THOSE READING THE LIST THROUGH NABBLE THIS WAS ALREADY POSTED THERE BUT NOT FORWARDED TO THE LIST FOR SOME UNKNOWN REASON *** I have a dataset that looks like: $ V1: factor with 4 levels $ V2: factor with 4 levels $ V3: factor with 2 levels $ V4: num (summing up to 100 within V3 levels) $ V5: num (nr of cases for each unique combination of V1*V2*V3 levels) Quite new to

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Being italians when writing comments/instructions we use accented letters - like ?, ?, ?, etc.... when running R scripts using such characters I get and error saying: invalid multibyte character in parser I have been looking at the help and searched the r-help archives but I haven't find anything that I could intelligibly apply to my case. Can anyone suggest a fix for this error? Thanks,

Hi, This must be an easy one but so far I haven't find a way out... I have a data frame such as: $ v1 : Factor w/ 5 levels $ v2 : Factor w/ 2 levels $ v3 : Class 'difftime' atomic [1:6666] basically v1 and v2 are factors, while v3 is a variable containing the duration of certain activities (values ranging from 11 to 45000 sec, no missing values) How can I get a table

Hi, I have been trying to get the label under the total column - i.e. a mean value of columns 2 to 6 - in a barplot I generate with this script: t1 <- tapply(A, B, sum) t1[8] <- mean(t1[2:6]) t1 <- as.table(t1) barplot(t1, ylim=c(0,3000)) mtext("Var1", side = 1, line = 3) mtext("Var2", side = 2, line = 3) I have been trying to use axis(1, at=1:8,

Hello, I have a multi-year dataset (see below) with date, a data value and a flag for the data value. I want to find the monthly median for each month in this dataset and then plot it. If anyone has suggestions they would be greatly apperciated. It should be noted that there are some dates with no values and they should be removed. Thanks Emily > print ( str(data$flow$daily) )

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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

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

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 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)

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

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

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.

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 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

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")

Dear R-users, I have a factor variable within my data frame which I derive week after week from a POSIXct variable using the cut(var,"weeks") command I have found in the chron package. The levels() command gives me: [1] "2009-03-30 00:00:00" "2009-04-06 00:00:00" "2009-04-13 00:00:00" "2009-04-20 00:00:00" "2009-04-27 00:00:00"