This one should work: awk ‘BEGIN{ OFS = “\t” }NR==1{ for ( i = 1; i <= NF; i++) if ( $i ~”Pval” ) a[i]; print $0}NR>1{for ( i in a ) if ( $i > 0.5 ) $i = “NA”; print $0 }’ file Output: Rv_FC Rv_Pval Gc_FC Gc_Pval 2 0.001 3 NA 3 … Read more
Try this: df <- read.table(text = “v1 v2 v3 v4 v5 0 1 1 1 1 0 1 1 0 1 1 0 1 1 0”, skip = 1) df ll <- vector(mode = “list”, length = ncol(df)-1) ll <- lapply(2:ncol(df), function(ncols){ tmp <- t(apply(df, 1, function(rows) combn(x = rows, m = ncols, prod))) if(ncols … Read more
NOTE: Your question as it stands is not appropriate for this site. You should include a minimum reproducible example. Include code that you’ve tried and where it’s causing you trouble. At this point, you’re likely to continue getting down votes, and your question may potentially get closed. Be aware of that. Please visit the help … Read more
Simplest answer: Purchase more RAM. If you work in R with large datasets often, it’s worth it. If you don’t have enough memory to load your files, you may not have enough to manipulate them as you want either. Let’s assume that you could hold this data in RAM and manipulate it as you wish … Read more
Next time please include the head this can be done using head(Store_sales) ProductID category sales product 1 101 Bakery 9468 White bread 2 102 Personal Care 9390 Everday Female deodorant 3 103 Cereal 9372 Weetabix 4 104 Produce 9276 Apple 5 105 Meat 9268 Chicken Breasts 6 106 Bakery 9252 Pankcakes I reproduced relevant fields … Read more
Why not extract that column and take the mean?? > m <- matrix(1:10, nrow=5) > mean(m[,1]) # mean of the first column [1] 3 # mean of the second column conditional on values of column 3 are bigger than 4 > mean(x[x[,3]>4,2]) [1] 5 I don’t understand what you mean with “…second column where the … Read more
Yahoo made some changes, so you can’t access the download without the respective “crumb”, see Question few days ago. Unless you want to manually copy-paste this Download-URL for each stock, I’d highly suggest you to use the quantmod package. It works, after applying a short fix (which will probably soon be included in a new … Read more
use this a = matrix(c(1,2,3,4,5,2,3,3,4,1,5,6,77,8,9,10,9,3,4,17), byrow = T, ncol = 5) which(a[, 3] == 3 & a[ ,4] == 4) [1] 1 2 4 0 solved How to find the row number of a matrix when its two columns match the given values in r
I’d do it in several steps: Split on logical operators into separate inequalities. Change double inequalities like -3>x1>=-1.45 into two inequalities. Change “=” to “==”, and put it all together. For example: a1 <- strsplit(a, “&”, fixed = TRUE)[[1]] a1a <- gsub(” “, “”, a1) # get rid of spaces a2 <- gsub(“([-0-9.]+[<>=]+)([[:alpha:]]+[[:alnum:]]*)([<>=]+.+)”, “\\1\\2 & … Read more
The commenters are right. It doesn’t matter how long the list of numbers is. R can do the operation in one go. x <- c(25.01, 25.8, 25.4) x-15 [1] 10.01 10.80 10.40 If you do run into a situation with many wide-ranging numbers and need to literally keep what comes after the decimal and pick … Read more
Using seq function, you need to following two steps: Step-1 Generate sequence from 2 to 10 using following code: a<-seq(from=2,to = 10) Step-2 Generate sequence from 10 to 2 using following code: b<-seq(from=9,to = 2) Now, combine above two results using following code: data<-c(a,b) The output should as follow: > data [1] 2 3 4 … Read more
I have been using the following snippet: library(data.table) filepaths <- list.files(dir) resultFilename <- “/path/to/resultFile.txt” for (i in 1:length(filepaths)) { content <- fread(filepaths, header = FALSE, sep = “,”) ### some manipulation for the content results <- content[1] fwrite(results, resultFilename, col.names = FALSE, quote = FALSE, append = TRUE) } finalData <- fread(resultFilename, header = FALSE, … Read more
df <- transform(df, N = (df$N – min(df$N)) / (max(df$N) – min(df$N)) or as @jogo commented: df <- transform(df, N = (N – min(N)) / (max(N) – min(N)) 1 solved Normalization of only one column in data frame
This saves a bitmap file on the harddrive: bmp(“spam.bmp”) plot(sin, -pi, 2*pi) dev.off() Googling R save bitmap get’s you this answer. solved Using R, how to capture the bitmap of a plot?
library(dplyr) customer <- customer %>% rowwise() %>% mutate(age = age(as.Date(dateofbirth))) 2 solved How to add new column in R during for loop?