Loop through files in R and select rows by string - r

Question

I have a large number of CSV files. I need to extract relevant data from each file, and compile all of the relevant data into a new file.
I have been copying/pasting the code below and changing relevant details (e.g., file name) to repeat the same process for many CSV files. After that, I use cbind()/write.xlsx() to combine all of the relevant data and write it to an excel file. I need a more efficient method to accomplish this task.
How can I:
incorporate a loop that imports a large number of CSV files (to replace #1 below)
select relevant rows based on a string instead of entering specific row numbers
(to replace # 2 below)
combine all of the relevant data into a single data frame with each file's data in one column
library(tidyr)
# 1 - import raw data files
file1 <- read.csv ("1.csv", header = FALSE, sep = "\n")
# 2 - select relevant rows
file1 <- as.data.frame(file1[c(41:155),])
colnames(file1) <- c("file1")
#separate components of each line from raw csv file / isolate data
temp1 <- separate(file1, file1, into = c("Text", "IntNum", "Data", sep = "\\s"))
temp1 <- temp1$Data
temp1 <- as.data.frame(temp1)

Answer 1

If the number of relevant rows in each file is the same, you could do it like this. Option 1 shows a solution using a loop, option 2 shows a solution using sapply.
In a first step I generate three csv-files to make the code reproducible. The start row in each file is defined by "start", the end row by "end". I then get a list with the names of these files with dir().
#make csv-files, target vector always same length (3)
set.seed(1)
for (i in 1:3) {
df <- data.frame(x = c(rep(0, sample(1:10,1)), "begin",
paste0("dat", i),
"end",rep(0, sample(1:10, 1))))
write.csv(df, file = paste0("file", i, ".csv"), quote = FALSE, row.names = FALSE)
}
#get list of file names
allFiles <- dir(pattern = glob2rx("*.csv"))
Option 1 - loop
For the loop you could first initialize a result data frame ("outDF") with the number of columns set to the number of csv-files and the number of rows set to the length of the target vector in each file ("start" to "end"). You can then loop over the files and fill the data frame. The start and end rows can be indexed using which().
#initialise result data frame
outDF <- data.frame(matrix(nrow = 3, ncol = length(allFiles),
dimnames = list(NULL, allFiles)))
#loop over csv files
for (iFile in allFiles) {
idat <- read.csv(iFile, stringsAsFactors = FALSE) #read csv
outDF[, iFile] <- idat[which(idat$x == "start"):which(idat$x == "end"),]
}
Option 2 - sapply
Instead of a loop you could use sapply with a custom function to extract the relevant rows in each file. This returns a matrix which you could then transform into a dataframe.
out <- sapply(allFiles, FUN = function(x) {
idat <- read.csv(x, stringsAsFactors = FALSE)
return(idat[which(idat$x == "start"):which(idat$x == "end"),])
})
outDF <- as.data.frame(out)
If the number of rows between "start" and "end" differs between files, the above options won´t work. In this case you could generate a data frame by first using lapply() (similar to option 2) to generate a result list (with different lengths of the list elements) and then padding shorter lists with NAs before transforming the result into a dataframe again.
#make csv-files with with target vector of different lengths (3:12)
set.seed(1)
for (i in 1:3) {
df <- data.frame(x = c(rep(0, sample(1:10,1)), "start",
rep(paste0("dat", i), sample(1:10,1)),
"end",rep(0, sample(1:10, 1))))
write.csv(df, file = paste0("file", i, ".csv"), quote = FALSE, row.names = FALSE)
}
#lapply
out <- lapply(allFiles, FUN = function(x) {
idat = read.csv(x, stringsAsFactors = FALSE)
return(idat[which(idat$x == "start"):which(idat$x == "end"),])
})
out <- lapply(out, `length<-`, max(lengths(out)))
outDF <- do.call(cbind, out)