Extracting file numbers from file names in r and looping through files - r

I have a folder full of .txt files that I want to loop through and compress into one data frame, but each .txt file is data for one subject and there are no columns in the text files that indicate subject number or time point in the study (e.g. 1-5). I need to add a line or two of code into my loop that looks for strings of four numbers (i.e. each file is labeled something like: "4325.5_ERN_No_Startle") and just creates a column with 4325 and another column with 5 that will appear for every data point for that subject until the loop gets to the next one. I have been looking for awhile but am still coming up empty, any suggestions?
I also have not quite gotten the loop to work:
path = "/Users/me/Desktop/Event Codes/ERN task/ERN text files transferred"
out.file <- ""
file <- ""
file.names <- dir(path, pattern =".txt")
for(i in 1:length(file.names)){
file <- read.table(file.names[i],header=FALSE, fill = TRUE)
out.file <- rbind(out.file, file)
}
which runs okay until I get this error message part way through:
Error in read.table(file.names[i], header = FALSE, fill = TRUE) :
no lines available in input

Consider using regex to parse the file name for study period and subject, both of which are then binded in a lapply of list.files:
path = "path/to/text/files"
# ANY TXT FILE WITH PATTERN OF 4 DIGITS FOLLOWED BY A PERIOD AND ONE DIGIT
file.names <- list.files(path, pattern="*[0-9]{4}\\.[0-9]{1}.*txt", full.names=TRUE)
# IMPORT ALL FILES INTO A LIST OF DATAFRAMES AND BINDS THE REGEX EXTRACTS
dfList <- lapply(file.names, function(x) {
if (file.exists(x)) {
data.frame(period=regmatches(x, gregexpr('[0-9]{4}', x))[[1]],
subject=regmatches(x, gregexpr('\\.[0-9]{1}', x))[[1]],
read.table(x, header=FALSE, fill=TRUE),
stringsAsFactors = FALSE)
}
})
# COMBINE EACH DATA FRAME INTO ONE
df <- do.call(rbind, dfList)
# REMOVE PERIOD IN SUBJECT (NEEDED EARLIER FOR SPECIAL DIGIT)
df['subject'] <- sapply(df['subject'],
function(x) gsub("\\.", "", x))

You can try to use tryCatchwhich basically would give you a NULL instead of an error.
file <- tryCatch(read.table(file.names[i],header=FALSE, fill = TRUE), error=function(e) NULL))

Related

Appending a list in a loop (R)

I want to use a loop to read in multiple csv files and append a list in R.
path = "~/path/to/csv/"
file.names <- dir(path, pattern =".csv")
mylist=c()
for(i in 1:length(file.names)){
datatmp <- read.csv(file.names[i],header=TRUE, sep=";", stringsAsFactors=FALSE)
listtmp = datatmp[ ,6]
finallist <- append(mylist, listtmp)
}
finallist
For each csv file, the desired column has a different length.
In the end, I want to get the full appended list with all values in that certain column from all csv files.
I am fairly new to R, so I am not sure what I'm missing...
There are four errors in your approach.
First, file.names <- dir(path, pattern =".csv") will extract just file names, without path. So, when you try to import then, read.csv() doesn't find.
Building the path
You can build the right path including paste0():
path = "~/path/to/csv/"
file.names <- paste0(path, dir(path, pattern =".csv"))
Or file.path(), which add slashes automaticaly.
path = "~/path/to/csv"
file.names <- file.path(path, dir(path, pattern =".csv"))
And another way to create the path, for me more efficient, is that suggested in the answer commented by Tung.
file.names <- list.files(path = "~/path/to/csv", recursive = TRUE,
pattern = "\\.csv$", full.names = TRUE)
This is better because in addition to being all in one step, you can use within a directory containing multiple files of various formats. The code above will match all .csv files in the folder.
Importing, selecting and creating the list
The second error is in mylist <- c(). You want a list, but this creates a vector. So, the correct is:
mylist <- list()
And the last error is inside the loop. Instead of create other list when appending, use the same object created before the loop:
for(i in 1:length(file.names)){
datatmp <- read.csv(file.names[i], sep=";", stringsAsFactors=FALSE)
listtmp = datatmp[, 6]
mylist <- append(mylist, list(listtmp))
}
mylist
Another approach, easier and cleaner, is looping with lapply(). Just this:
mylist <- lapply(file.names, function(x) {
df <- read.csv(x, sep = ";", stringsAsFactors = FALSE)
df[, 6]
})
Hope it helps!

Using lapply to apply a function over read-in list of files and saving output as new list of files

I'm quite new at R and a bit stuck on what I feel is likely a common operation to do. I have a number of files (57 with ~1.5 billion rows cumulatively by 6 columns) that I need to perform basic functions on. I'm able to read these files in and perform the calculations I need no problem but I'm tripping up in the final output. I envision the function working on 1 file at a time, outputting the worked file and moving onto the next.
After calculations I would like to output 57 new .txt files named after the file the input data first came from. So far I'm able to perform the calculations on smaller test datasets and spit out 1 appended .txt file but this isn't what I want as a final output.
#list filenames
files <- list.files(path=, pattern="*.txt", full.names=TRUE, recursive=FALSE)
#begin looping process
loop_output = lapply(files,
function(x) {
#Load 'x' file in
DF<- read.table(x, header = FALSE, sep= "\t")
#Call calculated height average a name
R_ref= 1647.038203
#Add column names to .las data
colnames(DF) <- c("X","Y","Z","I","A","FC")
#Calculate return
DF$R_calc <- (R_ref - DF$Z)/cos(DF$A*pi/180)
#Calculate intensity
DF$Ir_calc <- DF$I * (DF$R_calc^2/R_ref^2)
#Output new .txt with calcuated columns
write.table(DF, file=, row.names = FALSE, col.names = FALSE, append = TRUE,fileEncoding = "UTF-8")
})
My latest code endeavors have been to mess around with the intial lapply/sapply function as so:
#begin looping process
loop_output = sapply(names(files),
function(x) {
As well as the output line:
#Output new .csv with calcuated columns
write.table(DF, file=paste0(names(DF), "txt", sep="."),
row.names = FALSE, col.names = FALSE, append = TRUE,fileEncoding = "UTF-8")
From what I've been reading the file naming function during write.table output may be one of the keys I don't have fully aligned yet with the rest of the script. I've been viewing a lot of other asked questions that I felt were applicable:
Using lapply to apply a function over list of data frames and saving output to files with different names
Write list of data.frames to separate CSV files with lapply
to no luck. I deeply appreciate any insights or paths towards the right direction on inputting x number of files, performing the same function on each, then outputting the same x number of files. Thank you.
The reason the output is directed to the same file is probably that file = paste0(names(DF), "txt", sep=".") returns the same value for every iteration. That is, DF must have the same column names in every iteration, therefore names(DF) will be the same, and paste0(names(DF), "txt", sep=".") will be the same. Along with the append = TRUE option the result is that all output is written to the same file.
Inside the anonymous function, x is the name of the input file. Instead of using names(DF) as a basis for the output file name you could do some transformation of this character string.
example.
Given
x <- "/foo/raw_data.csv"
Inside the function you could do something like this
infile <- x
outfile <- file.path(dirname(infile), gsub('raw', 'clean', basename(infile)))
outfile
[1] "/foo/clean_data.csv"
Then use the new name for output, with append = FALSE (unless you need it to be true)
write.table(DF, file = outfile, row.names = FALSE, col.names = FALSE, append = FALSE, fileEncoding = "UTF-8")
Using your code, this is the general idea:
require(purrr)
#list filenames
files <- list.files(path=, pattern="*.txt", full.names=TRUE, recursive=FALSE)
#Call calculated height average a name
R_ref= 1647.038203
dfTransform <- function(file){
colnames(file) <- c("X","Y","Z","I","A","FC")
#Calculate return
file$R_calc <- (R_ref - file$Z)/cos(file$A*pi/180)
#Calculate intensity
file$Ir_calc <- file$I * (file$R_calc^2/R_ref^2)
return(file)
}
output <- files %>% map(read.table,header = FALSE, sep= "\t") %>%
map(dfTransform) %>%
map(write.table, file=paste0(names(DF), "txt", sep="."),
row.names = FALSE, col.names = FALSE, append = TRUE,fileEncoding = "UTF-8")

Combine csv files with common file identifier

I have a list of approximately 500 csv files each with a filename that consists of a six-digit number followed by a year (ex. 123456_2015.csv). I would like to append all files together that have the same six-digit number. I tried to implement the code suggested in this question:
Import and rbind multiple csv files with common name in R but I want the appended data to be saved as new csv files in the same directory as the original files are currently saved. I have also tried to implement the below code however the csv files produced from this contain no data.
rm(list=ls())
filenames <- list.files(path = "C:/Users/smithma/Desktop/PM25_test")
NAPS_ID <- gsub('.+?\\([0-9]{5,6}?)\\_.+?$', '\\1', filenames)
Unique_NAPS_ID <- unique(NAPS_ID)
n <- length(Unique_NAPS_ID)
for(j in 1:n){
curr_NAPS_ID <- as.character(Unique_NAPS_ID[j])
NAPS_ID_pattern <- paste(".+?\\_(", curr_NAPS_ID,"+?)\\_.+?$", sep = "" )
NAPS_filenames <- list.files(path = "C:/Users/smithma/Desktop/PM25_test", pattern = NAPS_ID_pattern)
write.csv(do.call("rbind", lapply(NAPS_filenames, read.csv, header = TRUE)),file = paste("C:/Users/smithma/Desktop/PM25_test/MERGED", "MERGED_", Unique_NAPS_ID[j], ".csv", sep = ""), row.names=FALSE)
}
Any help would be greatly appreciated.
Because you're not doing any data manipulation, you don't need to treat the files like tabular data. You only need to copy the file contents.
filenames <- list.files("C:/Users/smithma/Desktop/PM25_test", full.names = TRUE)
NAPS_ID <- substr(basename(filenames), 1, 6)
Unique_NAPS_ID <- unique(NAPS_ID)
for(curr_NAPS_ID in Unique_NAPS_ID){
NAPS_filenames <- filenames[startsWith(basename(filenames), curr_NAPS_ID)]
output_file <- paste0(
"C:/Users/nwerth/Desktop/PM25_test/MERGED_", curr_NAPS_ID, ".csv"
)
for (fname in NAPS_filenames) {
line_text <- readLines(fname)
# Write the header from the first file
if (fname == NAPS_filenames[1]) {
cat(line_text[1], '\n', sep = '', file = output_file)
}
# Append every line in the file except the header
line_text <- line_text[-1]
cat(line_text, file = output_file, sep = '\n', append = TRUE)
}
}
My changes:
list.files(..., full.names = TRUE) is usually the best way to go.
Because the digits appear at the start of the filenames, I suggest substr. It's easier to get an idea of what's going on when skimming the code.
Instead of looping over the indices of a vector, loop over the values. It's more succinct and less likely to cause problems if the vector's empty.
startsWith and endsWith are relatively new functions, and they're great.
You only care about copying lines, so just use readLines to get them in and cat to get them out.
You might consider something like this:
##will take the first 6 characters of each file name
six.digit.filenames <- substr(filenames, 1,6)
path <- "C:/Users/smithma/Desktop/PM25_test/"
unique.numbers <- unique(six.digit.filenames)
for(j in unique.numbers){
sub <- filenames[which(substr(filenames,1,6) == j)]
data.for.output <- c()
for(file in sub){
##now do your stuff with these files including read them in
data <- read.csv(paste0(path,file))
data.for.output <- rbind(data.for.output,data)
}
write.csv(data.for.output,paste0(path,j, '.csv'), row.names = F)
}

rbind txt files from online directory (R)

I am trying to get concatenate text files from url but i don't know how to do this with the html and the different folders?
This is the code i tried, but it only lists the text files and has a lot of html code like this How do I fix this so that I can combine the text files into one csv file?
library(RCurl)
url <- "http://weather.ggy.uga.edu/data/daily/"
dir <- getURL(url, dirlistonly = T)
filenames <- unlist(strsplit(dir,"\n")) #split into filenames
#append the files one after another
for (i in 1:length(filenames)) {
file <- past(url,filenames[i],delim='') #concatenate for urly
if (i==1){
cp <- read_delim(file, header=F, delim=',')
}
else{
temp <- read_delim(file,header=F,delim=',')
cp <- rbind(cp,temp) #append to existing file
rm(temp)# remove the temporary file
}
}
here is a code snippet that I got to work for me. I like to use rvest over RCurl, just because that's what I've learned. In this case, I was able to use the html_nodes function to isolate each file ending in .txt. The result table has the times saved as character strings, but you could fix that later. Let me know if you have any questions.
library(rvest)
library(readr)
url <- "http://weather.ggy.uga.edu/data/daily/"
doc <- xml2::read_html(url)
text <- rvest::html_text(rvest::html_nodes(doc, "tr td a:contains('.txt')"))
# define column types of fwf data ("c" = character, "n" = number)
ctypes <- paste0("c", paste0(rep("n",11), collapse = ""))
data <- data.frame()
for (i in 1:2){
file <- paste0(url, text[1])
date <- as.Date(read_lines(file, n_max = 1), "%m/%d/%y")
# Read file to determine widths
columns <- fwf_empty(file, skip = 3)
# Manually expand `solar` column to be 3 spaces wider
columns$begin[8] <- columns$begin[8] - 3
data <- rbind(data, cbind(date,read_fwf(file, columns,
skip = 3, col_types = ctypes)))
}

Read in multiple txt files and create a list of it to access each file by accessing the list element in R

Being relatively new to R programming I am struggling with a huge data set of 16 text files (, seperated) saved in one dierctory. All the files have same number of columns and the naming convention, for example file_year_2000, file_year_2001 etc. I want to create a list in R where i can access each file individually by accessing the list elementts. By searching through the web i found some code and tried the following but as a result i get one huge list (16,2 MB) where the output is just strange. I would like to have 16 elements in the list each represting one file read from the directory. I tried the following code but it does not work as i want:
path = "~/.../.../.../Data_1999-2015"
list.files(path)
file.names <- dir(path, pattern =".txt")
length(file.names)
df_list = list()
for( i in length(file.names)){
file <- read.csv(file.names[i],header=TRUE, sep=",", stringsAsFactors=FALSE)
year = gsub('[^0-9]', '', file)
df_list[[year]] = file
}
Any suggestions?
Thanks in advance.
Just to give more details
path = "~/.../.../.../Data_1999-2015"
list.files(path)
file.names <- dir(path, pattern =".txt")
length(file.names)
df_list = list()
for(i in seq(length(file.names))){
year = gsub('[^0-9]', '', file.names[i])
df_list[[year]] = read.csv(file.names[i],header=TRUE, sep=",", stringsAsFactors=FALSE)
}
Maybe it would be worth joining the data frames into one big data frame with an additional column being the year?
I assume that instead of "access each file individually" you mean you want to access individually data in each file.
Try something like this (untested):
path = "~/.../.../.../Data_1999-2015"
file.names <- dir(path, pattern =".txt")
df_list = vector("list", length(file.names))
# create a list of data frames with correct length
names(df_list) <- rep("", length(df_list))
# give it empty names to begin with
for( i in seq(along=length(file.names))) {
# now i = 1,2,...,16
file <- read.csv(file.names[i],header=TRUE, sep=",", stringsAsFactors=FALSE)
df_list[[i]] = file
# save the data
year = gsub('[^0-9]', '', file.names[i])
names(df_list)[i] <- year
}
Now you can use either df_list[[1]] or df_list[["2000"]] for year 2000 data.
I am uncertain if you are reading yout csv files in the right directory. If not, use
file <- read.csv(paste0(path, file.names[i], sep="/"),header=TRUE, sep=",", stringsAsFactors=FALSE)
when reading the file.

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