Loading Multiple .txt files where columns are separated by | character in R - r

I am attempting to load multiple text files into R and in each of the files, the columns are divided using the "|" character.
To give a sense of what the file structure looks like, a given row will look like:
congression printer|-182.431552949032
In this file I want to separate the congressional printer string from the numerical characters.
When using the following code:
folder <- '~/filepath'
file_list <- list.files(path=folder, pattern="*.txt")
data <-
do.call('rbind',
lapply(file_list,
function(x)
read.table(paste(folder, x, sep= ""),
header = TRUE, row.names = NULL)))
It'll load in the data as:
[1] [2]
congression printer|-182.431552949032
Is there away to correct this later using the tidyr::separate() function or by hedging the problem in the beginning? When trying to just put sep ="|" in the code above, that just impacts how my text files are found so that doesn't really work.

Things are always easier (and more powerful) with data.table :
library(data.table)
folder <- '~/filepath'
pathsList <- list.files(path=folder, pattern="*.txt", full.names = T)
rbindlist(lapply(pathsList, fread))

this works too:
folder <- '~/filepath'
file_list <- list.files(path=folder, pattern="*.txt")
data <-
do.call('rbind',
lapply(file_list,
function(x)
read.table(paste0(folder, x), sep = "|",
header = TRUE, row.names = NULL)))

Related

Naming a dataframe like the path

I have a lot of CSV that need to be standardized. I created a dictionary for doing so and so far the function that I have looks like this:
inputpath <- ("input")
files<- paste0(inputpath, "/",
list.files(path = inputpath, pattern = '*.gz',
full.names = FALSE))
standardizefunctiontofiles = lapply(files, function(x){
DF <- read_delim(x, delim = "|", na="")
names(DF) <- dictionary$final_name[match(names(DF), dictionary$old_name)]
})
Nonetheless, the issue that I have is that when I read the CSV and turn them into a dataframe they lose their path and therefore I can't not write each of them as a CSV that matches the input name. What I would normally do would be:
output_name <- str_replace(x, "input", "output")
write_delim(x, "output_name", delim = "|")
I was thinking that a way of solving this would be to make this step:
DF <- read_delim(x, delim = "|", na="")
so that the DF gets the name of the path but I haven't find any solution for that.
Any ideas on how to solve this issue for being able to apply a function and writing each of them as a standardized CSV?
I don't completely understand the question. But as far as I understood you want to overwrite CSV files you are reading with a new CSV file that contains the information of a modified (and correct) data frame.
I think you have two alternatives
Option 1) When reading data, store both CSV as a data frame and path as a string within a list.
This would be something like
file_list <- list()
for (i in seq_along(files)) {
file_list[[i]] <- list(df = read_delim(files[[i]], delim = "|", na = ""),
path = files[[i]])
}
Then, when you write the corrected data frames, you can use the paths in the second element of the list within the list file_list. Note that in order to get the path as a string you will need to do something like file_list[[1]][["path"]]
Option 2) Use assign
for (i in seq_along(files)) {
assign(files[[i]], read_delim(files[[i]], delim = "|", na = ""))
}
Option 3) Use do.call and the fact that <- is a function!
for (i in seq_along(files)) {
do.call("<-", list(files[[i]], read_delim(files[[i]], delim = "|", na = "")))
}
I hope this is useful!!
NB) None of the functions are implemented as efficiently as possible. They just introduce the idea.

Rename multiple datasets in R

I have almost 400 dataframes loaded in R. But the names still have the .csv extension.
I read the data with this code
Files <- list.files(pattern="\\.csv$")
for (i in 1:length(Files)){
assign(Files[i],
read.csv(Files[i],
sep = ";",
header = T))
}
Is there a way to remove the .cvs extension while importing the datasets?
Thanks a lot!
Here is a way that doesn't use assign, which is likely much better practice. You can keep the file names as the element names of the list.
library(tidyverse)
files <- list.files(pattern="\\.csv$")
df_list <- map(files, read_csv2)
names(df_list) <- str_remove(files, "\\.csv$")
Try this:
Files <- list.files(pattern="\\.csv$")
for (i in 1:length(Files)){
assign(gsub("\\..*","",Files)[i], # replace your this line of code
read.csv(Files[i],
sep = ";",
header = T))
}
You may want to add an extra gsub step:
Files <- list.files(pattern="\\.csv$")
File.name <- gsub("\\.csv$", "", Files)
for (i in 1:length(Files)){
assign(File.name[i],
read.csv(Files[i],
sep = ";",
header = T))
}

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)
}

Extract data from text files using for loop

I have 40 text files with names :
[1] "2006-03-31.txt" "2006-06-30.txt" "2006-09-30.txt" "2006-12-31.txt" "2007-03-31.txt"
[6] "2007-06-30.txt" "2007-09-30.txt" "2007-12-31.txt" "2008-03-31.txt" etc...
I need to extract one specific data, i know how to do it individually but this take a while:
m_value1 <- `2006-03-31.txt`$Marknadsvarde_tot[1]
m_value2 <- `2006-06-30.txt`$Marknadsvarde_tot[1]
m_value3 <- `2006-09-30.txt`$Marknadsvarde_tot[1]
m_value4 <- `2006-12-31.txt`$Marknadsvarde_tot[1]
Can someone help me with a for loop which would extract the data from a specific column and row through all the different text files please?
Assuming your files are all in the same folder, you can use list.files to get the names of all the files, then loop through them and get the value you need. So something like this?
m_value<-character() #or whatever the type of your variable is
filelist<-list.files(path="...", all.files = TRUE)
for (i in 1:length(filelist)){
df<-read.table(myfile[i], h=T)
m_value[i]<-df$Marknadsvarde_tot[1]
}
EDIT:
In case you have imported already all the data you can use get:
txt_files <- list.files(pattern = "*.txt")
for(i in txt_files) { x <- read.delim(i, header=TRUE) assign(i,x) }
m_value<-character()
for(i in 1:length(txt_files)) {
m_value[i] <- get(txt_files[i])$Marknadsvarde_tot[1]
}
You could utilize the select-parameter from fread of the data.table-package for this:
library(data.table)
file.list <- list.files(pattern = '.txt')
lapply(file.list, fread, select = 'Marknadsvarde_tot', nrow = 1, header = FALSE)
This will result in a list of datatables/dataframes. If you just want a vector with all the values:
sapply(file.list, function(x) fread(x, select = 'Marknadsvarde_tot', nrow = 1, header = FALSE)[[1]])
temp = list.files(pattern="*.txt")
library(data.table)
list2env(
lapply(setNames(temp, make.names(gsub("*.txt$", "", temp))),
fread), envir = .GlobalEnv)
Added data.table to an existing answer at Importing multiple .csv files into R
After you get all your files you can get data from the data.tables using DT[i,j,k] where i will be your condition

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