to parse json, i can use this approach
library("rjson")
json_file <- "https://api.coindesk.com/v1/bpi/currentprice/USD.json"
json_data <- fromJSON(paste(readLines(json_file), collapse=""))
but what if i want work with set of json files
it located
json_file<-"C:/myfolder/"
How to parse in to data.frame all json files in this folder? (there 1000 files)?
A lot of missing info, but this will probably work..
I used pblapply to get a nice progress-bar (since you are mentioning >1000 files).
I never used the solution below for JSON-files (no experience wit JSON), but it works flawless on .csv and .xls files ( of course with different read-functions).. so I expect it to work with JSON also.
library(data.table)
library(pbapply)
library(rjson)
folderpath <- "C:\\myfolder\\"
filefilter <- "*.json$"
#set paramaters as needed
f <- list.files( path = folderpath,
pattern = filefilter,
full.names = TRUE,
recursive = FALSE )
#read all files to a list
f.list <- pblapply( f, function(x) fromJSON( file = x ) )
#join lists together
dt <- data.table::rbindlist( f.list )
Related
I'm processing files through an application using R. The application requires a simple inputfile, outputfilename specification as parameters. Using the below code, this works fine.
input <- "\"7374.txt\""
output <- "\"7374_cleaned.txt\""
system2("DataCleaner", args = c(input, output))
However I wish to process a folder of .txt files, rather then have to do each one individually. If i had access to the source code i would simply alter the application to accept a folder rather then an individual file, but unfortunately i don't. Is it possible to somehow do this in R? I had tried starting to create a loop,
input <- dir(pattern=".txt")
but i don't know how i could insert a vector in as an argument without the regex included as part of that? Also i would then need to be able to paste '_cleaned' on to the end of the outputfile names? Many thanks in advance.
Obviously, I can't test it because I don't have your DataCleaner program but how about this...
# make some files
dir.create('folder')
x = sapply(seq_along(1:5), function(f) {t = tempfile(tmpdir = 'folder', fileext = '.txt'); file.create(t); t})
# find the files
inputfiles = list.files(path = 'folder', pattern = 'txt', full.names = T)
# remove the extension
base = tools::file_path_sans_ext(inputfiles)
# make the output file names
outputfiles = paste0(base, '_cleaned.txt')
mysystem <- function(input, output) {
system2('DataCleaner', args = c(input, output))
}
lapply(seq_along(1:length(inputfiles)), function(f) mysystem(inputfiles[f], outputfiles[f]))
It uses lapply to iterate over all the members of the input and output files and calls the system2 function.
How can I read many CSV files and make each of them into data tables?
I have files of 'A1.csv' 'A2.csv' 'A3.csv'...... in Folder 'A'
So I tried this.
link <- c("C:/A")
filename<-list.files(link)
listA <- c()
for(x in filename) {
temp <- read.csv(paste0(link , x), header=FALSE)
listA <- list(unlist(listA, recursive=FALSE), temp)
}
And it doesn't work well. How can I do this job?
Write a regex to match the filenames
reg_expression <- "A[0-9]+"
files <- grep(reg_expression, list.files(directory), value = TRUE)
and then run the same loop but use assign to dynamically name the dataframes if you want
for(file in files){
assign(paste0(file, "_df"),read.csv(file))
}
But in general introducing unknown variables into the scope is bad practice so it might be best to do a loop like
dfs <- list()
for(index in 1:length(files)){
file <- files[index]
dfs[index] <- read.csv(file)
}
Unless each file is a completely different structure (i.e., different columns ... the number of rows does not matter), you can consider a more efficient approach of reading the files in using lapply and storing them in a list. One of the benefits is that whatever you do to one frame can be immediately done to all of them very easily using lapply.
files <- list.files(link, full.names = TRUE, pattern = "csv$")
list_of_frames <- lapply(files, read.csv)
# optional
names(list_of_frames) <- files # or basename(files), if filenames are unique
Something like sapply(list_of_frames, nrow) will tell you how many rows are in each frame. If you have something more complex,
new_list_of_frames <- lapply(list_of_frames, function(x) {
# do something with 'x', a single frame
})
The most immediate problem is that when pasting your file path together, you need a path separator. When composing file paths, it's best to use the function file.path as it will attempt to determine what the path separator is for operating system the code is running on. So you want to use:
read.csv(files.path(link , x), header=FALSE)
Better yet, just have the full path returned when listing out the files (and can filter for .csv):
filename <- list.files(link, full.names = TRUE, pattern = "csv$")
Combining with the idea to use assign to dynamically create the variables:
link <- c("C:/A")
files <-list.files(link, full.names = TRUE, pattern = "csv$")
for(file in files){
assign(paste0(basename(file), "_df"), read.csv(file))
}
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)
}
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.
I have a multiple-step file download process I would like to do within R. I have got the middle step, but not the first and third...
# STEP 1 Recursively find all the files at an ftp site
# ftp://prism.oregonstate.edu//pub/prism/pacisl/grids
all_paths <- #### a recursive listing of the ftp path contents??? ####
# STEP 2 Choose all the ones whose filename starts with "hi"
all_files <- sapply(sapply(strsplit(all_paths, "/"), rev), "[", 1)
hawaii_log <- substr(all_files, 1, 2) == "hi"
hi_paths <- all_paths[hawaii_log]
hi_files <- all_files[hawaii_log]
# STEP 3 Download & extract from gz format into a single directory
mapply(download.file, url = hi_paths, destfile = hi_files)
## and now how to extract from gz format?
For part 1, RCurl might be helpful. The getURL function retrieves one or more URLs; dirlistonly lists the contents of the directory without retrieving the file. The rest of the function creates the next level of url
library(RCurl)
getContent <- function(dirs) {
urls <- paste(dirs, "/", sep="")
fls <- strsplit(getURL(urls, dirlistonly=TRUE), "\r?\n")
ok <- sapply(fls, length) > 0
unlist(mapply(paste, urls[ok], fls[ok], sep="", SIMPLIFY=FALSE),
use.names=FALSE)
}
So starting with
dirs <- "ftp://prism.oregonstate.edu//pub/prism/pacisl/grids"
we can invoke this function and look for things that look like directories, continuing until done
fls <- character()
while (length(dirs)) {
message(length(dirs))
urls <- getContent(dirs)
isgz <- grepl("gz$", urls)
fls <- append(fls, urls[isgz])
dirs <- urls[!isgz]
}
we could then use getURL again, but this time on fls (or elements of fls, in a loop) to retrieve the actual files. Or maybe better open a url connection and use gzcon to decompress and process on the file. Along the lines of
con <- gzcon(url(fls[1], "r"))
meta <- readLines(con, 7)
data <- scan(con, integer())
I can read the contents of the ftp page if I start R with the internet2 option. I.e.
C:\Program Files\R\R-2.12\bin\x64\Rgui.exe --internet2
(The shortcut to start R on Windows can be modified to add the internet2 argument - right-click /Properties /Target, or just run that at the command line - and obvious on GNU/Linux).
The text on that page can be read like this:
download.file("ftp://prism.oregonstate.edu//pub/prism/pacisl/grids", "f.txt")
txt <- readLines("f.txt")
It's a little more work to parse out the Directory listings, then read them recursively for the underlying files.
## (something like)
dirlines <- txt[grep("Directory <A HREF=", txt)]
## split and extract text after "grids/"
split1 <- sapply(strsplit(dirlines, "grids/"), function(x) rev(x)[1])
## split and extract remaining text after "/"
sapply(strsplit(split1, "/"), function(x) x[1])
[1] "dem" "ppt" "tdmean" "tmax" "tmin"
It's about here that this stops seeming very attractive, and gets a bit laborious so I would actually recommend a different option. There would no doubt be a better solution perhaps with RCurl, and I would recommend learning to use and ftp client for you and your user. Command line ftp, anonymous logins, and mget all works pretty easily.
The internet2 option was explained for a similar ftp site here:
https://stat.ethz.ch/pipermail/r-help/2009-January/184647.html
ftp.root <- where are the files
dropbox.root <- where to put the files
#=====================================================================
# Function that downloads files from URL
#=====================================================================
fdownload <- function(sourcelink) {
targetlink <- paste(dropbox.root, substr(sourcelink, nchar(ftp.root)+1,
nchar(sourcelink)), sep = '')
# list of contents
filenames <- getURL(sourcelink, ftp.use.epsv = FALSE, dirlistonly = TRUE)
filenames <- strsplit(filenames, "\n")
filenames <- unlist(filenames)
files <- filenames[grep('\\.', filenames)]
dirs <- setdiff(filenames, files)
if (length(dirs) != 0) {
dirs <- paste(sourcelink, dirs, '/', sep = '')
}
# files
for (filename in files) {
sourcefile <- paste(sourcelink, filename, sep = '')
targetfile <- paste(targetlink, filename, sep = '')
download.file(sourcefile, targetfile)
}
# subfolders
for (dirname in dirs) {
fdownload(dirname)
}
}