I have an R function from a package that I need to pass a file path as an argument but it's expecting a csv and my file is an xlsx.
I've looked at the code for the function an it is using read.csv to load the file but unfortunately I can't make any changes to the package.
Is there a good way to read in the xlsx and pass it to the function without writing it to a csv and having the function read it back in?
I came across the text argument for read.csv here:
Is there a way to use read.csv to read from a string value rather than a file in R?
This seems like might be part way there but as I said I am unable to alter the function.
Maybe you could construct your own function checking if the file is xlsx, and in this case create a temporary csv file, feed it to your function, and delete it. Something like
yourfunction = function(path){
read.csv(path)
head(path)
}
library(readxl)
modified_function = function(path){
if(grepl{"\\.xlsx",path}){
tmp <- read_xlsx(path)
tmp_path <- paste0(gsub("\\.xlsx","",path),"_tmp.csv")
write.csv(tmp,file = tmp_path)
output <- yourfunction(tmp_path)
file.remove(tmp_path)
}else{
output <- yourfunction(path)
}
return(output)
}
If it is of help, here you can see how to modify only one function of a package: How to modify a function of a library in a module
Related
I am using the purrr:walk to read multiple excel files and it failed. I have 3 questions:
(1) I used the function list.files to read the excel file list in one folder. But the returned values also included the subfolders. I tried set value for the parameters recursive= and include.dirs=, but it didn't work.
setwd(file_path)
files<-as_tibble(list.files(file_path,recursive=F,include.dirs=F)) %>%
filter(str_detect(value,".xlsx"))
files
(2) When I used the following piece of code, it can run without any error or warning message, but there is no returned data.
###read the excel data
file_read <- function(value1) {
print(value1)
file1<-read_excel(value1,sheet=1)
}
walk(files$value,file_read)
When I used the following, it worked. Not sure why.
test<-read_excel(files$value,sheet=1)
(3) In Q2, actually I want to create file1 to file6, suppose there are 6 excel files. How can I dynamically assign the dataset name?
list.files has pattern argument where you can specify what kind of files you are looking for. This will help you avoid filter(str_detect(value,".xlsx")) step. Also list.files only returns the files that are included in the main directory (file_path) and not it's subdirectory unless you specify recursive = TRUE.
library(readxl)
setwd(file_path)
files <- list.files(pattern = '\\.xlsx')
In the function you need to return the object.
file_read <- function(value1) {
data <- read_excel(value1,sheet=1)
return(data)
}
Now you can use map/lapply to read the files.
result <- purrr::map(files,file_read)
I am repeatedly applying a function to read and process a bunch of csv files. Each time it runs, the function creates a data frame (this.csv.data) and uses save() to write it to a .RData file with a unique name. Problem is, later when I read these .RData files using load(), the loaded variable names are not unique, because each one loads with the name this.csv.data....
I'd like to save them with unique tags so that they come out properly named when I load() them. I've created the following code to illustrate .
this.csv.data = list(data=c(1:9), unique_tag = "some_unique_tag")
assign(this.csv.data$unique_tag,this.csv.data$data)
# I want to save the data,
# with variable name of <unique_tag>,
# at a file named <unique_tag>.dat
saved_file_name <- paste(this.csv.data$unique_tag,"RData",sep=".")
save(get(this.csv.data$unique_tag), saved_file_name)
but the last line returns:
"Error in save(get(this_unique_tag), file = data_tag) :
object ‘get(this_unique_tag)’ not found"
even though the following returns the data just fine:
get(this.csv.data$unique_tag)
Just name the arguments you use. With your code the following works fine:
save(list = this.csv.data$unique_tag, file=saved_file_name)
My preference is to avoid the name in the RData file on load:
obj = local(get(load('myfile.RData')))
This way you can load various RData files and name the objects whatever you want, or store them in a list etc.
You really should use saveRDS/readRDS to serialize your objects.
save and load are for saving whole environments.
saveRDS(this.csv.data, saved_file_name)
# later
mydata <- readRDS(saved_file_name)
you can use
save.image("myfile.RData")
This worked for me:
env <- new.env()
env[[varname]] <- object_to_save
save(list=c(varname), envir=env, file='out.Rda')
You could probably do it without a new env (but I didn't try this):
.GlobalEnv[[varname]] <- object_to_save
save(list=c(varname), envir=.GlobalEnv, file='out.Rda')
You might even be able to remove the envir variable.
I want to read csv file from google cloud storage with a function similar to
read.csv.
I used library googleCloudStorageR and I can't find a function for that. I don't want to download it, I just want to read it in environment like a data frame.
If you download a .csv file, googleCloudStorageR will by default put it into a data.frame for you via write.csv - you can turn off the behaviour by specifying saveToDisk
# will make a data.frame
gcs_get_object("mtcars.csv")
# save to disk as a CSV
gcs_get_object("mtcars.csv", saveToDisk = "mtcars.csv")
You can specify your own parse function by supplying it via parseFunction
## default gives a warning about missing column name.
## custom parse function to suppress warning
f <- function(object){
suppressWarnings(httr::content(object, encoding = "UTF-8"))
}
## get mtcars csv with custom parse function.
gcs_get_object("mtcars.csv", parseFunction = f)
I’ve tried running a sample csv file with the as.data.frame() function.
In order to run this code snippet make sure you install (install.packages("data.table")) and included the library library(“data.table”)
Also be sure that you include the fread() within the as.data.frame() function in order to read the file from it’s location.
Here is the code snippet I ran and managed to display the data frame for my data set:
library(“data.table”)
MyData <- as.data.frame(fread(file="$FILE_PATH",header=TRUE, sep = ','))
print(MyData)
Reading Data with TensorFlow:
There is one other way you can read a csv from your cloud storage with the TensorFlow API. I would assume you are accessing this data from a bucket? Firstly, you would need to install the “readr” and “cloudml” packages for these functionalities to work. Then you would need to use gs_data_dir(“gs://your-bucket-name”) along with specifying the file path file.path(data_dir, “something.csv”). You would then want to read data from the file path with read_csv(file.path(data_dir, “something.csv”)). If you want it formatted as a data frame it should look something like this.
library(“data.table”)
library(cloudml)
library(readr)
data_dir <- gs_data_dir(“gs://your-bucket-name”)
MyData <- as.data.frame(read_csv(file.path(data_dir, “something.csv”)))
print(MyData)
Make sure you have properly authenticated access to your storage
More information in this link
I have a simple syntax question: Is there a way to specify the path in which to write a csv file within the .csv function itself?
I always do the following:
setwd("C:/Users/user/Desktop")
write.csv(dt, "my_file.csv", row.names = F)
However, I would like to skip the setwd() line and include it directly in the write.csv() function. I can't find a path setting in the write.csv documentation file. Is it possible to do this exclusively in write.csv without using write.table() or having to download any packages?
I am writing around 300 .csv files in a script that runs auomatically everyday. The loop runs slower when using write.table() than when using write.csv(). The whole reason I want to include the path in the write.csv() function is to see if I can decrease the time it takes to execute any further.
I typically set my "out" path in the beginning and then just use paste() to create the full filename to save to.
path_out = 'C:\\Users\\user\\Desktop\\'
fileName = paste(path_out, 'my_file.csv',sep = '')
write.csv(dt,fileName)
or all within write.csv()
path_out = 'C:\\Users\\user\\Desktop\\'
write.csv(dt,paste(path_out,'my_file.csv',sep = ''))
There is a specialized function for this: file.path:
path <- "C:/Users/user/Desktop"
write.csv(dt, file.path(path, "my_file.csv"), row.names=FALSE)
Quoting from ?file.path, its purpose is:
Construct the path to a file from components in a platform-independent way.
Some of the few things it does automatically (and paste doesn't):
Using a platform-specific path separator
Adding the path separator between path and filename (if it's not already there)
Another way might be to build a wrapper function around the write.csv function and pass the arguments of the write.csv function in your wrapper function.
write_csv_path <- function(dt,filename,sep,path){
write.csv(dt,paste0(path,filename,sep = sep))
}
Example
write_csv_path(dt = mtcars,filename = "file.csv",sep = "",path = ".\\")
In my case this works fine,
create a folder -> mmult.datas
copy its directory-> C:/Users/seyma/TP/tp.R/tp.R5 - Copy
give a name of your .csv -> df.Bench.csv
do not forget the write your data.frame -> df
write.csv(df, file ="C:/Users/seyma/TP/tp.R/tp.R5 - Copy/mmult.datas/df.Bench.csv")
for more you can check the link
I'm writing a loop script which involves reading a file from a workbook (using the package XLConnect). The challenge is that the file names contain characters (representing time) that I want to ignore.
For example, here are 3 paths to those files:
G://User//Documents//daily_data//Op_Schedule_20160520_132025.xlsx
G://User//Documents//daily_data//Op_Schedule_20160521_142805.xlsx
G://User//Documents//daily_data//Op_Schedule_20160522_103052.xlsx
I need to import hundreds of those files. I can easily account for the character string representing the date (e.g. 20160522), but not the time.
Is there a way to tell R to ignore some characters located in the file path? Here is how I was thinking of writing my script (the "???" is where i need help). I know a loop is probably not the most efficient way, but i'm open to suggestions, should you have any:
require(XLConnect)
path= "G://User//Documents//daily_data//Op_Schedule_"
wd.seq = format(seq(as.Date("2014-01-01"),as.Date("2016-12-31"),"days"),format="%Y%m%d")
scheduleList = rep(list(matrix(1,1,1)),length(wd.seq))
for(i in 1:length(wd.seq)) {
wb = loadWorkbook(file= paste0(path,wd.seq[i],"???",".xlxs"))
scheduleList[[i]] = readWorksheet(wb,sheet='=SCHEDULE', header = TRUE)
}
`
Thanks for reading and suggestions, if any.
Mathieu
I don't know if this is helpful, but if you want to read all the files in a certain directory (which it seems to me is what you're after), you can read all the filenames into a list using the list.files() function, for example
fileList <- list.files(""G://User//Documents//daily_data//")
And then load the xlsx files looping through the list with a for loop
for(i in fileList) {
loadWorkbook(file = i)
}
I haven't used the XLConnect function before so that exact code probably doesn't work, but the loop will iterate through all the files in that directory and so you can construct your loading call using the i variable for the filename (it won't be an absolute path though, so you might need to use paste to add the first part of the filepath)
I realize there might be other files in the directory that are not excel files, you could use grepl to select only files containg "OP_Schedule_"
fileListClean <- fileList[grepl("Op_Schedule_",fileList)]
or perhaps only selecting .xlsx files in the directory:
fileListClean <- fileList[grepl(".xlsx",fileList)]
Edit to fit your reply:
Since you need to fit it to a sequence, you can do it as you did earlier:
wd.seq = format(seq(as.Date("2014-01-01"),as.Date("2016-12-31"),"days"),format="%Y%m%d")
wd.seq2 <- paste("Op_Schedule_", wd.seq, sep = "")
And then use grepl to only pick files starting with that extensions:
fileListClean <- fileList[grepl(paste(wd.seq2, collapse = "|"), fileList)]
Full disclosure: The last part i got from this SO answer: grep using a character vector with multiple patterns