As usual we can save output file use syntax like this
write.csv(sst, file = "new_sst.csv", row.names = TRUE, sep = ',', col.names = TRUE)
then the result is new_sst.csv in our directory
My question is how can we define free name using file.choose() to save the output file in our directory. I mean if we've done the same process to save different file input and file output so we don't need write the syntax several times.
I would write a for loop:
for (i in 1:10) {....
}
And inside that loop, you can call wildcard patterns using glob2rx to call input file names from your working directory, do your computations, and then paste() to create new file names using for the new file-write.
Here's an example of how to use paste()
http://stackoverflow.com/questions/13243873/is-it-appropriate-to-create-a-filename-using-paste-inside-of-file-path
Ok, given you have a list of data.frames in L, and you want to save all the data.frames in the working directory as
"new_sst(1).csv", "new_sst(2).csv", ""new_sst(3).csv" ...
the you can use a for loop or the function mapply():
# just to make it clear, I define a tiny function to create the numbered filename
# and to save a given data.frame under this name as CSV in the working directory:
SaveWithNumber = function(data,filename,number) {
# data: data.frame
# filename: common part of the file name without extension
# number: integer to be added after the filename
filename = paste0(filename,"(",number,").csv")
write.csv( filename, row.names = TRUE, sep = ',', col.names = TRUE)
}
# either the loop solution:
for (i in 1:length(L)) SaveWithNumber(L[[i]], "new_sst", i)
# or via mapply():
mapply(SaveWithNumber, L, "new_sst", 1:length(L))
thanks you Damian and Jochen, both of you give a good comment,
how about like when we working with office (word, excel, etc). If we want to save as file we can choose and write specific name in the directory we want, just like that but i want do it in R
OR
for my case, everyday i will run this script and i want to give the output name "sst20140123.csv" that tell us the output file is generated in 23 January 2014, according to day the script have been run
you can get the date part of the file name in the format you wish with
format(Sys.time(), "%y%m%d")
Changing my function from above:
SaveWithNumber = function(data,filename,number) {
datestr = format(Sys.time(), "%y%m%d")
filename = paste0("sst",datestr,".csv")
write.csv( filename, row.names = TRUE, sep = ',', col.names = TRUE)
}
hi Jochen, thanks for your quick respond..
For my case it's still not working, this is my full script, may be you could help me..
the objective of this scripts is to filter sst based on qual_sst index, and save the output file with another name like "sst20140125.csv"
# Read sst.data and QC
sst.data =3]= 3 is bad data
subset.par.sst
There was a typo in my post. in the function write.file() I forgot to give the data (your "sst.data" data.frame). But you might have recognized this.
If you have the "sst.data" and you want so save this under a filename like "D:/sst20140125.csv", where 2014125 is the date (yyyymd), then all you need to do is to generate this filename automatically of the current date, for instance by
datestr = format(Sys.time(), "%y%m%d")
filename = paste0("D:/sst",datestr,".csv")
Now you can use this filename to save the sst.data with
write.csv(sst.data,filename, sep=',', row.names = FALSE)
That's all.
Thanks you Jochen, it's working like what i want, just make little change on %y become %Y than i get output file sst20140206.csv
datestr = format(Sys.time(), "%Y%m%d")
one more, how about if we want to change the date (yymmdd) to julian calender, just like sst012.csv, that "012" is julian calendar for 12 january 2014
You now the date and time functions (see the help files) and you can have a look at the functions for string manipulation (like substr, nchar, paste,...). I am sure you will find a solution.
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