CODEWITHSUNDEEP
rr-markdown
Catherine Laing
Catherine Laing

Reputation: 657

Allocating specific filenames to set of different output files in rmarkdown

I am creating a set of PDFs that each include different information. I want to name the files according to the people who will be receiving the information. I am using for loops to run this:

 for (i in 1:nrow(files)){
   rmarkdown::render(input = "files_script.Rmd",
                     output_format = "pdf_document",
                     output_file = paste("Information_", i, ".pdf", sep=''),
                     output_dir = "directoryname/")
 }

The script works just fine, but the output files are simply numbered in my directory. I'd like to name them according to names in a column of my dataframe.

My data looks like this:

 Feedback Name    Surname
  xyz      John     Smith
  abc      Karen    Jones

I'd like to name the files by surname.

Upvotes: 0

Views: 49

Answers (1)

JBGruber
JBGruber

Reputation: 12478

So your names are in two columns in a data.frame (I called it df here). You could use the for loop like this:

for (name in paste0(df$Name, "_", df$Surname)) {
  rmarkdown::render(input = "files_script.Rmd",
                    output_format = "pdf_document",
                    output_file = paste0("Information_", name, ".pdf"),
                    output_dir = "directoryname/")
}

I replaced paste(sep = "") with paste0() since it is doing the same thing.

Not sure though why you would do it that way. files_script.Rmd is always the same file. So it would be faster to render it once and then just change the name.

Update

Since you said you want to keep the i as for loop variable, here is how you can do that:

names <- paste0(df$Name, "_", df$Surname)
for (i in seq_along(names)) {
  rmarkdown::render(input = "files_script.Rmd",
                    output_format = "pdf_document",
                    output_file = paste0("Information_", names[i], ".pdf"),
                    output_dir = "directoryname/")
}

seq_along is a safer way of doing 1:length(names) in this case (see this)

Upvotes: 2

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