On Thu, 8 Jul 2021 01:27:48 +0000 (UTC)
Kai Yang via R-help <r-help@r-project.org> wrote:
> Hello List,
> I have a one column data frame to store file name with extension. I
> want to create new column to keep file name only without extension. I
> tried to use strsplit("name1.csv", "\\.")[[1]] to do that, but it
> just retain the first row only and it is a vector. how can do this
> for all of rows and put it into a new column? thank you, Kai
Your example is confusing/garbled. You are applying strsplit() to a
single character string namely "name1.csv". Your intent presumably is
to apply it to a *vector* (say "v") of file names.
A syntax which would work is
sapply(strsplit(v,"\\."),function(x){x[1]})
E.g.
v <- c("clyde.txt","irving.tex","melvin.pdf","fred.csv")
sapply(strsplit(v,"\\."),function(x){x[1]})
which gives the output
> [1] "clyde" "irving" "melvin" "fred"
Note that the output of strplit() is a *list* the i-th entry of which
is a vector consisting of the "split" of the i-th entry of the vector
to which strsplit() is applied. In your example (corrected, so that it
makes sense, by replacing the string "names.csv" by my vector "v") you
get
strsplit(v,"\\.")[[1]]
[1] "clyde" "txt"
the result of splitting "clyde.txt"; you want the first entry of this
result, i.e. "clyde".
My "sapply()" construction produces the first entry of each entry of the
list produced by strsplit().
It is useful to get your thoughts clear, understand what you are doing
and understand what the functions that you are using do. (Read the
help!)
cheers,
Rolf Turner
--
Honorary Research Fellow
Department of Statistics
University of Auckland
Phone: +64-9-373-7599 ext. 88276
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