Find file name from full file path

Is there a way to extract the file name from the file full path (part of a file path) without the hassle of manipulating string?

The equivalent in Java would be:

File f = new File ("C:/some_dir/a")
f.getName() //output a
f.getFullAbs开发者_运维问答olutePath() //output c:/some_dir/a

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Use

basename("C:/some_dir/a.ext")
# [1] "a.ext"
dirname("C:/some_dir/a.ext")
# [1] "C:/some_dir"

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The tidyverse equivalent lives in the fs package. {fs} makes use of libuv under the hood.

library("fs")

path_file("/some/path/to/file.xyz")
#> [1] "file.xyz"

path_dir("/some/path/to/file.xyz")
#> [1] "/some/path/to"

^{Created on 2020-02-19 by the reprex package (v0.3.0)}

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@Honeybear. The function that removes the extension from the filename you could use is the function from the {tools} R package

tools::file_path_sans_ext("ABCD.csv")
## [1] "ABCD"

See this post in SO

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While trying to find the fastest method to extract a filename from a path in R I found that using sub with the regex ".*/" was ~an order of magnitude faster than basename (if speed is an issue).

files<-paste0("http://some/ppath/to/som/cool/file/",1:1000,".flac")

sub(".*/", "", files,perl = T)