Use R fs::dir_ls to match the beginning of file name?

user_6736902 0
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I'm trying to use fs::dir_ls() to return the same results as the list.files() example below. Ultimately, I'm just trying to return files that start with a specific pattern.

path <- "./path/to/files"
pattern <- "^ABC_.*\\.csv$"

# list files returns the expected output 
list.files(path = path, pattern = pattern, full.names = T)

# [1] "path/to/files/ABC_1312.csv" 
# [2] "path/to/files/ABC_ACAB.csv"

# dir_ls does not return any matching files
fs::dir_ls(path = path, regexp = pattern)

# character(0)

I think the issue here is that the scope of each method's pattern argument differs. The list.files() pattern is only applied to the basename() of the file path, whereas, the dir_ls() regexp argument is being applied to the full path. As a result, the ^ regex is being applied to the start of the path, instead of the beginning of each file. Is there a way to limit the scope of dir_ls() to only match patterns on the basename() of each file similar to list.files()? Any other insights are appreciated!

fs r regex
1 Answers

See this issue on GitHub:

you need to modify your regular expression to match the full path then, or use a filtering function that only looks at the basename.

Use

pattern <- paste0(.Platform$file.sep, "ABC_.*\\.csv$)

You can also do something like

regexp = fs::path(path, pattern)
user_113298900
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