CODEWITHSUNDEEP
rdirectoryfile-manipulation
timothyjgraham
timothyjgraham

Reputation: 1212

List directories that begin with a pattern

I have a directory with thousands of sub-directories.

Each subdirectory begins with a URL name, for example: /australia.gov.au_about-australia

I want to get a list of all the sub-directories that begin with a certain string, e.g. "australia.gov.au".

It appears that the list.dirs function does not allow for pattern matching?

I have tried the following, to no avail:

testSite <- "australia.gov.au"
list.files(paste0("main-directory/",paste0("^[",testSite,"]")), 
    full.names = TRUE, recursive=TRUE, ignore.case = TRUE)`

Upvotes: 1

Views: 552

Answers (2)

Ben Bolker
Ben Bolker

Reputation: 226162

Following @MrFlick's answer for the first part, but simplifying the rest slightly:

 re <- paste0("^", gsub(".", "\\.", testSite, fixed=TRUE))
 grep(re,list.dirs(),value=TRUE)

Upvotes: 1

MrFlick
MrFlick

Reputation: 206197

You can use Filter to filter your directory list

testSite <- "australia.gov.au"
Filter(function(x) grepl(paste0("^", gsub(".", "\\.", testSite, fixed=TRUE)), x),
    list.dirs())

We do some extra work to convert your URL to a regular expression to do the matching.

Upvotes: 3

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