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I have a directory with a large number of files. I don't see a ls switch to provide the count. Is there some command line magic to get a count of files?

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12 Answers 12

up vote 57 down vote accepted

Using a broad definition of "file"

ls | wc -l

(note that it doesn't count hidden files and assumes that file names don't contain newline characters).

To include hidden files (except . and ..) and avoid problems with newline characters, the canonical way is:

find . ! -name . -prune -print | grep -c /

Or recursively:

find .//. ! -name . -print | grep -c //
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wc is a "word count" program. The -l switch causes it to count lines. In this case, it's counting the lines in the output from ls. This is the always the way I was taught to get a file count for a given directory, too. – Sandy Aug 24 '10 at 6:07
please add note that ls does ls -1 if the output is a pipe. – lesmana Aug 24 '10 at 16:47
that doesn't get everything in a directory - you've missed dot files, and collect a couple extra lines, too. An empty directory will still return 1 line. And if you call ls -la, you will get three lines in the directory. You want ls -lA | wc -l to skip the . and .. entries. You'll still be off-by-one, however. – warren Aug 25 '10 at 15:14
An empty directory returns 0 for me – James Roth Sep 24 '13 at 2:16
A corrected approach, that would not double count files with newlines in the name, would be this: ls -q | wc -l - though note that hidden files will still not be counted by this approach, and that directories will be counted. – godlygeek Mar 3 '15 at 22:30

For narrow (I hope) definition of file:

 find -maxdepth 1 -type f | wc -l
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And you can of course omit the -maxdepth 1 for counting files recursively (or adjust it for desired max search depth). – user7089 Aug 3 '14 at 16:25
If you have a file whose name contains a newline, this approach will incorrectly count it twice. – godlygeek Mar 3 '15 at 22:20
A corrected approach, that would not double count files with newlines in the name, would be this: find -maxdepth 1 -type f -printf "\n" | wc -l – godlygeek Mar 3 '15 at 22:27
+1 Allow for hidden files and ignores directories – Michael Durrant Mar 12 '15 at 23:56

If you know the current directory contains at least one non-hidden file:

set -- *; echo "$#"

This is obviously generalizable to any glob.

In a script, this has the sometimes unfortunate side effect of overwriting the positional parameters. You can work around that by using a subshell or with a function (Bourne/POSIX version) like:

count_words () {
  eval 'shift; '"$1"'=$#'
count_words number_of_files *
echo "There are $number_of_files non-dot files in the current directory"

An alternative solution is $(ls -d -- * | wc -l). If the glob is *, the command can be shortened to $(ls | wc -l). Parsing the output of ls always makes me uneasy, but here it should work as long as your file names don't contain newlines, or your ls escapes them. And $(ls -d -- * 2>/dev/null | wc -l) has the advantage of handling the case of a non-matching glob gracefully (i.e., it returns 0 in that case, whereas the set * method requires fiddly testing if the glob might be empty).

If file names may contain newline characters, an alternative is to use $(ls -d ./* | grep -c /).

Any of those solutions that rely on passing the expansion of a glob to ls may fail with a argument list too long error if there are a lot of matching files.

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Do you really want to create 13,923 positional parameters? And you should make your local variable local or eliminate it: eval $1=$# or just use echo $# and do number_of_files=$(count_words *). – Dennis Williamson Aug 24 '10 at 1:16
@Dennis: part of the point was to avoid forking. I guess that's not a 21st century concern. Ok, I admit I don't care about non-POSIX shells any more, so I could have avoided the temporary variable. – Gilles Aug 24 '10 at 7:14
Why did you subtract one from $# (you hadn't done that prior to the edit)? – Dennis Williamson Aug 24 '10 at 22:12
@Dennis: I'm still avoiding a fork (well, it does make a difference on machines with a slow CPU such as routers) and passing a variable name as $1. So what I want to count is the number of parameters that aren't the first parameter. (I can't use shift because I need to keep the variable name around.) (Umm, now if you'd asked about the first line...) – Gilles Aug 24 '10 at 22:42
@Dennis: come to think of it, I can use shift if I time it right. – Gilles Aug 24 '10 at 22:50
ls -1 | wc -l


$ ls --help | grep -- '  -1'
    -1                         list one file per line


$ wc --help | grep -- '  -l'
    -l, --lines            print the newline counts

PS: Note ls -<number-one> | wc -<letter-l>

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Most versions of ls do -1 automatically when the output is to a pipe. – Dennis Williamson Aug 24 '10 at 1:01
@Dennis that's interesting I didn't know that an application could tell its output was going to a pipe. – xenoterracide Aug 24 '10 at 5:35
I +'ed this version since it is more explicit. Though, yes ls does use -1 if it's piped (try it: ls | cat), I find the -1 syntax more explicit. – gabe. Aug 24 '10 at 18:13
@xenoterracide: In Bash: [[ -p /dev/stdin ]] && echo "stdin is from a pipe" – Dennis Williamson Aug 25 '10 at 22:42
In my tests it was significantly faster to also provide the -f option to avoid ls sorting the filenames. Unfortunately you still get the wrong answer if your filenames contain newlines. – Samuel Edwin Ward Jan 8 '13 at 20:42

Here's another technique along the lines of the one Gilles posted:

word_count () { local c=("$@"); echo "${#c[@]}"; }
file_count=$(word_count *)

which creates an array with 13,923 elements (if that's how many files there are).

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What's the point of that c array? word_count() { echo "$#"; } would be enough. The point of @Gilles solution is to store the count in a returned variable to avoid having to use command substitution (which involves a fork and pipe in shells other than ksh93). – Stéphane Chazelas Mar 1 at 16:27

After installing the tree command, just type:


If you want hidden files too:

tree -a

If you are using Debian / Mint / Ubuntu Linux, type the following command to install the tree command:

sudo apt-get install tree

The option -L is used for specifying the maximum display level of the directory tree. The tree command does not only count the number of files, but also the number of directories, considering as many levels of the directory tree as you like.

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No pipe, no string copy, no fork, just plain bash one liner

$ fcount() { local f i=0; for f in *; do let i++; done; echo $i; }; fcount
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Probably the most complete answer using ls/wc pair is

ls -Aq | wc -l

if you want to count dot files, and

ls -q | wc -l


  • -A is to count dot files, but omit . and ...
  • -q make ls replace nongraphic characters, specifically newline character, with ?, making output 1 line for each file

To get one-line output from ls in terminal (i.e. without piping it into wc), -1 option has to be added.

(behaviour of ls tested with coreutils 8.23)

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As you said, -1 is not needed. As to "it handles newlines in filenames sensibly with console output", this is because of the -q switch (that you should use instead of -b because it's portable) which "Forces each instance of non-printable filename characters and <tab> characters to be written as the <question-mark> ( '?' ) character. Implementations may provide this option by default if the output is to a terminal device." So e.g. ls -Aq | wc -l to count all files/dirs or ls -qp | grep -c / to count only non-hidden dirs etc... – don_crissti May 13 '15 at 11:08
Thanks for your input. Changed -b to -q. – Frax May 14 '15 at 15:05

Use the tree command, just:

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While using ls/wc pair if we are adding -U it will be much faster (do not sort ).

ls -AqU | wc -l
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-U is GNU-specific though. – Stéphane Chazelas Mar 1 at 16:25

Try this i hope this answer will help you

echo $((`ls -l | wc -l` -1 ))
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You can check with:

ls -l | grep -v ^l | wc -l
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