Find count of files matching a pattern in a directory in linux

Question

I am new to linux. I have a directory in linux with approx 250,000 files I need to find count of number of files matching a pattern.

I tried using following command :

ls -1 20061101-20131101_kh5x7tte9n_2010_* | wc -l

I got the following error message:

-bash: /bin/ls: Argument list too long
0

Please help. Thanks in advance

Answer 1

First of all it is better not to use ls according to this article !!!

and this problem can be solved in many ways. I will list some of the most elegant ones that come to my mind.

count=$(printf '%s\n' *pattern* | wc -l) 
#or
count=$(shopt -s nullglob; files=(*pattern*); echo ${#files[@]})
#or
count=$(file *pattern* | wc -l)
#or
count=$(stat -c "%n" *pattern* | wc -l)
#or
count=$(du -a *pattern* | wc -l)
#or
count=$(echo *pattern* | wc -w)

but last one gives the wrong number when the file names contain whitespace.

Answer 2

Just do:

find . -name "pattern_*" |wc -l

Answer 3

It might be better to use find for this:

find . -name "pattern_*" -printf '.' | wc -m

In your specific case:

find . -maxdepth 1 -name "20061101-20131101_kh5x7tte9n_2010_*" -printf '.' | wc -m

find will return a list of files matching the criteria. -maxdepth 1 will make the search to be done just in the path, no subdirectories (thanks Petesh!). -printf '.' will print a dot for every match, so that names with new lines won't make wc -m break.

Then wc -m will indicate the number of characters which will match the number of files.

---

Performance comparation of two possible options:

Let's create 10 000 files with this pattern:

$ for i in {1..10000}; do touch 20061101-20131101_kh5x7tte9n_201_$i; done

And then compare the time it takes to get the result with ls -1 ... or find ...:

$ time find . -maxdepth 1 -name "20061101-20131101_kh5x7tte9n_201_*" | wc -m
10000

real    0m0.034s
user    0m0.017s
sys     0m0.021s

$ time ls -1 | grep 20061101-20131101_kh5x7tte9n_201 | wc -m
10000

real    0m0.254s
user    0m0.245s
sys     0m0.020s

find is x5 times faster! But if we use ls -1f (thanks Petesh again!), then ls is even faster than find:

$ time ls -1f | grep 20061101-20131101_kh5x7tte9n_201 | wc -m
10000

real    0m0.023s
user    0m0.020s
sys     0m0.012s

Answer 4

The MacOS / OS X command line solution

If you are attempting to do this in the command line on a Mac you will soon find out that find does not support the -printf option.

To accomplish the same result as the solution proposed by fedorqui-supports-monica try this:

find . -name "pattern_*" -exec stat -f "." {} \; | wc -l

This will find all files matching the pattern you entered, print a . for each of them in a newline, then finally count the number of lines and output that number.

To limit your search depth to the current directory, add -maxdepth 1 to the command like so:

find . -maxdepth 1 -name "196288.*" -exec stat -f "." {} \; | wc -l

Answer 5

You should generally avoid ls in scripts and in fact, performing the calculation in a shell function will avoid the "argument list too long" error because there is no exec boundary and so the ARGV_MAX limit doesn't come into play.

number_of_files () {
    if [ -e "$1" ]; then
        echo "$#"
    else
        echo 0
    fi
}

The conditional guards against the glob not being expanded at all (which is the default out of the box; in Bash, you can shopt -s nullglob to make wildcards which don't match any files get expanded into the empty string).

Try it:

number_of_files 20061101-20131101_kh5x7tte9n_2010_*

Answer 6

ls -1 | grep '20061101-20131101_kh5x7tte9n_2010_*' | wc -l

Previous answer did not included quotes around search criteria neither * wildcard.

Answer 7

you got "argument too long" because shell expands your pattern to the list of files. try:

find  -maxdepth 1 -name '20061101-20131101_kh5x7tte9n_2010_*' |wc -l

please pay attention - pattern is enclosed in quotes to prevent shell expansion

Answer 8

Try this:

ls -1 | grep 20061101-20131101_kh5x7tte9n_2010_ | wc -l