How to remove the filename from wc -l output?

Question

I want to count the number of lines in a file using wc -l but only output the number, without the following space and filename. Edit from comments: The filename may have numbers in it.

Currently if I do:

wc -l /path/to/file

On a file with 176 lines, the output is:

176 /path/to/file

This is within a bash script, and the resulting number will be assigned as the value of a variable.

kos · Accepted Answer · 2016-03-03T21:28:45.290

13

I'd use:

<file wc -l

Which contarily to cat file | wc -l doesn't need to fork a shell and to run another process (and runs faster):

% time </tmp/ramdisk/file wc -l     
8000000
wc -l < /tmp/ramdisk/file  0,07s user 0,06s system 97% cpu 0,132 total
% time cat /tmp/ramdisk/file | wc -l
8000000
cat /tmp/ramdisk/file  0,01s user 0,16s system 80% cpu 0,204 total
wc -l  0,09s user 0,10s system 94% cpu 0,203 total

(/tmp/ramdisk/file was stored in a ramdisk to take I/O and caching out of the equation.)

However for small files indeed the difference is neglectable.

Yet another way would be:

wc -l file | cut -d ' ' -f 1

Which in my tests performs approximately the same as <file wc -l.

edited Mar 03 '16 at 21:28

answered Mar 03 '16 at 18:13

kos

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or you could use `wc -l < /etc/passwd` . Same thing, just different syntax :) – Sergiy Kolodyazhnyy Mar 03 '16 at 21:10
1

Cheers, I'd tried using `cut` previously, but got hung up on using `-c` which wasn't useful as the line count could be of variable length. – Arronical Mar 04 '16 at 09:47
Nice. I tried using cut as well, but clearly couldn't get the syntax right. – Elder Geek Mar 05 '16 at 03:14

Elder Geek · Answer 2 · 2016-03-03T17:21:42.003

5

You can do this with a simple one liner using cat. cat deluge1.log | wc -l

Answer too simple to require a long drawn out answer.

edited Mar 03 '16 at 17:21

answered Mar 03 '16 at 17:10

Elder Geek

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Ah, I forgot to mention that the filenames may have numbers in them! – Arronical Mar 03 '16 at 17:20
I've ended up using an intermediate variable, with bash expansion, so `line_count=$(wc -l ${file})` then `no_of_lines="${line_count/ */}` – Arronical Mar 03 '16 at 17:22
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Proving once again there are many ways to accomplish the same result. ;-) – Elder Geek Mar 03 '16 at 17:24
I definitely prefer the simple `cat` usage, to save on assigning more variables. Cheers for editing the question too, was just about to get round to that. – Arronical Mar 03 '16 at 17:28
@Arronical I'm glad it's working for you! – Elder Geek Mar 03 '16 at 17:29
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This is a [useless use of cat](http://porkmail.org/era/unix/award.html) , but it sure works :) – Sergiy Kolodyazhnyy Mar 03 '16 at 21:09

Shreeram Joshi · Answer 3 · 2022-08-01T17:14:55.580

0

another cleaner option wc -l filename | awk '{print $1}

example list of entries in a file

bash-4.1$ wc -l ../CBB_Nodes.txt | awk '{print $1}' 
398
bash-4.1$ 
without awk it would have been as usual with the file name. 

bash-4.1$ wc -l ../CBB_Nodes.txt
398 ../CBB_Nodes.txt
bash-4.1$

edited Aug 01 '22 at 17:14

answered Aug 01 '22 at 17:12

Shreeram Joshi

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score 0 · Answer 4 · answered Jul 06 '23 at 19:50

0

Another simple alternative that did the trick for me.

cat /c/dev/P2Working/.gitignore | wc -l

answered Jul 06 '23 at 19:50

Murray Foxcroft

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How to remove the filename from wc -l output?

4 Answers4