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Given the following ls result:

$ ls
Desktop  Documents  Downloads  Music  Pictures  Public  Templates  Videos

This will give the total number of characters for all files:

$ ls | wc -m

But how can I have calculate the number of characters per file name? For the same file list the result I'm after would be something like this:

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up vote 3 down vote accepted

This can be done in a very simple shell script:

for file in *; do echo -n "$file" | wc -m; done

Just loop through each file echoing the name to wc. The -n on the echo is so that it doesn't append a newline, which would erroneously increase the count by 1.

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Perfect. Thanks. Also thanks for the echo -n tip. I had noticed the wc was counting one extra character but had not yet started dwelling on that one. – Fabricio Jan 14 '14 at 1:00

You don't need to call out to wc, bash is perfectly capable: ${#var} is the length of the value of $var.

for f in *; do echo ${#f}; done


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While @Patrick's answer is perfectly fine if you have to do a similar task over a directory tree's worth of files you'll need to change your tactics slightly. One method for handling this is to use find & while.

find & while

$ depth=2
$ find . -maxdepth $depth -type f -print0 | sed 's|\./||g' | \
    while IFS= read -r -d '' file; do \
      f=$(basename "$file"); printf "%s: %s\n" "$file" "${#f}"; \
    done | column -s : -t
dir2/more files3.txt        15
some long spacey file.txt   25
dir1/more files1.txt        15
dir1/more files2.txt        15
file 1.txt                  10
file 2.txt                  10

The above will generate a list of files separated by \0 (i.e. NULLs). You can use the variable $depth to control how deep find will look. This list is then scrubbed so that any .\ characters are removed via sed.

Lastly we loop through this list and use a printf to print each file's name out along with its length, using Bash's built-in facility to count the length of a string, ${#var}. The printf will print the file + its path but only the size of the file.

The column -s : -t is just to pretty print it. It does so by splitting the output on the colon, :, and then splitting the output up into equidistant columns.

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There's no whitespace problem in Patrick's answer, not even newlines. Since the variable is quoted, there's no issue with horizontal whitespace: var=$'hello\nworld';echo -n "$var" | wc -m => 11 – glenn jackman Jan 14 '14 at 2:09
@glennjackman - what am I missing then? unix.stackexchange.com/questions/9496/…. I know I've switched back and forth on this most annoying topic. I tried his code and you're correct BTW. – slm Jan 14 '14 at 2:22
The difference is for f in * -- using the wild card avoids word splitting. The shell knows you're iterating over filenames. When you do for f in $(...) you're iterating over words in text output of some program. – glenn jackman Jan 14 '14 at 11:28
@glennjackman - thanks, that's what I kind of figured but a bit more research and re-relearning it I now see the diff. – slm Jan 14 '14 at 13:08
In more detail, I guess the wildcard is expanded into filenames, and those filenames are not subjected to a 2nd round of word-splitting expansion. – glenn jackman Jan 14 '14 at 14:27

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