Count line lengths in file using command line tools - bash

Question

Problem
If I have a long file with lots of lines of varying lengths, how can I count the occurrences of each line length?
Example:
file.txt
this
is
a
sample
file
with
several
lines
of
varying
length
Running count_line_lengths file.txt would give:
Length Occurences
1 1
2 2
4 3
5 1
6 2
7 2
Ideas?

Answer 1

This
counts the line lengths using awk, then
sorts the (numeric) line lengths using sort -n and finally
counts the unique line length values uniq -c.
$ awk '{print length}' input.txt | sort -n | uniq -c
1 1
2 2
3 4
1 5
2 6
2 7
In the output, the first column is the number of lines with the given length, and the second column is the line length.

Answer 2

Pure awk
awk '{++a[length()]} END{for (i in a) print i, a[i]}' file.txt
4 3
5 1
6 2
7 2
1 1
2 2

Answer 3

Using bash arrays:
#!/bin/bash
while read line; do
((histogram[${#line}]++))
done < file.txt
echo "Length Occurrence"
for length in "${!histogram[#]}"; do
printf "%-6s %s\n" "${length}" "${histogram[$length]}"
done
Example run:
$ ./t.sh
Length Occurrence
1 1
2 2
4 3
5 1
6 2
7 2

Answer 4

$ perl -lne '$c{length($_)}++ }{ print qq($_ $c{$_}) for (keys %c);' file.txt
Output
6 2
1 1
4 3
7 2
2 2
5 1

Answer 5

Try this:
awk '{print length}' FILENAME
Or next if you want the longest length:
awk '{ln=length} ln>max{max=ln} END {print FILENAME " " max}'
You can combine above command with find using -exec option.

Answer 6

You can accomplish this by using basic unix utilities only:
$ printf "%s %s\n" $(for line in $(cat file.txt); do printf $line | wc -c; done | sort -n | uniq -c | sed -E "s/([0-9]+)[^0-9]+([0-9]+)/\2 \1/")
1 1
2 2
4 3
5 1
6 2
7 2
How it works?
Here's the source file:
$ cat file.txt
this
is
a
sample
file
with
several
lines
of
varying
length
Replace each line of the source file with its length:
$ for line in $(cat file.txt); do printf $line | wc -c; done
4
2
1
6
4
4
7
5
2
7
6
Sort and count the number of length occurrences:
$ for line in $(cat file.txt); do printf $line | wc -c; done | sort -n | uniq -c
1 1
2 2
3 4
1 5
2 6
2 7
Swap and format the numbers:
$ printf "%s %s\n" $(for line in $(cat file.txt); do printf $line | wc -c; done | sort -n | uniq -c | sed -E "s/([0-9]+)[^0-9]+([0-9]+)/\2 \1/")
1 1
2 2
4 3
5 1
6 2
7 2

Answer 7

If you allow for the columns to be swapped and don't need the headers, something as easy as
while read line; do echo -n "$line" | wc -m; done < file | sort | uniq -c
(without any advanced tricks with sed or awk) will work. The output is:
1 1
2 2
3 4
1 5
2 6
2 7
One important thing to keep in mind: wc -c counts the bytes, not the characters, and will not give the correct length for strings containing multibyte characters. Therefore the use of wc -m.
References:
man uniq(1)
man sort(1)
man wc(1)