Given input file
z
b
a
f
g
a
b
...
I want to output the number of occurrences of each string, for example:
z 1
b 2
a 2
f 1
g 1
How can this be done in a bash script?
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6 Answers
You can sort the input and pass to uniq -c:
$ sort input_file | uniq -c
2 a
2 b
1 f
1 g
1 z
If you want the numbers on the right, use awk to switch them:
$ sort input_file | uniq -c | awk '{print $2, $1}'
a 2
b 2
f 1
g 1
z 1
Alternatively, do the whole thing in awk:
$ awk '
{
++count[$1]
}
END {
for (word in count) {
print word, count[word]
}
}
' input_file
f 1
g 1
z 1
a 2
b 2
Comments
cat text | sort | uniq -c
should do the job
Comments
Try:
awk '{ freq[$1]++; } END{ for( c in freq ) { print c, freq[c] } }' test.txt
Where test.txt would be your input file.
Comments
Here's a bash-only version (requires bash version 4), using an associative array.
#! /bin/bash
declare -A count
while read val ; do
count[$val]=$(( ${count[$val]} + 1 ))
done < your_intput_file # change this as needed
for key in ${!count[@]} ; do
echo $key ${count[$key]}
done
2 Comments
glenn jackman
requires bash version 4 for associative arrays
Dennis Williamson
Simpler:
(( count[$val]++ )). Also, you should almost always use -r with read. Always quote variables: for key in "${!count[@]}" and echo "$key ${count[$key]}".This might work for you:
cat -n file |
sort -k2,2 |
uniq -cf1 |
sort -k2,2n |
sed 's/^ *\([^ ]*\).*\t\(.*\)/\2 \1/'
This output the number of occurrences of each string in the order in which they appear.
Comments
You can use sort filename | uniq -c.
Have a look at the Wikipedia page on uniq.
answered Jan 28, 2012 at 10:14
Balaswamy Vaddeman
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