Bash script to find the frequency of every letter in a file

Question

I am trying to find out the frequency of appearance of every letter in the english alphabet in an input file. How can I do this in a bash script?

Answer 1

Here is a suggestion:

while read -n 1 c
do
    echo "$c"
done < "$INPUT_FILE" | grep '[[:alpha:]]' | sort | uniq -c | sort -nr

Answer 2

Just one awk command

awk -vFS="" '{for(i=1;i<=NF;i++)w[$i]++}END{for(i in w) print i,w[i]}' file

if you want case insensitive, add tolower()

awk -vFS="" '{for(i=1;i<=NF;i++)w[tolower($i)]++}END{for(i in w) print i,w[i]}' file

and if you want only characters,

awk -vFS="" '{for(i=1;i<=NF;i++){ if($i~/[a-zA-Z]/) { w[tolower($i)]++} } }END{for(i in w) print i,w[i]}' file

and if you want only digits, change /[a-zA-Z]/ to /[0-9]/

if you do not want to show unicode, do export LC_ALL=C

Answer 3

A solution with sed, sort and uniq:

sed 's/\(.\)/\1\n/g' file | sort | uniq -c

This counts all characters, not only letters. You can filter out with:

sed 's/\(.\)/\1\n/g' file | grep '[A-Za-z]' | sort | uniq -c

If you want to consider uppercase and lowercase as same, just add a translation:

sed 's/\(.\)/\1\n/g' file | tr '[:upper:]' '[:lower:]' | grep '[a-z]' | sort | uniq -c

Answer 4

My solution using grep, sort and uniq.

grep -o . file | sort | uniq -c

Ignore case:

grep -o . file | sort -f | uniq -ic