CODEWITHSUNDEEP
bashgrep
Ashley
Ashley

Reputation: 45

Find number of vowels in a word

How can a find all the vowels for a word in bash?

grep -o "a"  <<<$1 | wc l

This command finds only a, and I want to find aeiou.

For example:

Upvotes: 0

Views: 2485

Answers (3)

RomanPerekhrest
RomanPerekhrest

Reputation: 92854

bash approach with GNU expr command:

Input variables:

v1="computer"
v2="car"

expr length "${v1//[^aeuoi]}"
3

expr length "${v2//[^aeuoi]}"
1

  • ${v2//[^aeuoi]} - replacing/removing all non-vowel characters

  • expr length STRING - evaluates length of STRING

A more compatible variation would look like:

vowels="${v1//[^aeuoi]}"
echo "${#vowels}"
3

Upvotes: 1

RavinderSingh13
RavinderSingh13

Reputation: 133600

try:

echo "computer" | awk '{print gsub(/[aeiou]/,"")}'

So I am using echo here to print the word, sending its standard output to awk by pipe (|) as standard input, then performing global substitution of letters a,e,i, o,u with the empty string ("").
Since gsub returns the count of substitutions performed, it tells us how many vowel letters are present in the string.

Upvotes: 2

Benjamin W.
Benjamin W.

Reputation: 52291

You can combine -o (only retain matches) with wc -l and use a bracket expression to match all vowels:

$ grep -o '[aeiou]' <<< car | wc -l
1
$ grep -o '[aeiou]' <<< computer | wc -l
3

As a function (notice quotes around $1 to prevent word splitting and pathname expansion):

vowcount () { grep -o '[aeiou]' <<< "$1" | wc -l; }

Used like

$ vowcount alphabet
3

Upvotes: 3

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