Check number of words in a file that contain a specific letter

Question

Bash command to check number of words in a file that contain letter “a”

Answer 1

Suppose that we have this test file:

$ cat file
the cat in the hat
the quick brown dog
jack splat

With grep implementations that have adopted GNU's -o extension, we can retrieve all the words containing a:

$ grep -wo '[[:alnum:]]*a[[:alnum:]]*' file
cat
hat
jack
splat

We can count those words:

$ grep -wo '[[:alnum:]]*a[[:alnum:]]*' file | wc -l
4

Answer 2

POSIXly:

<file tr -s '[:space:]' '[\n*]' | grep -c a

Here, words are sequences of non-spacing characters.

Answer 3

Here's a Perl way:

 perl -0lnE 'say scalar grep(/a/,split(/\s/,$_));' file

And an awk way:

 awk '{for(i=1;i<=NF;i++){if($(i)~/a/){k++}}}END{print k}' file

Answer 4

With GNU grep:

 grep -oP '\b\w*a+\w*\b' file

Answer 5

awk 'BEGIN{RS="[[:space:][:punct:]]"; c=0}
     index($0,"a"){c++} 
     END{print c}'

Using a version of awk that supports multi-character Record Separator (RS), eg. GNU awk, you can cause awk to read one word per record.

Within that record, the index(in, string) function searches in for the first occurrence of string, and returns the 1-based character position of where it is found. If not found, index() returns 0. Thus the return value can be treated as a boolean condition test (0 = false, not zero = true). Note, this is not a regular expression search.

If a match is found, the variable c is incremented by 1 (c++)

The c=0 is required in the BEGIN{} block, for when c is never incremented - which would cause c to be null instead of 0. Another way to handle this issue would be to print 0+c (in the END{} block)