Print every word and its number of occurrences, using pure `bash`

Question

I have below given code. I want to print each word and its number of occurrences without using external utils such as wc, awk, tr, etc.

I am able to count the total number of words, but here I have also one issue: in the output I am not getting total word count, the output is less than what it should be.

What should I do?

#!/bin/bash
#v=1

echo -n "ENTER FILE NAME: "
read file
IFS=$'\n'
cnew_line=`echo -e "\n"`
cspace=`echo  " "`

if [ $# -ne 0 ] 
then

echo "You didn't entered a filename as a parameter"
exit

elif [ $# -eq 0 ] 
then
filename="$file"

num_line=0
num_word=0
num_char=0

while read -n1  w
do
if [ "$w" = "$cnew_line" ]
then
(( num_line++ ))
elif [ "$w" = "$cspace" ]
then

(( num_word++ ))

else
(( num_char++ ))
fi
done < "$filename"


echo "Line Number = $num_line"
echo "Word Number = $num_word"
echo "Character Number =$num_char"

fi

    enter code here

Doing this in pure Bash is an extremely inefficient and clunky use of tools. Can you explain why you want to do this in an environment which isn't a particularly good fit for the task? Also, your code lacks indentation and triggers a number of warnings on http://shellcheck.net/. You should get these things into shape before asking for help here. — tripleee, Dec 19 '17 at 05:54
Assinging a newline to `IFS` with `IFS=$'\n'` and then doing it again the wrong way on the next line suggests you don't really understand your own code. Hint: `cnew_line` does not end up containing a newline. — tripleee, Dec 19 '17 at 05:55

score 1 · Accepted Answer · answered Dec 19 '17 at 06:30

You could use an associative array for counting the words, a bit like this:

$ cat foo.sh
#!/bin/bash                                                                     

declare -A words

while read line
do
    for word in $line
    do
        ((words[$word]++))
    done
done

for i in "${!words[@]}"
do
    echo "$i:" "${words[$i]}"
done

Testing it:

$ echo this is a test is this | bash foo.sh
is: 2
this: 2
a: 1
test: 1

This answer was constructed pretty much from these fine answers: this and this. Don't forget to upvote them.

That code considers punctuation part of a word, and it can't handle apostrophes. Example: `echo "Seward's folly" | bash foo.sh` returns "*bad array subscript*". — agc, Dec 19 '17 at 15:27

agc · Answer 2 · 2017-12-19T16:45:38.683

Two improved versions of James Brown's answer, (which considers punctuation part of a word, and breaks on groups of double and single quotes):

Punctuation considered part of word:

#!/bin/bash
declare -A words

while read line ; do
    for word in ${line} ; do
        ((words[${word@Q}]++))
done ; done

for i in ${!words[@]} ; do
    echo ${i}: ${words[$i]}
done

Punctuation not part of word, (like wc):

#!/bin/bash
declare -A words

while read line ; do
    line="${line//[[:punct:]]}"
    for word in ${line} ;do 
        ((words[${word}]++))
done ; done

for i in ${!words[@]} ;do
    echo ${i}: ${words[$i]}
done

Tested code, with tricky quoted text:

fortune -m "swear" | bash foo.sh
man bash | ./foo.sh | sort -gr -k2 | head

Print every word and its number of occurrences, using pure `bash`

2 Answers2