Bash: How do I extract the count of all the "n" digit numbers in a string?

Question

I am trying to extract the total number of n digited numbers from a string using bash.

E.g. For a 3 digit number,

I am trying to extract 3 digited numbers 333, 334, 335 from this string #should return 3
I have 243 pens for sale #should return 1

Unfortunately, I will not be able to use sed or grep with perl-regexp.

Appreciate any suggestions!

Answer 1

You can use regular expressions in bash.

#! /bin/bash
cat <<EOF |
I am trying to extract 3 digited numbers 333, 334, 335 from this string #should return 3, but should ignore 12345
I have 243 pens for sale #should return 1
123 should work at text boundaries, too 123
EOF
while read line ; do
    c=0
    while [[ $line =~ ([^[:digit:]]|^)([0-9][0-9][0-9])([^[:digit:]]|$) ]] ; do
        m=${BASH_REMATCH[0]}
        line=${line#*$m}
        ((++c))
    done
    echo $c
done

The regex explained:

([^[:digit:]]|^)([0-9][0-9][0-9])([^[:digit:]]|$)
~~~~~~~~~~~~~                                     non-digit
             ~~                                   or the very beginning
                 ~~~~~~~~~~~~~~~                  three digits
                                  ~~~~~~~~~~~~    non-digit
                                              ~~  or the very end

As bash can't match the same string several times, we need to remove the already processed part from the string before trying another match.

Answer 2

echo "$str" | grep -o '\b[0-9]\{3\}\b' | wc -l

This way we match 3-digit numbers inside word bountaries, which are allowed to be re-used (e.g. if two numbers are separated by one char that is a bountary, like comma or space).

Or like this:

echo "$str" | grep -o '\<[[:digit:]]\{3\}\>' | wc -l

Answer 3

Using POSIX shell grammar only:

#!/usr/bin/env sh

# Should return 3
str1='I am trying to extract 3 digited numbers 333, 334, 335 from this string'

# Should return 1
str2='I have 243 pens for sale'

# should return 2
str3='This is 123 456'

_OIFS=$IFS
IFS=$IFS' ,.:;!?-_+=*#$§^&{}[]|`@"()\\/'\'

for str in "$str1" "$str2" "$str3"
do
  count=0
  for word in $str
  do
    case $word in
      [[:digit:]][[:digit:]][[:digit:]])
        count=$((count + 1 ))
        ;;
    esac
  done
  printf 'String:\n%s\n-> Count: %d\n\n' "$str" "$count"
done

IFS=$_OIFS

Output:

String:
I am trying to extract 3 digited numbers 333, 334, 335 from this string
-> Count: 3

String:
I have 243 pens for sale
-> Count: 1

String:
This is 123 456
-> Count: 2

Answer 4

Assuming the OP only wants exactly 3-digit numbers and is not interested in breaking longer numbers down into 3-digit segments, eg, the string 12345 will return a zero count as opposed to a 3 count ( 123 / 234 / 345 ).

---

Some sample data:

$ cat numbers.dat
I am trying to extract 3 digited numbers 333, 334, 335 from this string #should return 3
I have 243 pens for sale #should return 1
123 xyz
def 456
def 789-345 abc                    # should match 7-8-9 and 3-4-5
tester876tester                    # should match 8-7-6
testing9999testing                 # should not match 9-9-9-9

$ str=$(cat numbers.dat)           # load data into a variable

A 2-pass grep solution:

NOTE: borrowed thanasisp's word boundary flag (\b)

Find patterns of 3-digits with non-digit book ends (including front/end of line)

$ grep -Eo '(^|[^0-9]|\b)[0-9]{3}(\b|[^0-9]|$)' numbers.dat
# or
$ grep -Eo '(^|[^0-9]|\b)[0-9]{3}(\b|[^0-9]|$)' <<< "${str}"

 333,
 334,
 335
 243
123
 456
 789-
345
r876t

Now strip off the non-digits:

$ grep -Eo '(^|[^0-9]|\b)[0-9]{3}(\b|[^0-9]|$)' numbers.dat | grep -Eo '[0-9]{3}'
# or
$ grep -Eo '(^|[^0-9]|\b)[0-9]{3}(\b|[^0-9]|$)' <<< "${str}" | grep -Eo '[0-9]{3}'

333
334
335
243
123
456
789
345
876

Pipe to wc -l for a count:

$ grep -Eo '(^|[^0-9]|\b)[0-9]{3}(\b|[^0-9]|$)' numbers.dat | grep -Eo '[0-9]{3}' | wc -l
# or
$ grep -Eo '(^|[^0-9]|\b)[0-9]{3}(\b|[^0-9]|$)' <<< "${str}" | grep -Eo '[0-9]{3}' | wc -l

9

Storing count in a variable:

$ counter=$(grep -Eo '(^|[^0-9]|\b)[0-9]{3}(\b|[^0-9]|$)' numbers.dat | grep -Eo '[0-9]{3}' | wc -l)
# or
$ counter=$(grep -Eo '(^|[^0-9]|\b)[0-9]{3}(\b|[^0-9]|$)' <<< "${str}" | grep -Eo '[0-9]{3}' | wc -l)

$ echo "${counter}"

9

Answer 5

Could you please try following, written and tested with following link https://ideone.com/bh6zjR#stdin in shown samples. Since OP said in comments digits can't have anything else before/in between/after (apart from , I believe as per samples) so going with traversing all fields of current line and using regex to find matches for them.

awk '
{
  for(i=1;i<=NF;i++){
    if(match($i,/^[0-9]{3}[,]?$/)){
       count++
    }
  }
  print "Line " FNR " has " count " number of 3 digits."
  count=""
}
' Input_file

Output will be as follows.

Line 1 has 3 number of 3 digits.
Line 2 has 1 number of 3 digits.