Remove specific number of digits using sed

Question

John, 1234567
Bob, 2839211
Alex, 2817821
Mary, 9371281

I am currently trying to retrieve the first column with the last 4 digits of the second column using sed, so the output should look like this:

John, 4567
Bob, 9211
Alex, 7821
Mary, 1281

This is my command: 's/\(.*,\)\(.*\)//', I think that this command matches the first column until the comma and the second column until the end, but I am unsure on how to continue.

Answer 1

You can use

sed 's/^\([^,]*\), *[0-9]*\([0-9]\{4\}\).*/\1, \2/' file

See the online demo.

Details

• ^ - start of string
• \([^,]*\) - Group 1: any zero or more chars other than a comma
• , * - a comma and zero or more spaces
• [0-9]* - zero or more digits
• \([0-9]\{4\}\) - Group 2: four digits
• .* - the rest of the line
• \1, \2 - The replacement is: Group 1, ,, space and Group 2 value.

Answer 2

Just capture the last four digits of each line and delete any preceding digits:

$ sed 's/[0-9]*\([0-9]\{4\}\)$/\1/' input.txt
John, 4567
Bob, 9211
Alex, 7821
Mary, 1281

If using a version of sed that supports POSIX Extended Regular Expressions, it can be cleaned up a bit to

sed -E 's/[0-9]*([0-9]{4})$/\1/' input.txt