How to get max length of each column in unix?
awk, unix
Solution
This can be a general way to do it, so that you don't have to care about the number of fields you have: store the lengths in an array and keep checking if it is the maximum or not. Finally, loop through them and print the results.
awk -F'|' 'NR>1{for (i=1; i<=NF; i++) max[i]=(length($i)>max[i]?length($i):max[i])}
END {for (i=1; i<=NF; i++) printf "%d%s", max[i], (i==NF?RS:FS)}' file
See output:
$ awk -F'|' 'NR>1{for (i=1; i<=NF; i++) max[i]=(length($i)>max[i]?length($i):max[i])} END {for (i=1; i<=NF; i++) printf "%d%s", max[i], (i==NF?RS:FS)}' a
2|4|6
For variable number of columns, we can store the maximum amount of columns in for example `cols`:
$ awk -F'|' 'NR>1{cols=(cols<=NF?NF:cols); for (i=1; i<=NF; i++) max[i]=(length($i)>max[i]?length($i):max[i])} END {for (i=1; i<=cols; i++) printf "%d%s", max[i], (i==cols?RS:FS)}' a
2|4|6
Problem
Suppose, I have a source file like this. ``` ID|NAME|ADDRESS 1|ABC|PUNE 2|XYZA|MUMBAI 12|VB|NAGPUR ``` I want to get the maximum length of each column (excluding the header names). Output should be like this. 2|4|6 I have tried the command like this. tail +2 filename | cut -d"|" -f1 | awk '{ print length }' | sort -r | uniq This works for 1st column. Is there any option available in awk to get max length for each column?