CODEWITHSUNDEEP
shellsedawk
broun
broun

Reputation: 2593

awk script: find and print max value and file name containing max value

I am trying to find a file with in a dir that contains the largest number (at position 3rd row 3rd column). I want both the max value and the file name containing max value printed. This is what i have right now

find ./ -name sample-file\* -exec sed '3!d' {} \; | awk '{print $3}' | awk 'n<$1{n=$1}END{print n}'

This gets me the max value, but i also want the file name containing the max value. Print along with this.

Current output:

When run for dir1:

487987

When run for dir2:

9879889

I want the output to be like this

when run for dir1:

file16    487987

when run for dir2:

file23    9879889

Appreciate some inputs on this. Thanks

Upvotes: 0

Views: 602

Answers (1)

Etan Reisner
Etan Reisner

Reputation: 80931

awk script:

BEGIN {
    n = 0
    fn = ""
}

(FNR == 3) && ($3 > n) {
    n = $3
    fn = FILENAME
}

END {
    printf "%s: %s\n", fn, n
}

use as awk -f <file.awk> sample-file*.

Could probably be more efficient with nextfile after the fn assignment in the FNR block or similar mechanisms to short-circuit the rest of the other lines in each input file.

zcat and shell

declare n=0 fn=
for i in sample-file*; do
    t=$(zcat "$i" | awk 'NR == 3 {print $3; exit}')
    if [ $t -gt $n ]; then
        n=$t
        fn=$i
    fi
done
echo "$fn: $n"

Upvotes: 3

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