Finding the maximum of the values in a file

I am aware this is not perl:

awk '{for(i=1;i<=NF;i++) if($i>maxval) maxval=$i;}; END { print maxval;}' file

Other answers are quite nice and Perl / awk is the way to go.

Just for fun, you can also use this:

$ grep -o '[0-9]*' file | sort -rn | head -1
63

Explanation

• grep -o '[0-9]*' file prints all matches of numbers in the file. Each match will be printed in a different line, as per the -o flag.
• sort -rn sorts the list numerically and in reverse, so that the first number is the biggest.
• head -1 prints the first line.

By steps:

$ grep -o '[0-9]*' file
21
12
33
35
21
12
33
44
52
63
14
12
23
34
11
12
13
53
1
12
43
33
44

$ grep -o '[0-9]*' file | sort -rn 
63
53
52
44
44
43
35
34
33
33
33
23
21
21
14
13
12
12
12
12
12
11
1

Another perl solution:

$ perl -MList::Util=max -anle 'print max(@F)' file 
63

This will work with file contains one line, if you have muliple line in file, and want to find the maximum value over all lines, try:

$ perl -MList::Util=max -alne '$tmp = max @F; $max = $tmp if $max < $tmp; END { print $max }'

This works in Ubuntu, but didn't on MacOS:

echo "21 12 33 35 21 12 33 44 52 63 14 12 23 34 11 12 13 53 1 12 43 33 44" | grep -oE "[0-9]*" | sort -nr | head -n 1

First grep for patterns matching a number (add ., if you have decimals), then sort numeric in reverse order and then pick the first result.

Here's a faster Perl approach that does not require sorting:

$ echo '21 12 33 35 21 12 33 44 52 63 14 12 23 34 11 12 13 53 1 12 43 33 44' |
    perl -lane 'map{$m=$_ if $_>$m}@F; print $m'

or, as @Gnouc suggested in the comments:

$ echo '21 12 33 35 21 12 33 44 52 63 14 12 23 34 11 12 13 53 1 12 43 33 44' |
    perl -lane '$m<$_ and $m=$_ for @F; print $m'

No need for awk, grep, or perl.

{ 
    for n in 21 12 33 35 21 12 33 44 52 63 14 12 23 34 11 12 13 53 1 12 43 33 44
    do 
        echo $n
    done 
} | sort | tail -n1