Bash comparing array value to find max

Question

I'm trying to find the max in an array of integers in bash. I'm pretty new to bash. Here is what I have so far...

    max="${array[0]}"
    for ((i=0;i<${#array[@]};i++))
    do  
        if [ ${array[$i]} > $max ]
        then
            max="${array[$i]}"

        fi
    done

where the array is around 500 positive integers ex. 24 27 13 34 2 104 645 411 1042 38 5 24 120 236 2 33 6. Currently it is always returning the last integer in my array. Seems like it should be an easy fix, but I'm not sure what I'm missing. Thanks for any help.

glenn jackman · Accepted Answer · 2014-03-01 20:54:15Z

6

This test [ ${array[$i]} > $max ] is performing a lexical comparison, so 99 is greater than 100

You want one of these instead:

[[ ${array[$i]} -gt $max ]]   # numeric comparison operator
(( ${array[$i]} > $max ))     # arithmetic evaluation

Or, use standard tools which will probably be faster despite having to spawn a couple of extra processes:

max=$( printf "%d\n" "${array[@]}" | sort -n | tail -1 )

answered Mar 1, 2014 at 20:54

glenn jackman

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1 Comment

chepner Over a year ago

Sorting is overkill for finding the max, as it is an O(n lg n) operation compared to O(n) to find the max with a loop. Depending on the size of the list, though, you may not notice a difference, or the difference won't be worth worrying about.

chepner · Accepted Answer · 2014-03-01 22:07:12Z

1

Rather than iterating over the index, iterate over the items themselves. More specific to your actual problem, make sure you are doing an arithmetic comparison, not a string comparison.

max="${array[0]}"
for i in "${array[@]}"; do
    (( i > $max )) && max=$i
done

answered Mar 1, 2014 at 22:07

chepner

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