Bash array question

krock923 · 06-21-2006, 09:37 PM

Is there an easy way to get the most frequent item in an array? i.e. an array contains the following: 1,3,3,4,3,6,3 I want it to return 3.

Is there a better method than an array for doing this?

krock923 · 06-21-2006, 10:02 PM

Eh, I have a feeling I'll end up just using perl. Thanks anyway.

elluva · 06-22-2006, 03:25 AM

For these types of things, I always write a small utility in C, it's way faster and easier to maintain.

demon_vox · 06-22-2006, 11:39 AM

Hi,
I think it can be done in Bash (bah, I know because I just did it jajaja). So here I present a solution:

Code:

function moreFrequent {

  nums=$*

  # Builds an array using your numbers as an index
  # and the frequency as its value.
  # In Bash it doesnt matter if the index are not contiguos
  for NUM in ${nums[@]}; do

    freq[$NUM]=`expr ${freq[$NUM]} + 1 `;

  done


  # The for writes a list with the number and its frequency
  # Then it is sorted by the frequency (with the sort command),
  # is filtered the first line online (with the head command),
  # and finaly just cuts the frequency part (with the cut command).
  for NUM in echo ${nums[@]}; do

    echo "$NUM ${freq[$NUM]}"

  done | sort -n -k 2 -r | head -n 1 | cut -d ' ' -f 1

}


numbers=(1 3 3 4 3 6 3 6 4 4)

yourCoolNumber=`moreFrequent ${numbers[@]}`

echo "The more frequent numbers is $yourCoolNumber"

Perhaps this is not the best solution, but it's a cool function which may be useful for you.

Hope it is!

Greetings.