declare in array and print the values randomly from array in bash script

Question

Here is my bash script code

declare -a types=("m4.xlarge" "m5.12xlarge" "m5d.12xlarge" "m4.large" "m4.16xlarge" "t2.2xlarge" "c4.large" "c5.xlarge" "r4.2xlarge" "x1e.4xlarge" "h1.16xlarge" "i3.16xlarge" );
echo "array declared"


for i in {1..100}
do

for (( i=1; i<${arraylength}+1; i++ ))   
 do

#index=$( jot -r 1  0 $((${#expressions[@]} - 1)) )
    randominstancetype=$[$RANDOM % ${#types[@]}];

    #randominstancetype=$( shuf -i0-1 -n1 $((${#types[@]} )) );
    #randominstancepvtiptype=$[$RANDOM % ${#pvtip[@]}];
        #randominstancepubiptype=$[$RANDOM % ${#pubip[@]}];
done
 done

I am trying to declare array and then print the elements inside the array randomly for around 100 times. Currently the name of the elements are not getting displayed instead it displays as 3 5 8 etc.. Anyhelp will be appreciated.

Judging from your previous questions that are all unaccepted... If there is a satisfying answer that solves your problem, take a look at: What should I do when someone answers my question? Note that this will give reputation not only to the answerer but also to you. — PesaThe
– PesaThe, Commented Jul 14, 2018 at 14:06

PesaThe · Accepted Answer · 2018-07-10 13:27:48Z

1

$[...] is the old and deprecated version of $((...)). So what you are doing is just simple arithmetic expansion that expands back to the random index.

To access an element of the array with the generated index, use:

echo "${types[$RANDOM%${#types[@]}]}"

answered Jul 10, 2018 at 13:27

PesaThe

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Comments

hnicke · Accepted Answer · 2018-07-10 13:30:30Z

0

Try this snippet:

#!/bin/bash  
declare -a types=("m4.xlarge" "m5.12xlarge" "m5d.12xlarge" "m4.large" "m4.16xlarge" "t2.2xlarge" "c4.large" "c5.xlarge" "r4.2xlarge" "x1e.4xlarge" "h1.16xlarge" "i3.16xlarge" )
echo "array declared"

max_random=32767
type_count=${#types[@]}
factor=$(( max_random / type_count ))

for i in {1..1000}
do
    random_index=$(( $RANDOM / $factor ))
    random_instance_type=${types[$random_index]}
    echo $random_instance_type
done

answered Jul 10, 2018 at 13:30

hnicke

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Comments

kvantour · Accepted Answer · 2018-07-10 13:43:03Z

0

This will print a randomized order of your array types.

for j in {1..100}; do
  for i in $(shuf -i 0-$((${#types[*]}-1))); do
     printf "%s " "${types[i]}";
  done;
  printf "\n";
done

If you would allow repetitions, then you can do

for j in {1..100}; do
  for i in $(shuf -n ${#types[*]} -r -i 0-$((${#types[*]}-1))); do
     printf "%s " "${types[i]}";
  done;
  printf "\n";
done

The commands make use of shuf and its options :

-n, --head-count=COUNT: output at most COUNT lines

-i, --input-range=LO-HI: treat each number LO through HI as an input line

-r, --repeat: output lines can be repeated

_{source man shuf}

edited Jul 10, 2018 at 13:43

answered Jul 10, 2018 at 13:37

kvantour

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