Prepend every array element with a sequential number in Bash

Question

Given an array in Bash, is there a simple and efficient way to prepend numbers for each element sequentially?

Note: Adding commas bellow just to make arrays more readable!

Example, given:

my_array=(a, b, c, d, e)

Desired as result:

my_array=(1, a, 2, b, 3, c, 4, d, 5, e)

Or getting lines from a command result, where every line would be an element, have a number prepended before each item:

readarray -t my_array < <(my_command)

If there was a way to expand the array indexes along with the elements it would work for what I need, but I didn't found anything like this.

Is you intention to have comma in array elements? but I didn't found anything like this. Really? — KamilCuk
– KamilCuk, Commented Nov 10, 2020 at 22:50
@KamilCuk No, no comma. I just added the comma to make it more visible. — user5507535
– user5507535, Commented Nov 11, 2020 at 7:33

KamilCuk · Accepted Answer · 2020-11-10 22:54:59Z

4

The obvious solution should be fast:

my_array=(a, b, c, d, e)  tmp=()
for ((i=0;i<${#my_array[@]};++i)); do
   tmp+=("$((i+1))," "${my_array[i]}")
done
my_array=("${tmp[@]}")
declare -p my_array
# would output:
# declare -a my_array=([0]="1," [1]="a," [2]="2," [3]="b," [4]="3," [5]="c," [6]="4," [7]="d," [8]="5," [9]="e")

If the , behind numbers is not relevant, you may use a neat trick with sed in your readarray command:

readarray -t my_array < <(printf "%s\n" "${my_array[@]}" | sed =)

answered Nov 10, 2020 at 22:54

KamilCuk

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1 Comment

glenn jackman Over a year ago

Or, iterate over the array indices: for i in "${!my_array[@]}"

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