Remove multiple elements from array based on index - bash

Question

I would like to remove multiple elements from an array based on their indexes.
array=("a" "b" "c" "d")
indexes=(1 3)
Output should be
array=("a" "c")
I know how to remove an element from an array knowing the index of the element:
If $i is the index:
array=("${(#)array[1,$i-1]}" "${(#)array[$i+1,$#array]}")
But what if I have multiple elements to remove? If I loop over the array of indexes, once I have removed one element the other indexes won't correspond anymore to the elements to be removed. So how is it possible to do that?

Answer 1

Using BASH arrays you can do this easily:
# original array
array=("a" "b" "c" "d")
# indexes array
indexes=(1 3)
# loop through indexes array and delete from element array
for i in "${indexes[#]}"; do
unset "array[$i]"
done
# check content of original array
declare -p array
declare -a array=([0]="a" [2]="c")
As per Chepner's comments below if OP wants an array of contiguous indices then loop through differential array and populate a new array
# result array
out=()
# loop through differential array and populate result
for i in "${array[#]}"; do
out+=("$i")
done
declare -p out
declare -a out=([0]="a" [1]="c")

Answer 2

Assuming indices is sorted, keep a counter of how many items you have already removed, and subtract that from each index in your loop.
count=0
for i in $indices; do
c=$((i - count))
array=("${(#)array[1,$c-1]}" "${(#)array[$c+1,$#array]}")
count=$((count + 1))
done
In bash, the same approach looks like
count=0
for i in "${indices[#]}"; do
c=$((i - count))
array=( "${array[#]:0:c-1}" "${array[#]:c+1}" )
count=$((count + 1))
done