Remove multiple elements from array based on index

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?

Don't know about bash, but in other languages I'd loop backwards. — melpomene
– melpomene, Commented May 21, 2018 at 17:04
@chepner so I don't really know how to do that in pure bash :) or I've read it's complicated stackoverflow.com/questions/16860877/… — Sulli
– Sulli, Commented May 21, 2018 at 17:32

anubhava · Accepted Answer · 2018-05-21 17:59:49Z

3

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")

edited May 21, 2018 at 17:59

answered May 21, 2018 at 17:28

anubhava

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3 Comments

chepner Over a year ago

You don't actually need an associative array for indexes, since the keys are still integers.

chepner Over a year ago

This is definitely the simplest way to handle it; the only question is whether the resulting array should have contiguous indices or not.

Sulli Over a year ago

I do need contiguous indices indeed, I didn't even know indices could be non-contiguous. So it's actually a lot of code, I thought there might be an easier way but apparently not

chepner · Accepted Answer · 2018-05-21 17:35:08Z

0

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

edited May 21, 2018 at 17:35

answered May 21, 2018 at 17:30

chepner

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1 Comment

Sulli Over a year ago

c-1 is going to be negative on the first round if index=0

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