Compare and discard elements from array in bash

Question

I have two sorted arrays list1=( a b c d) list2=( a b d)

Suppose list1 is a constant list and list2 is created during script execution. I want to compare list2 elements with list1 and discard only those values from list2 which are not there in list1.

Example: if list2=(a b d e f), then I should update list2 as list2=(a b)as (d e f) isn't there in list1.

@Inian I took help from some of the other array comparisons related questions, but couldn't able to achieve what I intended. — BMD
– BMD, Commented Mar 26, 2020 at 11:06
can the actual array contents contains spaces?. Also post whatever half efforts you made so far. — Inian
– Inian, Commented Mar 26, 2020 at 11:06

Orion · Accepted Answer · 2020-03-26 11:14:20Z

1

list1=( a b c d) list2=( a b d) list2=($(echo ${list1[*]} ${list2[*]} | tr " " "\n" |sort | uniq -d)) echo ${list2[*]}

Here I convert the 2 lists to strings, separate by white space, sort the values and then find duplicates. The duplicates are then reassigned to the list2 array

edited Mar 26, 2020 at 11:14

answered Mar 26, 2020 at 11:07

Orion

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2 Comments

BMD Over a year ago

Thanks for this. Is there any way I can find if list2 is having any elements which is not there in list1. For example, let list2=( a b e ) then it should return [e] or simple Boolean flag sayng False as [e] is not there in list1

Prasanna Over a year ago

@BiswajitMaharana, given that your original list2 and new list2 could be same or different, you should be able to determine this boolean flag yourself. Give it a try.

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