Intersection of two arrays in BASH

Question

I have two arrays like this:

A=(vol-175a3b54 vol-382c477b vol-8c027acf vol-93d6fed0 vol-71600106 vol-79f7970e vol-e3d6a894 vol-d9d6a8ae vol-8dbbc2fa vol-98c2bbef vol-ae7ed9e3 vol-5540e618 vol-9e3bbed3 vol-993bbed4 vol-a83bbee5 vol-ff52deb2)
B=(vol-175a3b54 vol-e38d0c94 vol-2a19386a vol-b846c5cf vol-98c2bbef vol-7320102b vol-8f6226cc vol-27991850 vol-71600106 vol-615e1222)

The arrays are not sorted and might possibly even contain duplicated elements.

1. I would like to make the intersection of these two arrays and store the elements in another array. How would I do that?

2. Also, how would I get the list of elements that appear in B and are not available in A?

Solution 1

comm(1) is a tool that compares two lists and can give you the intersection or difference between two lists. The lists need to be sorted, but that's easy to achieve.

To get your arrays into a sorted list suitable for comm:

$ printf '%s\n' "${A[@]}" | LC_ALL=C sort

That will turn array A into a sorted list. Do the same for B.

To use comm to return the intersection:

$ comm -1 -2 file1 file2

-1 -2 says to remove entries unique to file1 (A) and unique to file2 (B) - the intersection of the two.

To have it return what is in file2 (B) but not file1 (A):

$ comm -1 -3 file1 file2

-1 -3 says to remove entries unique to file1 and common to both - leaving only those unique to file2.

To feed two pipelines into comm, use the "Process Substitution" feature of bash:

$ comm -1 -2 <(pipeline1) <(pipeline2)

To capture this in an array:

$ C=($(command))

Putting it all together:

# 1. Intersection
$ C=($(comm -12 <(printf '%s\n' "${A[@]}" | LC_ALL=C sort) <(printf '%s\n' "${B[@]}" | LC_ALL=C sort)))

# 2. B - A
$ D=($(comm -13 <(printf '%s\n' "${A[@]}" | LC_ALL=C sort) <(printf '%s\n' "${B[@]}" | LC_ALL=C sort)))

Solution 2

There is rather elegant and efficient approach to do that, using uniq — but, we will need to eliminate duplicates from each array, leaving only unique items. If you want to save duplicates, there is only one way "by looping through both arrays and comparing".

Consider we have two arrays:

A=(vol-175a3b54 vol-382c477b vol-8c027acf vol-93d6fed0 vol-71600106 vol-79f7970e vol-e3d6a894 vol-d9d6a8ae vol-8dbbc2fa vol-98c2bbef vol-ae7ed9e3 vol-5540e618 vol-9e3bbed3 vol-993bbed4 vol-a83bbee5 vol-ff52deb2)
B=(vol-175a3b54 vol-e38d0c94 vol-2a19386a vol-b846c5cf vol-98c2bbef vol-7320102b vol-8f6226cc vol-27991850 vol-71600106 vol-615e1222)

First of all, lets transform these arrays into sets. We will do it because there is mathematical operation intersection which is known like intersection of sets, and set is a collection of distinct objects, distinct or unique. To be honest, I don't know what is "intersection" if we speak about lists or sequences. Though we can pick out a subsequence from sequence, but this operation (selection) has slightly different meaning.

So, lets transform!

$ A=(echo ${A[@]} | sed 's/ /\n/g' | sort | uniq)
$ B=(echo ${B[@]} | sed 's/ /\n/g' | sort | uniq)

1. Intersection:

   $ echo ${A[@]} ${B[@]} | sed 's/ /\n/g' | sort | uniq -d
   

   If you want to store the elements in another array:

   $ intersection_set=$(echo ${A[@]} ${B[@]} | sed 's/ /\n/g' | sort | uniq -d)
   
   $ echo $intersection_set
   vol-175a3b54 vol-71600106 vol-98c2bbef
   

   uniq -d means show only duplicates (I think, uniq is rather fast because of its realisation: I guess that it is done with XOR operation).

2. Get the list of elements that appear in B and are not available in A, i.e. B\A

   $ echo ${A[@]} ${B[@]} | sed 's/ /\n/g' | sort | uniq -d | xargs echo ${B[@]} | sed 's/ /\n/g' | sort | uniq -u
   

   Or, with saving in a variable:

   $ subtraction_set=$(echo ${A[@]} ${B[@]} | sed 's/ /\n/g' | sort | uniq -d | xargs echo ${B[@]} | sed 's/ /\n/g' | sort | uniq -u)
   
   $ echo $subtraction_set
   vol-27991850 vol-2a19386a vol-615e1222 vol-7320102b vol-8f6226cc vol-b846c5cf vol-e38d0c94
   

   Thus, at first we have got intersection of A and B (which is simply the set of duplicates between them), say it is A/\B, and then we used operation of inverting intersection of B and A/\B (which is simply only unique elements), so we get B\A = ! (B /\ (A/\B)).

P.S. uniq was written by Richard M. Stallman and David MacKenzie.

Solution 3

You can get all elements that are in both A and B by looping through both arrays and comparing:

A=(vol-175a3b54 vol-382c477b vol-8c027acf vol-93d6fed0 vol-71600106 vol-79f7970e vol-e3d6a894 vol-d9d6a8ae vol-8dbbc2fa vol-98c2bbef vol-ae7ed9e3 vol-5540e618 vol-9e3bbed3 vol-993bbed4 vol-a83bbee5 vol-ff52deb2)
B=(vol-175a3b54 vol-e38d0c94 vol-2a19386a vol-b846c5cf vol-98c2bbef vol-7320102b vol-8f6226cc vol-27991850 vol-71600106 vol-615e1222)

intersections=()

for item1 in "${A[@]}"; do
    for item2 in "${B[@]}"; do
        if [[ $item1 == "$item2" ]]; then
            intersections+=( "$item1" )
            break
        fi
    done
done

printf '%s\n' "${intersections[@]}"

You can get all elements in B but not in A in a similar manner:

A=(vol-175a3b54 vol-382c477b vol-8c027acf vol-93d6fed0 vol-71600106 vol-79f7970e vol-e3d6a894 vol-d9d6a8ae vol-8dbbc2fa vol-98c2bbef vol-ae7ed9e3 vol-5540e618 vol-9e3bbed3 vol-993bbed4 vol-a83bbee5 vol-ff52deb2)
B=(vol-175a3b54 vol-e38d0c94 vol-2a19386a vol-b846c5cf vol-98c2bbef vol-7320102b vol-8f6226cc vol-27991850 vol-71600106 vol-615e1222)

not_in_a=()

for item1 in "${B[@]}"; do
    for item2 in "${A[@]}"; do
        [[ $item1 == "$item2" ]] && continue 2
    done

    # If we reached here, nothing matched.
    not_in_a+=( "$item1" )
done

printf '%s\n' "${not_in_a[@]}"

Solution 4

Ignoring efficiency, here is an approach:

declare -a intersect
declare -a b_only
for bvol in "${B[@]}"
do
    in_both=""
    for avol in "${A[@]}"
    do
        [ "$bvol" = "$avol" ] && in_both=Yes
    done
    if [ "$in_both" ]
    then
        intersect+=("$bvol")
    else
        b_only+=("$bvol")
    fi
done
echo "intersection=${intersect[*]}"
echo "In B only=${b_only[@]}"

View more solutions

Comments

[F. Hauri]

Of course, if Duplicate lines are useless, they could simply be dropped.

[clerksx]

This will only work if your values don't contain \n.

[Gilles 'SO- stop being evil']

Exercise: if you interchange A and B, is intersections always the same up to reordering?

[camh]

@ChrisDown: That's right. I always try to write shell scripts that are properly quoted and handle all chars, but I've given up on \n. I have NEVER seen it in a filename, and a large bunch of unix tools work with \n delimited records that you lose a lot if you try to handle \n as a valid char.

[clerksx]

I've seen it in filenames when using GUI file managers that do not properly sanitise input filenames that are copied from somewhere else (also, nobody said anything about filenames).

[clerksx]

@Gilles If the arrays may contain duplicate elements, no.

[Jason R. Mick]

To protect \n try this: arr1=( one two three "four five\nsix\nseven" ); arr2=( ${arr1[@]:1} "four five\\nsix" ); n1=${#arr1[@]}; n2=${#arr2[@]}; arr=( ${arr1[@]/ /'-_-'} ${arr2[@]/ /'-_-'} ); arr=( $( echo "${arr[@]}"|tr '\t' '-t-'|tr '\n' '-n-'|tr '\r' '-r-' ) ); arr1=( ${arr[@]:0:${n1}} ); arr2=( ${arr[@]:${n1}:${n2}} ); unset arr; printf "%0.s-" {1..10}; printf '\n'; printf '{'; printf " \"%s\" " "${arr1[@]}"; printf '}\n'; printf "%0.s-" {1..10}; printf '\n'; printf '{'; printf " \"%s\" " "${arr2[@]}"; printf '}\n'; printf "%0.s-" {1..10}; printf '\n\n'; unset arr1; unset arr2

[Sorpigal]

One should not set LC_ALL=C. Instead set LC_COLLATE=C for the same performance gain without other side effects. In order to obtain correct results you will also need to set the same collation for comm that was used for sort, e.g.: unset LC_ALL; LC_COLLATE=C ; comm -12 <(printf '%s\n' "${A[@]}" | sort) <(printf '%s\n' "${B[@]}" | sort)

[Matt Alexander]

If you have spaces in the names, you'll be in trouble doing echo ${A[@]} | sed 's/ /\n/g' | sort | uniq. Better to do IFS=$'\n'; printf %s "${A[@]}" | sort | uniq.

[Matt Alexander]

By the way, this is a really slick solution, using uniq -d.