Array intersection in bash [duplicate]

Answer 1

Taking @Raihan's answer and making it work with non-files (though FDs are created) I know it's a bit of a cheat but seemed like good alternative

Side effect is that the output array will be lexicographically sorted, hope thats okay (also don't kno what type of data you have, so I just tested with numbers, there may be additional work needed if you have strings with special chars etc)

result=($(comm -12 <(for X in "${array1[@]}"; do echo "${X}"; done|sort)  <(for X in "${array2[@]}"; do echo "${X}"; done|sort)))

Testing:

$ array1=(1 17 33 99 109)
$ array2=(1 2 17 31 98 109)

result=($(comm -12 <(for X in "${array1[@]}"; do echo "${X}"; done|sort)  <(for X in "${array2[@]}"; do echo "${X}"; done|sort)))

$ echo ${result[@]}
1 109 17

p.s. I'm sure there was a way to get the array to out one value per line w/o the for loop, I just forget it (IFS?)

Answer 2

The elements of list 1 are used as regular expression looked up in list2 (expressed as string: ${list2[*]} ):

list1=( 1 2 3 4   6 7 8 9 10 11 12)
list2=( 1 2 3   5 6   8 9    11 )

l2=" ${list2[*]} "                    # add framing blanks
for item in ${list1[@]}; do
  if [[ $l2 =~ " $item " ]] ; then    # use $item as regexp
    result+=($item)
  fi
done
echo  ${result[@]}

The result is

1 2 3 6 8 9 11

Answer 3

If it was two files (instead of arrays) you were looking for intersecting lines, you could use the comm command.

$ comm -12 file1 file2

Answer 4

Now that I understand what you mean by "array", I think -- first of all -- that you should consider using actual Bash arrays. They're much more flexible, in that (for example) array elements can contain whitespace, and you can avoid the risk that * and ? will trigger filename expansion.

But if you prefer to use your existing approach of whitespace-delimited strings, then I agree with RHT's suggestion to use Perl:

result=$(perl -e 'my %array2 = map +($_ => 1), split /\s+/, $ARGV[1];
                  print join " ", grep $array2{$_}, split /\s+/, $ARGV[0]
                 ' "$array1" "$array2")

(The line-breaks are just for readability; you can get rid of them if you want.)

In the above Bash command, the embedded Perl program creates a hash named %array2 containing the elements of the second array, and then it prints any elements of the first array that exist in %array2.

This will behave slightly differently from your code in how it handles duplicate values in the second array; in your code, if array1 contains x twice and array2 contains x three times, then result will contain x six times, whereas in my code, result will contain x only twice. I don't know if that matters, since I don't know your exact requirements.

Answer 5

Your answer won't work, for two reasons:

• $array1 just expands to the first element of array1. (At least, in my installed version of Bash that's how it works. That doesn't seem to be a documented behavior, so it may be a version-dependent quirk.)
• After the first element gets added to result, result will then contain a space, so the next run of result=$result" "$item1 will misbehave horribly. (Instead of appending to result, it will run the command consisting of the first two items, with the environment variable result being set to the empty string.) Correction: Turns out, I was wrong about this one: word-splitting doesn't take place inside assignments. (See comments below.)

What you want is this:

result=()
for item1 in "${array1[@]}"; do
    for item2 in "${array2[@]}"; do
        if [[ $item1 = $item2 ]]; then
            result+=("$item1")
        fi
    done
done