Intersection of two lists in Bash

I'm trying to write a simple script that will list the contents found in two lists. To simplify, let's use ls as an example. Imagine "one" and "two" are directories.

one=`ls one`
two=`ls two`
intersection $one $two

I'm still quite green in Bash, so feel free to correct how I am doing this. I just need some command that will print out all files in "one" and "two". They must exist in both. You might call this the "intersection" between "one" and "two".

Solution 1:

comm -12  <(ls 1) <(ls 2)

Solution 2:

Solution with comm

comm is great, but indeed it needs to work with sorted lists. And fortunately here we use ls which from the ls Bash man page:

Sort entries alphabetically if none of -cftuSUX nor --sort.

comm -12  <(ls one) <(ls two)

Alternative with sort

Intersection of two lists:

sort <(ls one) <(ls two) | uniq -d

Symmetric difference of two lists:

sort <(ls one) <(ls two) | uniq -u

Bonus

Play with it ;)

cd $(mktemp -d) && mkdir {one,two} && touch {one,two}/file_{1,2}{0..9} && touch two/file_3{0..9}

Solution 3:

Use the comm command:

ls one | sort > /tmp/one_list
ls two | sort > /tmp/two_list
comm -12 /tmp/one_list /tmp/two_list

"sort" is not really needed, but I always include it before using "comm" just in case.

Solution 4:

A less efficient (than comm) alternative:

cat <(ls 1 | sort -u) <(ls 2 | sort -u) | uniq -d

Solution 5:

Join is another good option depending on the input and desired output

join -j1 -a1 <(ls 1) <(ls 2)