How to get the first column of comm output?

问题

So I'm trying to get the first column of comm output using awk. I read that Tab was used as a separator for comm so I did:

awk -F"\t" '{print $1}' comm-result.txt

With comm-result.txt containing the output of:

comm -3 file1 file2

But this doesn't seem to work.

This commend takes also the space character as a separator and I get weird results when my files contains multiple spaces.

How can i only get the first column from comm?

回答1:

"So I'm trying to get the first column of comm output"

The first column of the "comm file1 file2" output contains lines unique to the file1. You can skip the post-processing by simply calling comm with -2 (suppress lines unique to file2) and -3 (suppress lines that appear in both files).

comm -2 -3 file1 file2   # will show only lines unique to file1

---

However, if you have no choice but to process a pre-run output of comm then as Carl mentioned, cut would be an option:

cut -f1 comm-results.txt

However, this result in empty lines for cases where column 1 is empty. To deal with this, perhaps awk may be more suitable:

awk -F"\t" '{if ($1) print $1}' comm-results.txt
     ----    ----------------
      |                     |
   Use tab as delimiter     |
                            +-- only print if not empty

回答2:

cut(1) is probably a better choice than awk for this problem.

回答3:

You can use comm with -2 and -3 (as already explained above), or use comm with grep like:

grep -o '^\S\+' <(comm file1 file2)

so the output won't contain any trailing spaces. This is useful for non-comm commands.

来源：https://stackoverflow.com/questions/8299553/how-to-get-the-first-column-of-comm-output