extract columns with Bash or awk

Question

I have a folder with 1000 text files. I would like to extract fourth and fifth columns from each file and save it to an another folder with the same filenames. How can I do this with awk or Bash?

You should at least put an example of the kind of file that you're trying to parse. — Goles
– Goles, Commented Oct 23, 2012 at 12:02

ДМИТРИЙ МАЛИКОВ · Accepted Answer · 2012-10-23 11:49:27Z

5

Since you haven't specified about the field separator, I am assuming as the default(space) field separator.

awk '{print $4,$5>"/tmp/another_directory/"FILENAME}' *

edited Oct 23, 2012 at 11:49

ДМИТРИЙ МАЛИКОВ

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answered Oct 23, 2012 at 11:47

Vijay

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Thor · Accepted Answer · 2012-10-23 13:12:12Z

0

With GNU find and GNU sed, assuming space separated columns and the files to process are in the current directory:

mkdir tmp/
find . -maxdepth 1 -type f -print0 |
  xargs -0 -l /bin/sh -c 'sed -r "s/([^ ] *){3}([^ ]) *([^ ]*).*/\2 \3/" "$1" > "tmp/$1"' ''

Note the last pair of single quotes are important, they're a placeholders so the filename gets passed correctly to sh. The processed files are saved to tmp/ in the current directory.

answered Oct 23, 2012 at 13:12

Thor

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