Select a particular column using awk or cut or perl

Question

I have a requirement to select the 7th column from a tab delimited file. eg:

cat filename | awk \'{print $7}\'

The issue is that the data in th

Answer 1

If the data is unambiguously tab-separated, then cut will cut on tabs, not spaces:

cut -f7 filename

You can certainly do that with awk, too:

awk -F'\t' '{ print $7 }'

Answer 2

Judging by the format of your input file, you can get away with delimiting on - instead of spaces:

awk 'BEGIN{FS="-"} {print $2}' filename

• FS stands for Field Separator, just think of it as the delimiter for input.
• Given that we are now delimiting on -, your 7th field before now becomes the 2nd field.
• Save a cat! Specify input file filename as an argument to awk instead.

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Alternatively, if your data fields are separated by tabs, you can do it more explicitly as follows:

awk 'BEGIN{FS="\t"} {print $7}' filename

And this will resolve the issue since Out Global Doc Mark looks to be separated by spaces.

Answer 3

This might work for you (GNU sed):

sed -r 's/(([^\t]*)\t?){7}.*/\2/' file

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This substitute command selects everything in the line and returns the 7th non-tab. In sed the last thing grouped by (...) will be returned in the lefthand side of the substitution by using a back-reference. In this case the first back-reference would return both the non-tab characters and the tab character (if present N.B. the ? meta-character which either one or none of the proceeding pattern).The .* just swallows up what was left on the line if any.

Answer 4

If fields are separated by tabs and your concern is that some fields contain spaces, there is no problem here, just:

cut -f 7

(cut defaults to tab delimited fields.)