How to cut first n and last n columns?

Question

How can I cut off the first n and the last n columns from a tab delimited file?

I tried this to cut first n column. But I have no idea to combine first and last n column

cut -f 1-10 -d "<CTR>v <TAB>" filename 

Answer 1

Cut can take several ranges in -f:

Columns up to 4 and from 7 onwards:

cut -f -4,7- 

or for fields 1,2,5,6 and from 10 onwards:

cut -f 1,2,5,6,10- 

etc

Answer 2

The first part of your question is easy. As already pointed out, cut accepts omission of either the starting or the ending index of a column range, interpreting this as meaning either “from the start to column n (inclusive)” or “from column n (inclusive) to the end,” respectively:

$ printf 'this:is:a:test' | cut -d: -f-2 this:is $ printf 'this:is:a:test' | cut -d: -f3- a:test 

It also supports combining ranges. If you want, e.g., the first 3 and the last 2 columns in a row of 7 columns:

$ printf 'foo:bar:baz:qux:quz:quux:quuz' | cut -d: -f-3,6- foo:bar:baz:quux:quuz 

However, the second part of your question can be a bit trickier depending on what kind of input you’re expecting. If by “last n columns” you mean “last n columns (regardless of their indices in the overall row)” (i.e. because you don’t necessarily know how many columns you’re going to find in advance) then sadly this is not possible to accomplish using cut alone. In order to effectively use cut to pull out “the last n columns” in each line, the total number of columns present in each line must be known beforehand, and each line must be consistent in the number of columns it contains.

If you do not know how many “columns” may be present in each line (e.g. because you’re working with input that is not strictly tabular), then you’ll have to use something like awk instead. E.g., to use awk to pull out the last 2 “columns” (awk calls them fields, the number of which can vary per line) from each line of input:

$ printf '/a\n/a/b\n/a/b/c\n/a/b/c/d\n' | awk -F/ '{print $(NF-1) FS $(NF)}' /a a/b b/c c/d