Printing a file in columns with arbitrary amount of rows

Question

I have this file:

line1
line2
line3
line4
line5
line6
line7
line8
line9

And I want to print it across 3 columns with this structure so that first column has lines 1-2-3, second column has the lines 4-5 and the last column has the lines 6-7-8-9 (or the rest of the file):

line1 line4 line6
line2 line5 line7
line3       line8
            line9

Essentially I want to print a file in a way that the output is distributed to N columns and every column has predetermined (and possibly different from each other) amount of lines. If possible, I want to preserve leading/trailing spaces in lines. How can I do that?

I fiddled around with columns and pr, but I couldn't even come close.

Something like this could work: `pr -m <(sed -n 1,3p file) <(sed -n 4,5p file) <(sed -n 6,9p file)` — pLumo, Oct 05 '18 at 12:47
That solves my problem, thanks! Maybe add as an answer so I can accept it? (And can you explain how it works?) — isamert, Oct 05 '18 at 12:55

score 3 · Accepted Answer · answered Oct 05 '18 at 12:58

3

You could use the merge functionality of pr, which merges multiple files to columns.

Something like this would work:

pr -J -m <(sed -n 1,3p file) <(sed -n 4,5p file) <(sed -n 6,9p file)

answered Oct 05 '18 at 12:58

pLumo

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score 0 · Answer 2 · answered Oct 05 '18 at 13:26

0

Another option is using paste:

paste <(sed '1,3!d' infile) <(sed '4,5!d' infile) <(sed '6,$!d' infile)

answered Oct 05 '18 at 13:26

αғsнιη

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Printing a file in columns with arbitrary amount of rows

2 Answers2