How do I convert a single column to 3 columns?

Question

I have large text file which contains data, formatted like this:

1
2
3
4
5
6
7
8
9
10

I am trying to convert it to this:

1           2             3
4           5             6
7           8             9
10

I tried awk:

'{ if (NR%2) {printf "%40s\n", $0} else {printf "%80s\n", $0} }' file.txt

Answer 1

A solution with paste

seq 10 | paste - - -
1       2       3
4       5       6
7       8       9
10

paste is a Unix standard tool, and the standard guarantees that this works for at least 12 columns.

Answer 2

The columns tool can do this:

$ seq 10 | columns -W 16 -c 3
1     2     3
4     5     6
7     8     9
10

-W 16 is just to set the line width to something small.

columns is not a Unix standard tool. It is part of GNU AutoGen.

Some versions of the more common column command may be able to set the number of columns with -c, but modern versions seem to have changed its meaning to set the line width by number of characters.

There's also pr as suggested by mpez0 in a comment:

$ seq 10 | pr -aT3
1           2           3
4           5           6
7           8           9
10

-aT3 is short for --across --omit-pagination --columns=3.

pr is in coreutils and POSIX, though -T/--omit-pagination seems to be GNU-specific.

Answer 3

Try like

seq 1 10 | awk '{printf "%40s", $0} !(NR%3) {printf "\n"}'
                                       1                                       2                                       3
                                       4                                       5                                       6
                                       7                                       8                                       9
                                      10

Answer 4

The tool for this job is rs ("reshape"):

$ seq 10 | rs 0 3
1   2   3
4   5   6
7   8   9
10  

We can change the column separator to a tab:

$ seq 10 | rs -C 0 3
1       2       3
4       5       6
7       8       9
10

Or right-align:

$ seq 10 | rs -j 0 3
 1   2   3
 4   5   6
 7   8   9
10

rs is not a Unix standard tool. It is widely available, however. It was invented in 4.2BSD, and so is in all of the modern BSDs. There are ports of it to Linux-based operating systems, such as to Debian, for example.

Answer 5

This will move the string width alternately between 40, 80, 120 for each line of input:

awk '{ m = (NR-1) % 3; i = (m+1) * 40; printf "%*s\n", i, $0 }'

Variables:

• m - line number modulo 3 (i.e. 0, 1, 2 repeating)
• i - the indent for the given value of m

In the absence of any other instruction, I've continued to use your own printf formatting so that each line of input is formatted as its own line of output, and each will be right-justified in the available space.

If you want three 40-column entries per line as shown in your example, rather than in your code, you could use this (change the 40s to -40s if you want left-justified text):

awk '{ printf "%40s", $0 } !(NR % 3) { printf "\n" }'

Answer 6

Tried with Below command , Tested and Worked fine

command

for ((i=0;i<=10;i++)); do awk -v i="$i" 'NR>i && NR<(i+4)' o| perl -pne "s/\n/ /g";echo -e  '\n'; i=$(($i+2)); done

output

1 2 3 

4 5 6 

7 8 9 

10 

Answer 7

Using Python

#!/usr/bin/python
k=open('filename','r')
p=k.readlines()
for i in range(0,10,3):
    print str(p[i:i+3]).replace("[","").replace("]","").replace(",","").replace("\\n","").replace("'","")

output

1 2 3
4 5 6
7 8 9
10

Answer 8

Personally, I would prefer the format

1  4  7
2  5  8
3  6  9

which can be achieved by

mkdir temp
cat file | while read FN; do
  touch "temp/$FN"
done
ls temp

though you might have to adjust window width to make it three columns. And you can’t redirect to a file; have to cut-n-paste. Redirection tells ls to do a single column.