How do I construct a list with numbers like this?

Question

How do I construct a list like this:

1 2 3 4 5
6 7 8 9 10
11 12 13 14 15

Where I run command 15 or something. Or if I specify 100 it would make it with 100 numbers or 10000 and it would make it like this but 10000 numbers.

It should be five numbers on each line (as seen above).

Answer 1

you simply do

seq 1 n | xargs -n 5 echo

n being the number you want to reach

If your OS has bash but not seq, here is an alternative (thx to @cuonglm and @jimmyj for their remarks)

echo {1..n} | xargs -n5

(you may have to be careful when reaching very high number with that one, depending on the OS and bash version, and if bash actually tried to expand first or in that case is clever enough to feed little by little without trying to fit the whole 1..n as a string in memory and feed that to echo...)

And thanks to cuonglm and StephaneChazelas, I add an alternative that is very, very less CPU heavy than my first xargs solution (in which xargs calls /bin/echo, instead of being able to use the shell's builtin, every 5 numbers) (it's probably similar to the 2nd one where xargs doesn't invoke echo) :

printf '%s %s %s %s %s\n' {1..n}

That 2nd and 3rd solution differs from the 1st in that the shell have first to expand 1..n, before printf (or xargs) can start printing, if I'm not mistaken... so it starts later (especially if n is big)... And could reach some limits (line length, or memory, depending on the implementation and the OS) if n is very big.

Answer 2

Pure coreutils:

$ seq 15 | paste - - - - - 
1   2   3   4   5
6   7   8   9   10
11  12  13  14  15

Pure perl:

$ perl -e '@a=1..15; while($i<=$#a){print "@a[$i..$i+4]\n";$i+=5}'
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15

GNU grep (shamelessly stolen from @1_CR):

$ echo {1..15} | grep -oP '(\d+ ){4}\d+'
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15

If you don't mind leading 0s:

$ echo {01..15} | fold -sw 16
01 02 03 04 05 
06 07 08 09 10 
11 12 13 14 15

And if you do:

$ echo {01..15} | fold -sw 16 | sed 's/0\([1-9]\) /\1 /g'
1 2 3 4 5 
6 7 8 9 10 
11 12 13 14 15

---

Any of these can be made into a function that takes a number as input and prints the corresponding list. For example,

printnums(){
 seq $1 | paste - - - - - 
}

You can then run

$ printnums 30
1   2   3   4   5
6   7   8   9   10
11  12  13  14  15
16  17  18  19  20
21  22  23  24  25
26  27  28  29  30

Answer 3

With printf and brace expansion:

printf '%s %s %s %s %s\n' {1..15}

Answer 4

With zsh:

$ print -aC5 {1..15}
1   2   3   4   5
6   7   8   9   10
11  12  13  14  15

$ autoload zargs # if not in ~/.zshrc
$ zargs -n 5 {1..15} -- echo
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15

$ printf '%s %s %s %s %s\n' {1..15}
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15

(that latter one also works in recent versions of ksh93 or bash).

Answer 5

using looping :

for x in {1..15};do echo -n "$x ";if [ $(($x%5)) -eq 0 ];then echo ; fi done

output:

1 2 3 4 5 
6 7 8 9 10 
11 12 13 14 15 

Answer 6

With awk you can say:

seq 15 | awk 'ORS=NR%5?FS:RS'

It returns:

1 2 3 4 5
6 7 8 9 10
11 12 13 14 15

Basically, it changes the output record separator (line separator) by either a field separator (space, as default) or a record separator (new line, as default). So if the number of line is multiple of 5, a new line is added; otherwise, a space.

I wrote a broader explanation in a similar question: bash/sed/awk/etc remove every other newline.

Answer 7

You can use seq, tr and sed:

seq 15 | tr '\n' ' ' | sed 's/\([0-9]* [0-9]* [0-9]* [0-9]* [0-9]* \)/\1\
/g'

Or as an optimised combination of the answers by Olivier, Benoit and me (in his comments):

seq 15 | xargs -n 5

Answer 8

All the elegant solutions were taken...so with GNU awk and bash

echo {1..15} | awk -v RS='[[:space:]]' '{ORS=NR % numcols?RT:"\n"; print}' numcols=5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15

or with GNU grep

echo {1..15} | grep -oE '([[:digit:]]+[[:blank:]]+){4}[[:digit:]]+'
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15