Reconsider spacing of columns
Clash Royale CLAN TAG#URR8PPP
up vote
1
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I have a file with the following format
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
I want instead to produce a file with the format below
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
awk text-formatting printf
edited Aug 11 at 18:23
Jeff Schaller
32.4k849110
asked Aug 11 at 16:49
Dimitris Mintis
111
add a comment |Â
up vote
1
down vote
favorite
I have a file with the following format
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
I want instead to produce a file with the format below
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
awk text-formatting printf
edited Aug 11 at 18:23
Jeff Schaller
32.4k849110
asked Aug 11 at 16:49
Dimitris Mintis
111
So instead of defining space with awk like this 'print $1" "$2" "$3"..' to define characters (or real numbers) for each column)
– Dimitris Mintis
Aug 11 at 16:54
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a file with the following format
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
I want instead to produce a file with the format below
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
awk text-formatting printf
edited Aug 11 at 18:23
Jeff Schaller
32.4k849110
asked Aug 11 at 16:49
Dimitris Mintis
111
I have a file with the following format
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
I want instead to produce a file with the format below
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
awk text-formatting printf
awk text-formatting printf
edited Aug 11 at 18:23
Jeff Schaller
32.4k849110
asked Aug 11 at 16:49
Dimitris Mintis
111
edited Aug 11 at 18:23
Jeff Schaller
32.4k849110
asked Aug 11 at 16:49
Dimitris Mintis
111
edited Aug 11 at 18:23
Jeff Schaller
32.4k849110
edited Aug 11 at 18:23
Jeff Schaller
32.4k849110
edited Aug 11 at 18:23
Jeff Schaller
32.4k849110
32.4k849110
asked Aug 11 at 16:49
Dimitris Mintis
111
asked Aug 11 at 16:49
Dimitris Mintis
111
asked Aug 11 at 16:49
Dimitris Mintis
111
111
So instead of defining space with awk like this 'print $1" "$2" "$3"..' to define characters (or real numbers) for each column)
– Dimitris Mintis
Aug 11 at 16:54
add a comment |Â
So instead of defining space with awk like this 'print $1" "$2" "$3"..' to define characters (or real numbers) for each column)
– Dimitris Mintis
Aug 11 at 16:54
So instead of defining space with awk like this 'print $1" "$2" "$3"..' to define characters (or real numbers) for each column)
– Dimitris Mintis
Aug 11 at 16:54
So instead of defining space with awk like this 'print $1" "$2" "$3"..' to define characters (or real numbers) for each column)
– Dimitris Mintis
Aug 11 at 16:54
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
awk using printf+%f
Try this. By using printf with a width modified (e.g. %2d, the field is padded as required). Similarly, % 2.8f formats a floating point number here.
$ awk 'printf "%2d%7d%7d % 2.8f % 2.7f % 2.7fn",$1,$2,$3,$4,$5,$6' file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
$
Try it online!
awk using printf+tabs
The tab (t) approach also suggested makes the code a lot cleaner, but it doesn't achieve the alignment quite as you request.
$ awk 'print $1"t"$2"t"$3"t"$4"t"$5"t"$6' file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
Codeless version, column
Again, not quite the output you request, but simple to achieve using column.
$ column -t file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
$
answered Aug 11 at 17:28
steve
12.9k22149
1
awesome!!! I appreciate this a lot!!!
– Dimitris Mintis
Aug 11 at 17:42
add a comment |Â
up vote
1
down vote
Read the file with awk and output the values with printf.
awk ' printf ("%2dt%st%st%11st%st%sn", $1, $2, $3, $4, $5, $6); ' file
Adapt the printf format string as you need.
answered Aug 11 at 16:58
RalfFriedl
3,6401522
1
@steve The was just atmissing fromt.
– RalfFriedl
Aug 11 at 20:10
1
@steve Of course, you are correct.
– RalfFriedl
Aug 11 at 22:07
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
awk using printf+%f
Try this. By using printf with a width modified (e.g. %2d, the field is padded as required). Similarly, % 2.8f formats a floating point number here.
$ awk 'printf "%2d%7d%7d % 2.8f % 2.7f % 2.7fn",$1,$2,$3,$4,$5,$6' file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
$
Try it online!
awk using printf+tabs
The tab (t) approach also suggested makes the code a lot cleaner, but it doesn't achieve the alignment quite as you request.
$ awk 'print $1"t"$2"t"$3"t"$4"t"$5"t"$6' file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
Codeless version, column
Again, not quite the output you request, but simple to achieve using column.
$ column -t file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
$
answered Aug 11 at 17:28
steve
12.9k22149
1
awesome!!! I appreciate this a lot!!!
– Dimitris Mintis
Aug 11 at 17:42
add a comment |Â
up vote
3
down vote
awk using printf+%f
Try this. By using printf with a width modified (e.g. %2d, the field is padded as required). Similarly, % 2.8f formats a floating point number here.
$ awk 'printf "%2d%7d%7d % 2.8f % 2.7f % 2.7fn",$1,$2,$3,$4,$5,$6' file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
$
Try it online!
awk using printf+tabs
The tab (t) approach also suggested makes the code a lot cleaner, but it doesn't achieve the alignment quite as you request.
$ awk 'print $1"t"$2"t"$3"t"$4"t"$5"t"$6' file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
Codeless version, column
Again, not quite the output you request, but simple to achieve using column.
$ column -t file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
$
answered Aug 11 at 17:28
steve
12.9k22149
1
awesome!!! I appreciate this a lot!!!
– Dimitris Mintis
Aug 11 at 17:42
add a comment |Â
up vote
3
down vote
up vote
3
down vote
awk using printf+%f
Try this. By using printf with a width modified (e.g. %2d, the field is padded as required). Similarly, % 2.8f formats a floating point number here.
$ awk 'printf "%2d%7d%7d % 2.8f % 2.7f % 2.7fn",$1,$2,$3,$4,$5,$6' file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
$
Try it online!
awk using printf+tabs
The tab (t) approach also suggested makes the code a lot cleaner, but it doesn't achieve the alignment quite as you request.
$ awk 'print $1"t"$2"t"$3"t"$4"t"$5"t"$6' file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
Codeless version, column
Again, not quite the output you request, but simple to achieve using column.
$ column -t file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
$
answered Aug 11 at 17:28
steve
12.9k22149
awk using printf+%f
Try this. By using printf with a width modified (e.g. %2d, the field is padded as required). Similarly, % 2.8f formats a floating point number here.
$ awk 'printf "%2d%7d%7d % 2.8f % 2.7f % 2.7fn",$1,$2,$3,$4,$5,$6' file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
$
Try it online!
awk using printf+tabs
The tab (t) approach also suggested makes the code a lot cleaner, but it doesn't achieve the alignment quite as you request.
$ awk 'print $1"t"$2"t"$3"t"$4"t"$5"t"$6' file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
Codeless version, column
Again, not quite the output you request, but simple to achieve using column.
$ column -t file
1 1 1 0.02484183 57.2400000 24.6000000
2 1 2 -0.16091000 56.5200000 24.5600000
3 1 1 0.02484183 55.5700000 24.4900000
4 1 1 0.02484183 56.4400000 25.5100000
5 1 2 0.08638400 56.8900000 23.3300000
6 1 1 0.01376475 57.8900000 23.5200000
7 1 1 0.01376475 57.0600000 22.5200000
8 1 2 0.17612200 55.8800000 22.9700000
9 1 1 0.02042900 56.1600000 22.0400000
10 1 2 -0.34768900 55.9300000 24.0700000
11 1 1 0.07439383 55.5100000 25.0200000
12 1 1 0.07439383 55.4600000 23.7100000
13 1 1 0.07439383 57.0000000 24.2400000
14 1 2 0.02151800 54.4400000 22.8700000
15 1 1 0.08793975 54.5000000 22.4000000
$
answered Aug 11 at 17:28
steve
12.9k22149
edited Aug 11 at 17:34
answered Aug 11 at 17:28
steve
12.9k22149
answered Aug 11 at 17:28
steve
12.9k22149
answered Aug 11 at 17:28
steve
12.9k22149
12.9k22149
1
awesome!!! I appreciate this a lot!!!
– Dimitris Mintis
Aug 11 at 17:42
add a comment |Â
1
awesome!!! I appreciate this a lot!!!
– Dimitris Mintis
Aug 11 at 17:42
1
1
awesome!!! I appreciate this a lot!!!
– Dimitris Mintis
Aug 11 at 17:42
awesome!!! I appreciate this a lot!!!
– Dimitris Mintis
Aug 11 at 17:42
add a comment |Â
up vote
1
down vote
Read the file with awk and output the values with printf.
awk ' printf ("%2dt%st%st%11st%st%sn", $1, $2, $3, $4, $5, $6); ' file
Adapt the printf format string as you need.
answered Aug 11 at 16:58
RalfFriedl
3,6401522
1
@steve The was just atmissing fromt.
– RalfFriedl
Aug 11 at 20:10
1
@steve Of course, you are correct.
– RalfFriedl
Aug 11 at 22:07
add a comment |Â
up vote
1
down vote
Read the file with awk and output the values with printf.
awk ' printf ("%2dt%st%st%11st%st%sn", $1, $2, $3, $4, $5, $6); ' file
Adapt the printf format string as you need.
answered Aug 11 at 16:58
RalfFriedl
3,6401522
1
@steve The was just atmissing fromt.
– RalfFriedl
Aug 11 at 20:10
1
@steve Of course, you are correct.
– RalfFriedl
Aug 11 at 22:07
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Read the file with awk and output the values with printf.
awk ' printf ("%2dt%st%st%11st%st%sn", $1, $2, $3, $4, $5, $6); ' file
Adapt the printf format string as you need.
answered Aug 11 at 16:58
RalfFriedl
3,6401522
Read the file with awk and output the values with printf.
awk ' printf ("%2dt%st%st%11st%st%sn", $1, $2, $3, $4, $5, $6); ' file
Adapt the printf format string as you need.
answered Aug 11 at 16:58
RalfFriedl
3,6401522
edited Aug 11 at 22:07
answered Aug 11 at 16:58
RalfFriedl
3,6401522
answered Aug 11 at 16:58
RalfFriedl
3,6401522
answered Aug 11 at 16:58
RalfFriedl
3,6401522
3,6401522
1
@steve The was just atmissing fromt.
– RalfFriedl
Aug 11 at 20:10
1
@steve Of course, you are correct.
– RalfFriedl
Aug 11 at 22:07
add a comment |Â
1
@steve The was just atmissing fromt.
– RalfFriedl
Aug 11 at 20:10
1
@steve Of course, you are correct.
– RalfFriedl
Aug 11 at 22:07
1
1
@steve The was just a
t missing from t.– RalfFriedl
Aug 11 at 20:10
@steve The was just a
t missing from t.– RalfFriedl
Aug 11 at 20:10
1
1
@steve Of course, you are correct.
– RalfFriedl
Aug 11 at 22:07
@steve Of course, you are correct.
– RalfFriedl
Aug 11 at 22:07
add a comment |Â
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So instead of defining space with awk like this 'print $1" "$2" "$3"..' to define characters (or real numbers) for each column)
– Dimitris Mintis
Aug 11 at 16:54