AWK: Print first field of every line as identifier for every following field in the corresponding line
I have a input file like this with a blank as field seperator
AAABBB: 243.234.12.2 123.3.2 231.5.12 451.192.29.9
BBXDS: 324.22.32.5 235.235.283 234.239. 234.23.23.1
DDF: 23.12.59.09 98.39.239.29 394.293.2.2
The output should look like this:
AAABBB: 243.234.12.2
AAABBB: 123.3.2
AAABBB: 231.5.12
AAABBB: 451.192.29.9
BBXDS: 324.22.32.5
BBXDS: 235.235.283
BBXDS: 234.239.
.....
The first field of every line is a identifier and should be printed as a new line in front of every column in the corresponding line until the end of the line.
awk
asked Dec 28 '18 at 16:32
T-OneT-One
606
add a comment |
I have a input file like this with a blank as field seperator
AAABBB: 243.234.12.2 123.3.2 231.5.12 451.192.29.9
BBXDS: 324.22.32.5 235.235.283 234.239. 234.23.23.1
DDF: 23.12.59.09 98.39.239.29 394.293.2.2
The output should look like this:
AAABBB: 243.234.12.2
AAABBB: 123.3.2
AAABBB: 231.5.12
AAABBB: 451.192.29.9
BBXDS: 324.22.32.5
BBXDS: 235.235.283
BBXDS: 234.239.
.....
The first field of every line is a identifier and should be printed as a new line in front of every column in the corresponding line until the end of the line.
awk
asked Dec 28 '18 at 16:32
T-OneT-One
606
Really close to a duplicate: unix.stackexchange.com/q/456907/117549
– Jeff Schaller
Dec 28 '18 at 16:39
add a comment |
I have a input file like this with a blank as field seperator
AAABBB: 243.234.12.2 123.3.2 231.5.12 451.192.29.9
BBXDS: 324.22.32.5 235.235.283 234.239. 234.23.23.1
DDF: 23.12.59.09 98.39.239.29 394.293.2.2
The output should look like this:
AAABBB: 243.234.12.2
AAABBB: 123.3.2
AAABBB: 231.5.12
AAABBB: 451.192.29.9
BBXDS: 324.22.32.5
BBXDS: 235.235.283
BBXDS: 234.239.
.....
The first field of every line is a identifier and should be printed as a new line in front of every column in the corresponding line until the end of the line.
awk
asked Dec 28 '18 at 16:32
T-OneT-One
606
I have a input file like this with a blank as field seperator
AAABBB: 243.234.12.2 123.3.2 231.5.12 451.192.29.9
BBXDS: 324.22.32.5 235.235.283 234.239. 234.23.23.1
DDF: 23.12.59.09 98.39.239.29 394.293.2.2
The output should look like this:
AAABBB: 243.234.12.2
AAABBB: 123.3.2
AAABBB: 231.5.12
AAABBB: 451.192.29.9
BBXDS: 324.22.32.5
BBXDS: 235.235.283
BBXDS: 234.239.
.....
The first field of every line is a identifier and should be printed as a new line in front of every column in the corresponding line until the end of the line.
awk
awk
asked Dec 28 '18 at 16:32
T-OneT-One
606
asked Dec 28 '18 at 16:32
T-OneT-One
606
asked Dec 28 '18 at 16:32
T-OneT-One
606
asked Dec 28 '18 at 16:32
T-OneT-One
606
asked Dec 28 '18 at 16:32
T-OneT-One
606
606
Really close to a duplicate: unix.stackexchange.com/q/456907/117549
– Jeff Schaller
Dec 28 '18 at 16:39
add a comment |
Really close to a duplicate: unix.stackexchange.com/q/456907/117549
– Jeff Schaller
Dec 28 '18 at 16:39
Really close to a duplicate: unix.stackexchange.com/q/456907/117549
– Jeff Schaller
Dec 28 '18 at 16:39
Really close to a duplicate: unix.stackexchange.com/q/456907/117549
– Jeff Schaller
Dec 28 '18 at 16:39
add a comment |
1 Answer
1
active
oldest
votes
Fairly straightforwardly:
awk '{ for(i=2; i <= NF; i++) print $1, $i}' < input
On every line, loop from 2 until the last field (Number of Fields), printing field 1 and the looped field.
edited Dec 28 '18 at 19:25
Isaac
11.3k11651
answered Dec 28 '18 at 16:37
Jeff SchallerJeff Schaller
39k1054125
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Fairly straightforwardly:
awk '{ for(i=2; i <= NF; i++) print $1, $i}' < input
On every line, loop from 2 until the last field (Number of Fields), printing field 1 and the looped field.
edited Dec 28 '18 at 19:25
Isaac
11.3k11651
answered Dec 28 '18 at 16:37
Jeff SchallerJeff Schaller
39k1054125
add a comment |
Fairly straightforwardly:
awk '{ for(i=2; i <= NF; i++) print $1, $i}' < input
On every line, loop from 2 until the last field (Number of Fields), printing field 1 and the looped field.
edited Dec 28 '18 at 19:25
Isaac
11.3k11651
answered Dec 28 '18 at 16:37
Jeff SchallerJeff Schaller
39k1054125
add a comment |
Fairly straightforwardly:
awk '{ for(i=2; i <= NF; i++) print $1, $i}' < input
On every line, loop from 2 until the last field (Number of Fields), printing field 1 and the looped field.
edited Dec 28 '18 at 19:25
Isaac
11.3k11651
answered Dec 28 '18 at 16:37
Jeff SchallerJeff Schaller
39k1054125
Fairly straightforwardly:
awk '{ for(i=2; i <= NF; i++) print $1, $i}' < input
On every line, loop from 2 until the last field (Number of Fields), printing field 1 and the looped field.
edited Dec 28 '18 at 19:25
Isaac
11.3k11651
answered Dec 28 '18 at 16:37
Jeff SchallerJeff Schaller
39k1054125
edited Dec 28 '18 at 19:25
Isaac
11.3k11651
edited Dec 28 '18 at 19:25
Isaac
11.3k11651
edited Dec 28 '18 at 19:25
Isaac
11.3k11651
11.3k11651
answered Dec 28 '18 at 16:37
Jeff SchallerJeff Schaller
39k1054125
answered Dec 28 '18 at 16:37
Jeff SchallerJeff Schaller
39k1054125
answered Dec 28 '18 at 16:37
Jeff SchallerJeff Schaller
39k1054125
39k1054125
add a comment |
add a comment |
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Really close to a duplicate: unix.stackexchange.com/q/456907/117549
– Jeff Schaller
Dec 28 '18 at 16:39