repeat a list of data in the same file/column
2
I have a list of data in a file such as
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
and I would like to repeat this sequence of data 35 times so to have an output file such as
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
.
.(x35 times)
.
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
text-processing awk
edited Dec 23 '18 at 18:58
Jeff Schaller♦
44.6k1162143
asked Dec 23 '18 at 17:51
Dimitris MintisDimitris Mintis
805
add a comment |
2
I have a list of data in a file such as
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
and I would like to repeat this sequence of data 35 times so to have an output file such as
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
.
.(x35 times)
.
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
text-processing awk
edited Dec 23 '18 at 18:58
Jeff Schaller♦
44.6k1162143
asked Dec 23 '18 at 17:51
Dimitris MintisDimitris Mintis
805
Does it have to be awk? with perl, you could do something likeperl -00 -ne 'print $_ x 35' file
– steeldriver
Dec 23 '18 at 17:59
Is that list the only contents of the original file ? Are the other things in the file that need to be filtered out ? Please clarify
– Sergiy Kolodyazhnyy
Dec 23 '18 at 22:23
In a cshell you can do simply:repeat 35 cat data_file
– Rakesh Sharma
Dec 26 '18 at 4:29
add a comment |
2
2
2
I have a list of data in a file such as
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
and I would like to repeat this sequence of data 35 times so to have an output file such as
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
.
.(x35 times)
.
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
text-processing awk
edited Dec 23 '18 at 18:58
Jeff Schaller♦
44.6k1162143
asked Dec 23 '18 at 17:51
Dimitris MintisDimitris Mintis
805
I have a list of data in a file such as
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
and I would like to repeat this sequence of data 35 times so to have an output file such as
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
.
.(x35 times)
.
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
text-processing awk
text-processing awk
edited Dec 23 '18 at 18:58
Jeff Schaller♦
44.6k1162143
asked Dec 23 '18 at 17:51
Dimitris MintisDimitris Mintis
805
edited Dec 23 '18 at 18:58
Jeff Schaller♦
44.6k1162143
asked Dec 23 '18 at 17:51
Dimitris MintisDimitris Mintis
805
edited Dec 23 '18 at 18:58
Jeff Schaller♦
44.6k1162143
edited Dec 23 '18 at 18:58
Jeff Schaller♦
44.6k1162143
edited Dec 23 '18 at 18:58
Jeff Schaller♦
44.6k1162143
44.6k1162143
asked Dec 23 '18 at 17:51
Dimitris MintisDimitris Mintis
805
asked Dec 23 '18 at 17:51
Dimitris MintisDimitris Mintis
805
asked Dec 23 '18 at 17:51
Dimitris MintisDimitris Mintis
805
805
Does it have to be awk? with perl, you could do something likeperl -00 -ne 'print $_ x 35' file
– steeldriver
Dec 23 '18 at 17:59
Is that list the only contents of the original file ? Are the other things in the file that need to be filtered out ? Please clarify
– Sergiy Kolodyazhnyy
Dec 23 '18 at 22:23
In a cshell you can do simply:repeat 35 cat data_file
– Rakesh Sharma
Dec 26 '18 at 4:29
add a comment |
Does it have to be awk? with perl, you could do something likeperl -00 -ne 'print $_ x 35' file
– steeldriver
Dec 23 '18 at 17:59
Is that list the only contents of the original file ? Are the other things in the file that need to be filtered out ? Please clarify
– Sergiy Kolodyazhnyy
Dec 23 '18 at 22:23
In a cshell you can do simply:repeat 35 cat data_file
– Rakesh Sharma
Dec 26 '18 at 4:29
Does it have to be awk? with perl, you could do something like
perl -00 -ne 'print $_ x 35' file– steeldriver
Dec 23 '18 at 17:59
Does it have to be awk? with perl, you could do something like
perl -00 -ne 'print $_ x 35' file– steeldriver
Dec 23 '18 at 17:59
Is that list the only contents of the original file ? Are the other things in the file that need to be filtered out ? Please clarify
– Sergiy Kolodyazhnyy
Dec 23 '18 at 22:23
Is that list the only contents of the original file ? Are the other things in the file that need to be filtered out ? Please clarify
– Sergiy Kolodyazhnyy
Dec 23 '18 at 22:23
In a cshell you can do simply:
repeat 35 cat data_file– Rakesh Sharma
Dec 26 '18 at 4:29
In a cshell you can do simply:
repeat 35 cat data_file– Rakesh Sharma
Dec 26 '18 at 4:29
add a comment |
4 Answers
4
active
oldest
votes
2
With simple for loop:
for i in {1..35}; do cat input_file >> new_file; done
answered Dec 23 '18 at 17:59
RomanPerekhrestRomanPerekhrest
23.3k12448
Consider implementing the loop viawhilewith counter. The{1..35}works inbashandksh( IIRC ) but not in POSIX/bin/sh.
– Sergiy Kolodyazhnyy
Dec 23 '18 at 22:24
add a comment |
2
For exactly that input data structure (i.e. no empty lines), and for awks that allow for it (i.e. the last line read is available in the END section), try
$ awk 'END {for (i=1; i=35; i++) print}' RS= file
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
.
.
.
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
EDIT: or, even shorter,
awk 'END {for (i=35; i--;) print}' RS= file
answered Dec 23 '18 at 22:31
RudiCRudiC
4,3241312
add a comment |
0
In awk this can be done via adding lines of file into array, and iterating through array via nested loop within END statement:
awk '{a[i++]=$0} END{ for(c=1;c<=35;c++) for(j=0;j<=i;j++) print a[j]}' data.txt
Of course, if the number of lines in file is very large it would be more appropriate to cat the file multiple times and append as shown in Roman's answer
answered Dec 23 '18 at 22:31
Sergiy KolodyazhnyySergiy Kolodyazhnyy
10.7k42764
add a comment |
0
yes input_file | sed 35q | xargs cat > output_file
or with awk:
awk 'BEGIN{while(ARGC<36)ARGV[ARGC++]=ARGV[1]}1' input_file
awk 'BEGIN{f=ARGV[1]; for(i=35;i--;){while((getline <f) > 0) print; close(f)}}' input_file
answered Dec 23 '18 at 18:04
Uncle BillyUncle Billy
9348
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
With simple for loop:
for i in {1..35}; do cat input_file >> new_file; done
answered Dec 23 '18 at 17:59
RomanPerekhrestRomanPerekhrest
23.3k12448
Consider implementing the loop viawhilewith counter. The{1..35}works inbashandksh( IIRC ) but not in POSIX/bin/sh.
– Sergiy Kolodyazhnyy
Dec 23 '18 at 22:24
add a comment |
2
With simple for loop:
for i in {1..35}; do cat input_file >> new_file; done
answered Dec 23 '18 at 17:59
RomanPerekhrestRomanPerekhrest
23.3k12448
Consider implementing the loop viawhilewith counter. The{1..35}works inbashandksh( IIRC ) but not in POSIX/bin/sh.
– Sergiy Kolodyazhnyy
Dec 23 '18 at 22:24
add a comment |
2
2
2
With simple for loop:
for i in {1..35}; do cat input_file >> new_file; done
answered Dec 23 '18 at 17:59
RomanPerekhrestRomanPerekhrest
23.3k12448
With simple for loop:
for i in {1..35}; do cat input_file >> new_file; done
answered Dec 23 '18 at 17:59
RomanPerekhrestRomanPerekhrest
23.3k12448
answered Dec 23 '18 at 17:59
RomanPerekhrestRomanPerekhrest
23.3k12448
answered Dec 23 '18 at 17:59
RomanPerekhrestRomanPerekhrest
23.3k12448
answered Dec 23 '18 at 17:59
RomanPerekhrestRomanPerekhrest
23.3k12448
23.3k12448
Consider implementing the loop viawhilewith counter. The{1..35}works inbashandksh( IIRC ) but not in POSIX/bin/sh.
– Sergiy Kolodyazhnyy
Dec 23 '18 at 22:24
add a comment |
Consider implementing the loop viawhilewith counter. The{1..35}works inbashandksh( IIRC ) but not in POSIX/bin/sh.
– Sergiy Kolodyazhnyy
Dec 23 '18 at 22:24
Consider implementing the loop via
while with counter. The {1..35} works in bash and ksh ( IIRC ) but not in POSIX /bin/sh.– Sergiy Kolodyazhnyy
Dec 23 '18 at 22:24
Consider implementing the loop via
while with counter. The {1..35} works in bash and ksh ( IIRC ) but not in POSIX /bin/sh.– Sergiy Kolodyazhnyy
Dec 23 '18 at 22:24
add a comment |
2
For exactly that input data structure (i.e. no empty lines), and for awks that allow for it (i.e. the last line read is available in the END section), try
$ awk 'END {for (i=1; i=35; i++) print}' RS= file
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
.
.
.
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
EDIT: or, even shorter,
awk 'END {for (i=35; i--;) print}' RS= file
answered Dec 23 '18 at 22:31
RudiCRudiC
4,3241312
add a comment |
2
For exactly that input data structure (i.e. no empty lines), and for awks that allow for it (i.e. the last line read is available in the END section), try
$ awk 'END {for (i=1; i=35; i++) print}' RS= file
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
.
.
.
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
EDIT: or, even shorter,
awk 'END {for (i=35; i--;) print}' RS= file
answered Dec 23 '18 at 22:31
RudiCRudiC
4,3241312
add a comment |
2
2
2
For exactly that input data structure (i.e. no empty lines), and for awks that allow for it (i.e. the last line read is available in the END section), try
$ awk 'END {for (i=1; i=35; i++) print}' RS= file
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
.
.
.
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
EDIT: or, even shorter,
awk 'END {for (i=35; i--;) print}' RS= file
answered Dec 23 '18 at 22:31
RudiCRudiC
4,3241312
For exactly that input data structure (i.e. no empty lines), and for awks that allow for it (i.e. the last line read is available in the END section), try
$ awk 'END {for (i=1; i=35; i++) print}' RS= file
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
.
.
.
-0.2947990000000000
0.1722064000000000
0.1722064000000000
0.1722064000000000
0.0315420000000000
EDIT: or, even shorter,
awk 'END {for (i=35; i--;) print}' RS= file
answered Dec 23 '18 at 22:31
RudiCRudiC
4,3241312
answered Dec 23 '18 at 22:31
RudiCRudiC
4,3241312
answered Dec 23 '18 at 22:31
RudiCRudiC
4,3241312
answered Dec 23 '18 at 22:31
RudiCRudiC
4,3241312
4,3241312
add a comment |
add a comment |
0
In awk this can be done via adding lines of file into array, and iterating through array via nested loop within END statement:
awk '{a[i++]=$0} END{ for(c=1;c<=35;c++) for(j=0;j<=i;j++) print a[j]}' data.txt
Of course, if the number of lines in file is very large it would be more appropriate to cat the file multiple times and append as shown in Roman's answer
answered Dec 23 '18 at 22:31
Sergiy KolodyazhnyySergiy Kolodyazhnyy
10.7k42764
add a comment |
0
In awk this can be done via adding lines of file into array, and iterating through array via nested loop within END statement:
awk '{a[i++]=$0} END{ for(c=1;c<=35;c++) for(j=0;j<=i;j++) print a[j]}' data.txt
Of course, if the number of lines in file is very large it would be more appropriate to cat the file multiple times and append as shown in Roman's answer
answered Dec 23 '18 at 22:31
Sergiy KolodyazhnyySergiy Kolodyazhnyy
10.7k42764
add a comment |
0
0
0
In awk this can be done via adding lines of file into array, and iterating through array via nested loop within END statement:
awk '{a[i++]=$0} END{ for(c=1;c<=35;c++) for(j=0;j<=i;j++) print a[j]}' data.txt
Of course, if the number of lines in file is very large it would be more appropriate to cat the file multiple times and append as shown in Roman's answer
answered Dec 23 '18 at 22:31
Sergiy KolodyazhnyySergiy Kolodyazhnyy
10.7k42764
In awk this can be done via adding lines of file into array, and iterating through array via nested loop within END statement:
awk '{a[i++]=$0} END{ for(c=1;c<=35;c++) for(j=0;j<=i;j++) print a[j]}' data.txt
Of course, if the number of lines in file is very large it would be more appropriate to cat the file multiple times and append as shown in Roman's answer
answered Dec 23 '18 at 22:31
Sergiy KolodyazhnyySergiy Kolodyazhnyy
10.7k42764
answered Dec 23 '18 at 22:31
Sergiy KolodyazhnyySergiy Kolodyazhnyy
10.7k42764
answered Dec 23 '18 at 22:31
Sergiy KolodyazhnyySergiy Kolodyazhnyy
10.7k42764
answered Dec 23 '18 at 22:31
Sergiy KolodyazhnyySergiy Kolodyazhnyy
10.7k42764
10.7k42764
add a comment |
add a comment |
0
yes input_file | sed 35q | xargs cat > output_file
or with awk:
awk 'BEGIN{while(ARGC<36)ARGV[ARGC++]=ARGV[1]}1' input_file
awk 'BEGIN{f=ARGV[1]; for(i=35;i--;){while((getline <f) > 0) print; close(f)}}' input_file
answered Dec 23 '18 at 18:04
Uncle BillyUncle Billy
9348
add a comment |
0
yes input_file | sed 35q | xargs cat > output_file
or with awk:
awk 'BEGIN{while(ARGC<36)ARGV[ARGC++]=ARGV[1]}1' input_file
awk 'BEGIN{f=ARGV[1]; for(i=35;i--;){while((getline <f) > 0) print; close(f)}}' input_file
answered Dec 23 '18 at 18:04
Uncle BillyUncle Billy
9348
add a comment |
0
0
0
yes input_file | sed 35q | xargs cat > output_file
or with awk:
awk 'BEGIN{while(ARGC<36)ARGV[ARGC++]=ARGV[1]}1' input_file
awk 'BEGIN{f=ARGV[1]; for(i=35;i--;){while((getline <f) > 0) print; close(f)}}' input_file
answered Dec 23 '18 at 18:04
Uncle BillyUncle Billy
9348
yes input_file | sed 35q | xargs cat > output_file
or with awk:
awk 'BEGIN{while(ARGC<36)ARGV[ARGC++]=ARGV[1]}1' input_file
awk 'BEGIN{f=ARGV[1]; for(i=35;i--;){while((getline <f) > 0) print; close(f)}}' input_file
answered Dec 23 '18 at 18:04
Uncle BillyUncle Billy
9348
edited Dec 24 '18 at 6:04
answered Dec 23 '18 at 18:04
Uncle BillyUncle Billy
9348
answered Dec 23 '18 at 18:04
Uncle BillyUncle Billy
9348
answered Dec 23 '18 at 18:04
Uncle BillyUncle Billy
9348
9348
add a comment |
add a comment |
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Does it have to be awk? with perl, you could do something like
perl -00 -ne 'print $_ x 35' file– steeldriver
Dec 23 '18 at 17:59
Is that list the only contents of the original file ? Are the other things in the file that need to be filtered out ? Please clarify
– Sergiy Kolodyazhnyy
Dec 23 '18 at 22:23
In a cshell you can do simply:
repeat 35 cat data_file– Rakesh Sharma
Dec 26 '18 at 4:29