repeat a list of data in the same file/column
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 at 18:58
Jeff Schaller
38.5k1053125
asked Dec 23 at 17:51
Dimitris Mintis
755
add a comment |
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 at 18:58
Jeff Schaller
38.5k1053125
asked Dec 23 at 17:51
Dimitris Mintis
755
Does it have to be awk? with perl, you could do something likeperl -00 -ne 'print $_ x 35' file
– steeldriver
Dec 23 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 at 22:23
In a cshell you can do simply:repeat 35 cat data_file
– Rakesh Sharma
20 hours ago
add a comment |
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 at 18:58
Jeff Schaller
38.5k1053125
asked Dec 23 at 17:51
Dimitris Mintis
755
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 at 18:58
Jeff Schaller
38.5k1053125
asked Dec 23 at 17:51
Dimitris Mintis
755
edited Dec 23 at 18:58
Jeff Schaller
38.5k1053125
asked Dec 23 at 17:51
Dimitris Mintis
755
edited Dec 23 at 18:58
Jeff Schaller
38.5k1053125
edited Dec 23 at 18:58
Jeff Schaller
38.5k1053125
edited Dec 23 at 18:58
Jeff Schaller
38.5k1053125
38.5k1053125
asked Dec 23 at 17:51
Dimitris Mintis
755
asked Dec 23 at 17:51
Dimitris Mintis
755
asked Dec 23 at 17:51
Dimitris Mintis
755
755
Does it have to be awk? with perl, you could do something likeperl -00 -ne 'print $_ x 35' file
– steeldriver
Dec 23 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 at 22:23
In a cshell you can do simply:repeat 35 cat data_file
– Rakesh Sharma
20 hours ago
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 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 at 22:23
In a cshell you can do simply:repeat 35 cat data_file
– Rakesh Sharma
20 hours ago
Does it have to be awk? with perl, you could do something like
perl -00 -ne 'print $_ x 35' file– steeldriver
Dec 23 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 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 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 at 22:23
In a cshell you can do simply:
repeat 35 cat data_file– Rakesh Sharma
20 hours ago
In a cshell you can do simply:
repeat 35 cat data_file– Rakesh Sharma
20 hours ago
add a comment |
4 Answers
4
active
oldest
votes
With simple for loop:
for i in {1..35}; do cat input_file >> new_file; done
answered Dec 23 at 17:59
RomanPerekhrest
22.8k12346
Consider implementing the loop viawhilewith counter. The{1..35}works inbashandksh( IIRC ) but not in POSIX/bin/sh.
– Sergiy Kolodyazhnyy
Dec 23 at 22:24
add a comment |
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 at 22:31
RudiC
4,1491312
add a comment |
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 at 22:31
Sergiy Kolodyazhnyy
8,27212152
add a comment |
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 at 18:04
Uncle Billy
1935
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
With simple for loop:
for i in {1..35}; do cat input_file >> new_file; done
answered Dec 23 at 17:59
RomanPerekhrest
22.8k12346
Consider implementing the loop viawhilewith counter. The{1..35}works inbashandksh( IIRC ) but not in POSIX/bin/sh.
– Sergiy Kolodyazhnyy
Dec 23 at 22:24
add a comment |
With simple for loop:
for i in {1..35}; do cat input_file >> new_file; done
answered Dec 23 at 17:59
RomanPerekhrest
22.8k12346
Consider implementing the loop viawhilewith counter. The{1..35}works inbashandksh( IIRC ) but not in POSIX/bin/sh.
– Sergiy Kolodyazhnyy
Dec 23 at 22:24
add a comment |
With simple for loop:
for i in {1..35}; do cat input_file >> new_file; done
answered Dec 23 at 17:59
RomanPerekhrest
22.8k12346
With simple for loop:
for i in {1..35}; do cat input_file >> new_file; done
answered Dec 23 at 17:59
RomanPerekhrest
22.8k12346
answered Dec 23 at 17:59
RomanPerekhrest
22.8k12346
answered Dec 23 at 17:59
RomanPerekhrest
22.8k12346
answered Dec 23 at 17:59
RomanPerekhrest
22.8k12346
22.8k12346
Consider implementing the loop viawhilewith counter. The{1..35}works inbashandksh( IIRC ) but not in POSIX/bin/sh.
– Sergiy Kolodyazhnyy
Dec 23 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 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 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 at 22:24
add a comment |
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 at 22:31
RudiC
4,1491312
add a comment |
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 at 22:31
RudiC
4,1491312
add a comment |
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 at 22:31
RudiC
4,1491312
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 at 22:31
RudiC
4,1491312
answered Dec 23 at 22:31
RudiC
4,1491312
answered Dec 23 at 22:31
RudiC
4,1491312
answered Dec 23 at 22:31
RudiC
4,1491312
4,1491312
add a comment |
add a comment |
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 at 22:31
Sergiy Kolodyazhnyy
8,27212152
add a comment |
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 at 22:31
Sergiy Kolodyazhnyy
8,27212152
add a comment |
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 at 22:31
Sergiy Kolodyazhnyy
8,27212152
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 at 22:31
Sergiy Kolodyazhnyy
8,27212152
answered Dec 23 at 22:31
Sergiy Kolodyazhnyy
8,27212152
answered Dec 23 at 22:31
Sergiy Kolodyazhnyy
8,27212152
answered Dec 23 at 22:31
Sergiy Kolodyazhnyy
8,27212152
8,27212152
add a comment |
add a comment |
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 at 18:04
Uncle Billy
1935
add a comment |
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 at 18:04
Uncle Billy
1935
add a comment |
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 at 18:04
Uncle Billy
1935
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 at 18:04
Uncle Billy
1935
edited 2 days ago
answered Dec 23 at 18:04
Uncle Billy
1935
answered Dec 23 at 18:04
Uncle Billy
1935
answered Dec 23 at 18:04
Uncle Billy
1935
1935
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 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 at 22:23
In a cshell you can do simply:
repeat 35 cat data_file– Rakesh Sharma
20 hours ago