Delete multiple columns using awk or sed
I have a database with 6037 space-separated columns and 450 rows like the one below:
1807 1452 1598 1 6.655713 A B A B ... 0
1808 1452 1763 1 9.362033 0 0 A B ... A
1809 1452 1527 2 6.728534 A B A A ... B
1810 1452 1367 2 9.4055 A B A A B ... A
... ... ... ... ... ... ... ... ... ...
1812 1452 1258 1 6.363032 0 0 A B ... B
I want to get a new database with only the first 676 columns.
Preferably, some form that uses awk or sed command.
text-processing sed awk
edited 2 hours ago
dessert
24.7k672105
asked 3 hours ago
andrecandrec
61
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add a comment |
I have a database with 6037 space-separated columns and 450 rows like the one below:
1807 1452 1598 1 6.655713 A B A B ... 0
1808 1452 1763 1 9.362033 0 0 A B ... A
1809 1452 1527 2 6.728534 A B A A ... B
1810 1452 1367 2 9.4055 A B A A B ... A
... ... ... ... ... ... ... ... ... ...
1812 1452 1258 1 6.363032 0 0 A B ... B
I want to get a new database with only the first 676 columns.
Preferably, some form that uses awk or sed command.
text-processing sed awk
edited 2 hours ago
dessert
24.7k672105
asked 3 hours ago
andrecandrec
61
New contributor
andrec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The delimiter is the space.
– andrec
2 hours ago
add a comment |
I have a database with 6037 space-separated columns and 450 rows like the one below:
1807 1452 1598 1 6.655713 A B A B ... 0
1808 1452 1763 1 9.362033 0 0 A B ... A
1809 1452 1527 2 6.728534 A B A A ... B
1810 1452 1367 2 9.4055 A B A A B ... A
... ... ... ... ... ... ... ... ... ...
1812 1452 1258 1 6.363032 0 0 A B ... B
I want to get a new database with only the first 676 columns.
Preferably, some form that uses awk or sed command.
text-processing sed awk
edited 2 hours ago
dessert
24.7k672105
asked 3 hours ago
andrecandrec
61
New contributor
andrec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a database with 6037 space-separated columns and 450 rows like the one below:
1807 1452 1598 1 6.655713 A B A B ... 0
1808 1452 1763 1 9.362033 0 0 A B ... A
1809 1452 1527 2 6.728534 A B A A ... B
1810 1452 1367 2 9.4055 A B A A B ... A
... ... ... ... ... ... ... ... ... ...
1812 1452 1258 1 6.363032 0 0 A B ... B
I want to get a new database with only the first 676 columns.
Preferably, some form that uses awk or sed command.
text-processing sed awk
text-processing sed awk
edited 2 hours ago
dessert
24.7k672105
asked 3 hours ago
andrecandrec
61
New contributor
andrec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
dessert
24.7k672105
asked 3 hours ago
andrecandrec
61
New contributor
andrec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
dessert
24.7k672105
edited 2 hours ago
dessert
24.7k672105
edited 2 hours ago
dessert
24.7k672105
24.7k672105
asked 3 hours ago
andrecandrec
61
New contributor
andrec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
andrecandrec
61
asked 3 hours ago
andrecandrec
61
61
New contributor
andrec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
andrec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
andrec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The delimiter is the space.
– andrec
2 hours ago
add a comment |
The delimiter is the space.
– andrec
2 hours ago
The delimiter is the space.
– andrec
2 hours ago
The delimiter is the space.
– andrec
2 hours ago
add a comment |
2 Answers
2
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oldest
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If the column delimiter in your file is a single character, e.g. a space, cut can do that easily:
cut -d' ' -f-676 <in >out
This prints only the space-separated columns from the first to the 676th.
If you need e.g. every whitespace character to count as a delimiter, a sed solution is:
sed -r 's/s+S+//677g' <in >out
This replaces every column (= at least one whitespace character followed by at least one non-whitespace character) beginning with the 677th with nothing. Using character groups you can specify any set of delimiters you need, e.g. for “4”, “#” and “K”:
sed -r 's/[4#K]+[^4#K]+//677g' <in >out
For a reasonable awk approach kindly refer to steeldriver’s answer, but here is another one looping over the columns and only printing them (separated by FS) if their number is <= 676:
awk '{for (i=1;i<=676;i++) {printf (i==1?"":FS)$i}; print ""}' <in >out
For a character group you have to specify the output field separator for the output, e.g. for [4#K] and "sep":
awk -F'[4#K]' '{for (i=1;i<=676;i++) {printf (i==1?"":"sep")$i}; print ""}' <in >outanswered 3 hours ago
dessertdessert
24.7k672105
add a comment |
For a single-character delimiter (such as space or comma) I would recommend using the cut command over either awk or sed.
However since you asked about awk specifically, I think a reasonable way to do it would be to decrement the field count:
awk -v last=676 '{while(NF>last) NF--} 1' datafile
Tested in GNU Awk (gawk) and mawk.
answered 2 hours ago
steeldriversteeldriver
69.8k11114186
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If the column delimiter in your file is a single character, e.g. a space, cut can do that easily:
cut -d' ' -f-676 <in >out
This prints only the space-separated columns from the first to the 676th.
If you need e.g. every whitespace character to count as a delimiter, a sed solution is:
sed -r 's/s+S+//677g' <in >out
This replaces every column (= at least one whitespace character followed by at least one non-whitespace character) beginning with the 677th with nothing. Using character groups you can specify any set of delimiters you need, e.g. for “4”, “#” and “K”:
sed -r 's/[4#K]+[^4#K]+//677g' <in >out
For a reasonable awk approach kindly refer to steeldriver’s answer, but here is another one looping over the columns and only printing them (separated by FS) if their number is <= 676:
awk '{for (i=1;i<=676;i++) {printf (i==1?"":FS)$i}; print ""}' <in >out
For a character group you have to specify the output field separator for the output, e.g. for [4#K] and "sep":
awk -F'[4#K]' '{for (i=1;i<=676;i++) {printf (i==1?"":"sep")$i}; print ""}' <in >outanswered 3 hours ago
dessertdessert
24.7k672105
add a comment |
If the column delimiter in your file is a single character, e.g. a space, cut can do that easily:
cut -d' ' -f-676 <in >out
This prints only the space-separated columns from the first to the 676th.
If you need e.g. every whitespace character to count as a delimiter, a sed solution is:
sed -r 's/s+S+//677g' <in >out
This replaces every column (= at least one whitespace character followed by at least one non-whitespace character) beginning with the 677th with nothing. Using character groups you can specify any set of delimiters you need, e.g. for “4”, “#” and “K”:
sed -r 's/[4#K]+[^4#K]+//677g' <in >out
For a reasonable awk approach kindly refer to steeldriver’s answer, but here is another one looping over the columns and only printing them (separated by FS) if their number is <= 676:
awk '{for (i=1;i<=676;i++) {printf (i==1?"":FS)$i}; print ""}' <in >out
For a character group you have to specify the output field separator for the output, e.g. for [4#K] and "sep":
awk -F'[4#K]' '{for (i=1;i<=676;i++) {printf (i==1?"":"sep")$i}; print ""}' <in >outanswered 3 hours ago
dessertdessert
24.7k672105
add a comment |
If the column delimiter in your file is a single character, e.g. a space, cut can do that easily:
cut -d' ' -f-676 <in >out
This prints only the space-separated columns from the first to the 676th.
If you need e.g. every whitespace character to count as a delimiter, a sed solution is:
sed -r 's/s+S+//677g' <in >out
This replaces every column (= at least one whitespace character followed by at least one non-whitespace character) beginning with the 677th with nothing. Using character groups you can specify any set of delimiters you need, e.g. for “4”, “#” and “K”:
sed -r 's/[4#K]+[^4#K]+//677g' <in >out
For a reasonable awk approach kindly refer to steeldriver’s answer, but here is another one looping over the columns and only printing them (separated by FS) if their number is <= 676:
awk '{for (i=1;i<=676;i++) {printf (i==1?"":FS)$i}; print ""}' <in >out
For a character group you have to specify the output field separator for the output, e.g. for [4#K] and "sep":
awk -F'[4#K]' '{for (i=1;i<=676;i++) {printf (i==1?"":"sep")$i}; print ""}' <in >outanswered 3 hours ago
dessertdessert
24.7k672105
If the column delimiter in your file is a single character, e.g. a space, cut can do that easily:
cut -d' ' -f-676 <in >out
This prints only the space-separated columns from the first to the 676th.
If you need e.g. every whitespace character to count as a delimiter, a sed solution is:
sed -r 's/s+S+//677g' <in >out
This replaces every column (= at least one whitespace character followed by at least one non-whitespace character) beginning with the 677th with nothing. Using character groups you can specify any set of delimiters you need, e.g. for “4”, “#” and “K”:
sed -r 's/[4#K]+[^4#K]+//677g' <in >out
For a reasonable awk approach kindly refer to steeldriver’s answer, but here is another one looping over the columns and only printing them (separated by FS) if their number is <= 676:
awk '{for (i=1;i<=676;i++) {printf (i==1?"":FS)$i}; print ""}' <in >out
For a character group you have to specify the output field separator for the output, e.g. for [4#K] and "sep":
awk -F'[4#K]' '{for (i=1;i<=676;i++) {printf (i==1?"":"sep")$i}; print ""}' <in >outanswered 3 hours ago
dessertdessert
24.7k672105
edited 2 hours ago
answered 3 hours ago
dessertdessert
24.7k672105
answered 3 hours ago
dessertdessert
24.7k672105
answered 3 hours ago
dessertdessert
24.7k672105
24.7k672105
add a comment |
add a comment |
For a single-character delimiter (such as space or comma) I would recommend using the cut command over either awk or sed.
However since you asked about awk specifically, I think a reasonable way to do it would be to decrement the field count:
awk -v last=676 '{while(NF>last) NF--} 1' datafile
Tested in GNU Awk (gawk) and mawk.
answered 2 hours ago
steeldriversteeldriver
69.8k11114186
add a comment |
For a single-character delimiter (such as space or comma) I would recommend using the cut command over either awk or sed.
However since you asked about awk specifically, I think a reasonable way to do it would be to decrement the field count:
awk -v last=676 '{while(NF>last) NF--} 1' datafile
Tested in GNU Awk (gawk) and mawk.
answered 2 hours ago
steeldriversteeldriver
69.8k11114186
add a comment |
For a single-character delimiter (such as space or comma) I would recommend using the cut command over either awk or sed.
However since you asked about awk specifically, I think a reasonable way to do it would be to decrement the field count:
awk -v last=676 '{while(NF>last) NF--} 1' datafile
Tested in GNU Awk (gawk) and mawk.
answered 2 hours ago
steeldriversteeldriver
69.8k11114186
For a single-character delimiter (such as space or comma) I would recommend using the cut command over either awk or sed.
However since you asked about awk specifically, I think a reasonable way to do it would be to decrement the field count:
awk -v last=676 '{while(NF>last) NF--} 1' datafile
Tested in GNU Awk (gawk) and mawk.
answered 2 hours ago
steeldriversteeldriver
69.8k11114186
edited 2 hours ago
answered 2 hours ago
steeldriversteeldriver
69.8k11114186
answered 2 hours ago
steeldriversteeldriver
69.8k11114186
answered 2 hours ago
steeldriversteeldriver
69.8k11114186
69.8k11114186
add a comment |
add a comment |
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The delimiter is the space.
– andrec
2 hours ago