Delete multiple columns using awk or sedsplit string with awk and delimiterFiltering a line with sedUsing Regex Breaking a text on the last digit using linux tools like sed, or awkEvaluate Expression within awkawk: matching strings to decimal values and summing themsed remove last 2 numeralsremove 2nd line of output using awkwrite a number of strings using awkBash help: awk columnsPipe stdout through sed or awk command BEFORE redirecting to file?
Multi tool use
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Delete multiple columns using awk or sed
split string with awk and delimiterFiltering a line with sedUsing Regex Breaking a text on the last digit using linux tools like sed, or awkEvaluate Expression within awkawk: matching strings to decimal values and summing themsed remove last 2 numeralsremove 2nd line of output using awkwrite a number of strings using awkBash help: awk columnsPipe stdout through sed or awk command BEFORE redirecting to file?
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 29 mins ago
dessert
24.7k672105
asked 2 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.
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 29 mins ago
dessert
24.7k672105
asked 2 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
59 mins 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 29 mins ago
dessert
24.7k672105
asked 2 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 29 mins ago
dessert
24.7k672105
asked 2 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 29 mins ago
dessert
24.7k672105
asked 2 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 29 mins ago
dessert
24.7k672105
edited 29 mins ago
dessert
24.7k672105
edited 29 mins ago
dessert
24.7k672105
24.7k672105
asked 2 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 2 hours ago
andrecandrec
61
asked 2 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
59 mins ago
add a comment |
The delimiter is the space.
– andrec
59 mins ago
The delimiter is the space.
– andrec
59 mins ago
The delimiter is the space.
– andrec
59 mins ago
add a comment |
2 Answers
2
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 2 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 1 hour 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
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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 2 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 2 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 2 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 2 hours ago
dessertdessert
24.7k672105
edited 1 hour ago
answered 2 hours ago
dessertdessert
24.7k672105
answered 2 hours ago
dessertdessert
24.7k672105
answered 2 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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
steeldriversteeldriver
69.8k11114186
edited 47 mins ago
answered 1 hour ago
steeldriversteeldriver
69.8k11114186
answered 1 hour ago
steeldriversteeldriver
69.8k11114186
answered 1 hour ago
steeldriversteeldriver
69.8k11114186
69.8k11114186
add a comment |
add a comment |
andrec is a new contributor. Be nice, and check out our Code of Conduct.
andrec is a new contributor. Be nice, and check out our Code of Conduct.
andrec is a new contributor. Be nice, and check out our Code of Conduct.
andrec is a new contributor. Be nice, and check out our Code of Conduct.
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rA28AQkU f9JcY Mb2nk8A5,0k9H,tXrvY1VkLHDX a8UR0 soHZnWV2f8oP1IlDwYkJPk47AOWR0 BW78s
The delimiter is the space.
– andrec
59 mins ago