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Delete multiple columns using awk or sed


split string with awk and delimiterUsing Regex Breaking a text on the last digit using linux tools like sed, or awkcsv file adding and removing characters from rowsCount the number of unique values based on two columns in a spreadsheetProblem extracting data from file using awkReplacing a Substring with sedawk - compare 2 files and print columns from both filesBash help: awk columnsRemoving multiple space using sedDelete 'N' no lines only on the Nth occurrence of a pattern in a file using the sed/awk command













1















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







share|improve this question









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
    52 mins ago















1















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







share|improve this question









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
    52 mins ago













1












1








1








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







share|improve this question









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






share|improve this question









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.











share|improve this question









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.









share|improve this question




share|improve this question








edited 23 mins ago









dessert

24.7k672105




24.7k672105






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




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
    52 mins ago

















  • The delimiter is the space.

    – andrec
    52 mins ago
















The delimiter is the space.

– andrec
52 mins ago





The delimiter is the space.

– andrec
52 mins ago










2 Answers
2






active

oldest

votes


















4
















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 >out





share|improve this answer
































    1














    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.






    share|improve this answer
























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      2 Answers
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      2 Answers
      2






      active

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      active

      oldest

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      active

      oldest

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      4
















      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 >out





      share|improve this answer





























        4
















        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 >out





        share|improve this answer



























          4












          4








          4









          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 >out





          share|improve this answer

















          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 >out






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          dessertdessert

          24.7k672105




          24.7k672105























              1














              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.






              share|improve this answer





























                1














                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.






                share|improve this answer



























                  1












                  1








                  1







                  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.






                  share|improve this answer















                  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.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 40 mins ago

























                  answered 1 hour ago









                  steeldriversteeldriver

                  69.8k11114186




                  69.8k11114186




















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