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Delete multiple columns using awk or sed



The Next CEO of Stack Overflowsplit 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










3















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.










share|improve this question




























    3















    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.










    share|improve this question


























      3












      3








      3








      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.










      share|improve this question
















      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















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 21 at 23:36









      dessert

      25.3k673107




      25.3k673107










      asked Mar 21 at 21:51









      andrecandrec

      161




      161




















          3 Answers
          3






          active

          oldest

          votes


















          7
















          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














            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





              Why not just NF = last; print instead of the loop?

              – wchargin
              Mar 22 at 4:36






            • 1





              @wchargin Doh! yes that's much better - wanna post it as an answer?

              – steeldriver
              Mar 22 at 4:49


















            3














            You could use



            mlr --nidx --fs ' ' --repifs cat inputFile.csv | cut -d ' ' -f-2


            In this way with mlr (https://github.com/johnkerl/miller/releases/tag/5.4.0) you manage field separators (if you have more than one spaces, they become one per field), and with cut you extract (in my example) the first two fields.



            From



            1807 1452 1598 1 6.655713 A B A B
            1808 1452 1763 1 9.362033 0 0 A B
            1809 1452 1527 2 6.728534 A B A A
            1810 1452 1367 2 9.4055 A B A A B


            to



            1807 1452
            1808 1452
            1809 1452
            1810 1452


            Some notes about Miller options:




            • --nidx is to set the format; this is a generic index-numbered table (the first field is 1, the second is 2, ecc..);


            • --fs to set the separator (here is a space);


            • --repifs means that multiple successive occurrences of the field separator count as one


            • cat passes input records directly to output.





            share|improve this answer

























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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              7
















              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





























                7
















                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



























                  7












                  7








                  7









                  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 Mar 21 at 22:54

























                  answered Mar 21 at 22:04









                  dessertdessert

                  25.3k673107




                  25.3k673107























                      4














                      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





                        Why not just NF = last; print instead of the loop?

                        – wchargin
                        Mar 22 at 4:36






                      • 1





                        @wchargin Doh! yes that's much better - wanna post it as an answer?

                        – steeldriver
                        Mar 22 at 4:49















                      4














                      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





                        Why not just NF = last; print instead of the loop?

                        – wchargin
                        Mar 22 at 4:36






                      • 1





                        @wchargin Doh! yes that's much better - wanna post it as an answer?

                        – steeldriver
                        Mar 22 at 4:49













                      4












                      4








                      4







                      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 Mar 21 at 23:19

























                      answered Mar 21 at 22:45









                      steeldriversteeldriver

                      70.6k11114187




                      70.6k11114187







                      • 1





                        Why not just NF = last; print instead of the loop?

                        – wchargin
                        Mar 22 at 4:36






                      • 1





                        @wchargin Doh! yes that's much better - wanna post it as an answer?

                        – steeldriver
                        Mar 22 at 4:49












                      • 1





                        Why not just NF = last; print instead of the loop?

                        – wchargin
                        Mar 22 at 4:36






                      • 1





                        @wchargin Doh! yes that's much better - wanna post it as an answer?

                        – steeldriver
                        Mar 22 at 4:49







                      1




                      1





                      Why not just NF = last; print instead of the loop?

                      – wchargin
                      Mar 22 at 4:36





                      Why not just NF = last; print instead of the loop?

                      – wchargin
                      Mar 22 at 4:36




                      1




                      1





                      @wchargin Doh! yes that's much better - wanna post it as an answer?

                      – steeldriver
                      Mar 22 at 4:49





                      @wchargin Doh! yes that's much better - wanna post it as an answer?

                      – steeldriver
                      Mar 22 at 4:49











                      3














                      You could use



                      mlr --nidx --fs ' ' --repifs cat inputFile.csv | cut -d ' ' -f-2


                      In this way with mlr (https://github.com/johnkerl/miller/releases/tag/5.4.0) you manage field separators (if you have more than one spaces, they become one per field), and with cut you extract (in my example) the first two fields.



                      From



                      1807 1452 1598 1 6.655713 A B A B
                      1808 1452 1763 1 9.362033 0 0 A B
                      1809 1452 1527 2 6.728534 A B A A
                      1810 1452 1367 2 9.4055 A B A A B


                      to



                      1807 1452
                      1808 1452
                      1809 1452
                      1810 1452


                      Some notes about Miller options:




                      • --nidx is to set the format; this is a generic index-numbered table (the first field is 1, the second is 2, ecc..);


                      • --fs to set the separator (here is a space);


                      • --repifs means that multiple successive occurrences of the field separator count as one


                      • cat passes input records directly to output.





                      share|improve this answer





























                        3














                        You could use



                        mlr --nidx --fs ' ' --repifs cat inputFile.csv | cut -d ' ' -f-2


                        In this way with mlr (https://github.com/johnkerl/miller/releases/tag/5.4.0) you manage field separators (if you have more than one spaces, they become one per field), and with cut you extract (in my example) the first two fields.



                        From



                        1807 1452 1598 1 6.655713 A B A B
                        1808 1452 1763 1 9.362033 0 0 A B
                        1809 1452 1527 2 6.728534 A B A A
                        1810 1452 1367 2 9.4055 A B A A B


                        to



                        1807 1452
                        1808 1452
                        1809 1452
                        1810 1452


                        Some notes about Miller options:




                        • --nidx is to set the format; this is a generic index-numbered table (the first field is 1, the second is 2, ecc..);


                        • --fs to set the separator (here is a space);


                        • --repifs means that multiple successive occurrences of the field separator count as one


                        • cat passes input records directly to output.





                        share|improve this answer



























                          3












                          3








                          3







                          You could use



                          mlr --nidx --fs ' ' --repifs cat inputFile.csv | cut -d ' ' -f-2


                          In this way with mlr (https://github.com/johnkerl/miller/releases/tag/5.4.0) you manage field separators (if you have more than one spaces, they become one per field), and with cut you extract (in my example) the first two fields.



                          From



                          1807 1452 1598 1 6.655713 A B A B
                          1808 1452 1763 1 9.362033 0 0 A B
                          1809 1452 1527 2 6.728534 A B A A
                          1810 1452 1367 2 9.4055 A B A A B


                          to



                          1807 1452
                          1808 1452
                          1809 1452
                          1810 1452


                          Some notes about Miller options:




                          • --nidx is to set the format; this is a generic index-numbered table (the first field is 1, the second is 2, ecc..);


                          • --fs to set the separator (here is a space);


                          • --repifs means that multiple successive occurrences of the field separator count as one


                          • cat passes input records directly to output.





                          share|improve this answer















                          You could use



                          mlr --nidx --fs ' ' --repifs cat inputFile.csv | cut -d ' ' -f-2


                          In this way with mlr (https://github.com/johnkerl/miller/releases/tag/5.4.0) you manage field separators (if you have more than one spaces, they become one per field), and with cut you extract (in my example) the first two fields.



                          From



                          1807 1452 1598 1 6.655713 A B A B
                          1808 1452 1763 1 9.362033 0 0 A B
                          1809 1452 1527 2 6.728534 A B A A
                          1810 1452 1367 2 9.4055 A B A A B


                          to



                          1807 1452
                          1808 1452
                          1809 1452
                          1810 1452


                          Some notes about Miller options:




                          • --nidx is to set the format; this is a generic index-numbered table (the first field is 1, the second is 2, ecc..);


                          • --fs to set the separator (here is a space);


                          • --repifs means that multiple successive occurrences of the field separator count as one


                          • cat passes input records directly to output.






                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited Mar 22 at 9:04

























                          answered Mar 22 at 7:13









                          aborrusoaborruso

                          20115




                          20115



























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