Print all but the first three columns

Question

Too cumbersome   awk   print     4    5    6    7    8    9    10    11    12    13   things

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Try this   awk     1      2      3     print  0

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Perl solution which does not add leading or trailing whitespace   perl -lane  splice  F 0 3  print join      F  file   The perl  F autosplit array starts at index 0 while awk fields start with  1    Perl solution for comma-delimited data   perl -F  -lane  splice  F 0 3  print join      F  file     Python solution   python -c  import sys  sys stdout write     join line split   3        n   for line in sys stdin    lt  file

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awk   for i 1 i lt 4 i     i    print   file

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use cut    cut -f4-13 file   or if you insist on awk and  13 is the last field    awk    1  2  3    print   file   else    awk   for i 4 i lt  13 i   printf   s    i printf   n    file

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Another way to avoid using the print statement      awk    1  2  3    sub    FS         file   In awk when a condition is true print is the default action

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Use cut   cut -d  lt The character between characters gt  -f  lt number of first column gt   lt number of last column gt   lt file name gt    e g   If you have file1 containing   car is nice equal bmw  Run   cut -d   -f1 3 file1  will print car is nice

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The correct way to do this is with an RE interval because it lets you simply state how many fields to skip  and retains inter-field spacing for the remaining fields   e g  to skip the first 3 fields without affecting spacing between remaining fields given the format of input we seem to be discussing in this question is simply     echo  1   2 3 4   5    6    awk   sub            3       1  4   5    6   If you want to accommodate leading spaces and non-blank spaces  but again with the default FS  then it s     echo    1   2 3 4   5    6    awk   sub     space         space       space      3       1  4   5    6   If you have an FS that s an RE you can t negate in a character set  you can convert it to a single char first  RS is ideal if it s a single char since an RS CANNOT appear within a field  otherwise consider SUBSEP   then apply the RE interval subsitution  then convert to the OFS  e g  if chains of    s separated the fields     echo  1   2 3 4   5    6    awk -F         gsub FS RS  sub      RS     RS     3       gsub RS OFS  1  4 5 6   Obviously if OFS is a single char AND it can t appear in the input fields you can reduce that to     echo  1   2 3 4   5    6    awk -F         gsub FS OFS   sub      OFS     OFS     3       1  4 5 6   Then you have the same issue as with all the loop-based solutions that reassign the fields - the FSs are converted to OFSs  If that s an issue  you need to look into GNU awks  patsplit   function

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I ve found this other possibility  maybe it could be useful also     awk  BEGIN  OFS ORS   t      for i 1  i lt 14  i    print  i      print  NF   n     your file   Note  1  For tabular data and from column  1 to  14

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This isn t very far from some of the previous answers  but does solve a couple of issues   cols sh      bin bash awk -v s  1   for i s  i lt  NF i    printf   -5s    i  print         Which you can now call with an argument that will be the starting column     echo  1 2 3 4 5 6 7 8 9 10 11 12 13 14      cols sh 3  3    4    5    6    7    8    9    10   11   12   13   14   Or     echo  1 2 3 4 5 6 7 8 9 10 11 12 13 14      cols sh 7  7    8    9    10   11   12   13   14   This is 1-indexed  if you prefer zero indexed  use i s   1 instead    Moreover  if you would like to have to arguments for the starting index and end index  change the file to      bin bash awk -v s  1 -v e  2   for i s  i lt  e i    printf   -5s    i  print         For example     echo  1 2 3 4 5 6 7 8 9 10 11 12 13 14      cols sh 7 9  7    8    9   The  -5s aligns the result as 5-character-wide columns  if this isn t enough  increase the number  or use  s  with a space  instead if you don t care about alignment

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As I was annoyed by the first highly upvoted but wrong answer I found enough to write a reply there  and here the wrong answers are marked as such  here is my bit  I do not like proposed solutions as I can see no reason to make answer so complex   I have a log where after  5 with an IP address can be more text or no text  I need everything from the IP address to the end of the line should there be anything after  5  In my case  this is actualy withn an awk program  not an awk oneliner so awk must solve the problem  When I try to remove the first 4 fields using the old nice looking and most upvoted but completely wrong answer   echo    7 27 10 16  Thu 11 57 18 37 244 182 218 one two three    awk    1  2  3  4     printf    s  n    0     it spits out wrong and useless response  I added    to demonstrate         37 244 182 218 one two three    Instead  if columns are fixed width until the cut point and awk is needed  the correct and quite simple answer is   echo    7 27 10 16  Thu 11 57 18 37 244 182 218 one two three    awk   printf    s  n   substr  0 28      which produces the desired output    37 244 182 218 one two three

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For me the most compact and compliant solution to the request is     a  1   2 t  t3     4   5   6 7  t 8 t       echo -e   a    awk -v n 3   while  i lt n   i    sub  1 FS           print  0     And if you have more lines to process as for instance file foo txt   don t forget to reset i to 0     awk -v n 3   i 0  while  i lt n   i    sub  1 FS           print  0   foo txt   Thanks your forum

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awk    1  2  3     0  0  1  1 1    Input  1 2 3 4 5 6 7   Output  4 5 6 7

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A solution that does not add extra leading or trailing whitespace   awk    for i 4  i lt NF  i    printf   s   i OFS  if NF  printf   s   NF  printf ORS        Example       echo  1 2 3 4 5 6 7      awk   for i 4 i lt NF i   printf  s   i OFS if NF printf  s   NF printf ORS       tr      -  4-5-6-7   Sudo O proposes an elegant improvement using the ternary operator NF ORS OFS    echo  1 2 3 4 5 6 7      awk    for i 4  i lt  NF  i    printf   s   i  i  NF ORS OFS         tr      -  4-5-6-7   EdMorton gives a solution preserving original whitespaces between fields     echo  1   2 3 4   5    6 7      awk    sub            3        1      tr      -  4---5----6-7   BinaryZebra also provides two awesome solutions   these solutions even preserve trailing spaces from original string     echo -e   1   2 t  t3     4   5   6 7  t 8 t       awk -v n 3    for   i 1  i lt  n  i      sub     FS      FS     FS         0      1       sed  s     g s  t - gt  g s      s        4   5   6 7 - gt  8- gt       echo -e   1   2 t  t3     4   5   6 7  t 8 t       awk -v n 3    print gensub    FS       FS     FS      n       1          sed  s     g s  t - gt  g s      s        4   5   6 7 - gt  8- gt      The solution given by larsr in the comments is almost correct     echo  1 2 3 4 5 6 7       awk   for  i 3 i lt  NF i      i-2   i  NF NF-2  print  0     tr       -  3-4-5-6-7   This is the fixed and parametrized version of larsr solution     echo  1 2 3 4 5 6 7       awk   for i n i lt  NF i     i- n-1    i NF NF- n-1  print  0   n 4   tr      -  4-5-6-7   All other answers before Sep-2013 are nice but add extra spaces    Example of answer adding extra leading spaces     echo  1 2 3 4 5 6 7       awk    1  2  3    1       tr       -  ---4-5-6-7  Example of answer adding extra trailing space     echo  1 2 3 4 5 6 7       awk   for i 4 i lt  13 i   printf   s    i printf   n         tr      -  4-5-6-7-------

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Options 1 to 3 have issues with multiple whitespace  but are simple   That is the reason to develop options 4 and 5  which process multiple white spaces with no problem  Of course  if options 4 or 5 are used with n 0 both will preserve any leading whitespace as n 0 means no splitting   Option 1  A simple cut solution  works with single delimiters      echo  1 2 3 4 5 6 7 8    cut -d    -f4- 4 5 6 7 8   Option 2  Forcing an awk re-calc sometimes solve the problem  works with some versions of awk  of added leading spaces     echo  1 2 3 4 5 6 7 8    awk     1  2  3     0  0   NF NF  4 5 6 7 8   Option 3  Printing each field formated with printf will give more control     echo      1    2  3     4   5   6 7     8        awk -v n 3    for  i n 1  i lt  NF  i    printf   s s   i i  NF RS OFS       4 5 6 7 8   However  all previous answers change all FS between fields to OFS  Let s build a couple of solutions to that   Option 4  A loop with sub to remove fields and delimiters is more portable  and doesn t trigger a change of FS to OFS     echo      1    2  3     4   5   6 7     8      awk -v n 3    for i 1 i lt  n i      sub     FS      FS     FS         0      1   4   5   6 7     8   NOTE  The     FS     is to accept an input with leading spaces   Option 5  It is quite possible to build a solution that does not add extra leading or trailing whitespace  and preserve existing whitespace using the function gensub from GNU awk  as this     echo      1    2  3     4   5   6 7     8      awk -v n 3    print gensub    FS       FS     FS      n       1      4   5   6 7     8    It also may be used to swap a field list given a count n     echo      1    2  3     4   5   6 7     8        awk -v n 3    a gensub    FS       FS     FS      n       1                   b gensub         a       1  1                   print    a       b                        4   5   6 7     8         1    2  3         Of course  in such case  the OFS is used to separate both parts of the line  and the trailing white space of the fields is still printed   Note1    FS    is used to allow leading spaces in the input line

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echo 1 2 3 4 5  awk    for  i 3  i lt  NF  i    print  i

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I can t believe nobody offered plain shell   while read -r a b c d  do echo   d   done  lt  file

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AWK printf-based solution that avoids   problem  and is unique in that it returns nothing  no return character  if there are less than 4 columns to print   awk  NF  gt  3   for i 4  i lt NF  i    printf   s      i    print   i       Testing     x  1 2 3  s 4 5 6    echo   x    awk  NF  gt  3   for i 4  i lt NF  i    printf   s      i    print   i      s 4 5 6   x  1 2 3    echo   x    awk  NF  gt  3   for i 4  i lt NF  i    printf   s      i    print   i       x  1 2 3     echo   x    awk  NF  gt  3   for i 4  i lt NF  i    printf   s      i    print   i

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Cut has a --complement flag that makes it easy  and fast  to delete columns   The resulting syntax is analogous with what you want to do -- making the solution easier to read understand   Complement also works for the case where you would like to delete non-contiguous columns      foo  1 2 3  s 5 6 7    echo   foo    cut --complement -d    -f1-3  s 5 6 7

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Pretty much all the answers currently add either leading spaces  trailing spaces or some other separator issue  To select from the fourth field where the separator is whitespace and the output separator is a single space using awk would be   awk   for i 4 i lt  NF i   printf   s   i  i  NF ORS OFS    file   To parametrize the starting field you could do   awk   for i n i lt  NF i   printf   s   i  i  NF ORS OFS    n 4 file   And also the ending field   awk   for i n i lt  m  m gt NF NF m  i   printf   s   i  i  m ORS OFS    n 4 m 10 file

[awk] Print all but the first three columns

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