Delete multiple columns using awk

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I have a database with 6037 columns and 450 rows as 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.

Thanks in advance.

linux bash sed awk

  • Using a loop would not be enough?? awk -F" " '{ for(i=1; i<=$qntColunas; i++) print $i }'

  • 1

    Thanks, I got it sorted.

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4 answers

1

Create a cols.awk script file: 

{
  for (i = 1; i < 676; i++)
    printf("%s ", $i)
  printf($676 "\n")
}

shell run the command: 

awk -f cols.awk seu_arquivo > novo_arquivo

That’s it.

  • Interesting, thank you.

1

Good night. I managed to resolve this issue with the script below:

awk '{for (i=1;i<=676;i++) {printf (i==1?"":FS)$i}; print ""}' <file.in >file.out

The input file is "file.in" with 6037 columns. The output file is "file.out" with the first 676 columns. To get other files with other amounts of initial columns just change the number 676 by other amounts of columns.

Thank you.

1

  • Thanks, I got it sorted.

1

I suggest:

awk 'NF=676'  arquivo > novo

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