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.

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

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.

  • 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.

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