Take nth column in a text file

Question

I have a text file:

1 Q0 1657 1 19.6117 Exp
1 Q0 1410 2 18.8302 Exp
2 Q0 3078 1 18.6695 Exp
2 Q0 2434 2 14.0508 Exp
2 Q0 3129 3 13.5495 Exp

I want to take the 2nd and 4th word of every line like this:

1657 19.6117
1410 18.8302
3078 18.6695
2434 14.0508
3129 13.5495

I'm using this code:

 nol=$(cat "/path/of/my/text" | wc -l)
 x=1
 while  [ $x -le "$nol" ]
 do
     line=($(sed -n "$x"p /path/of/my/text)
     echo ""${line[1]}" "${line[3]}""  >> out.txt
     x=$(( $x + 1 ))
 done

It works, but it is very complicated and takes a long time to process long text files.

Is there a simpler way to do this?

Answer 1

iirc :

cat filename.txt | awk '{ print $2 $4 }'

or, as mentioned in the comments :

awk '{ print $2 $4 }' filename.txt

Answer 2

You can use the cut command:

cut -d' ' -f3,5 < datafile.txt

prints

1657 19.6117
1410 18.8302
3078 18.6695
2434 14.0508
3129 13.5495

the

• -d' ' - mean, use space as a delimiter
• -f3,5 - take and print 3rd and 5th column

The cut is much faster for large files as a pure shell solution. If your file is delimited with multiple whitespaces, you can remove them first, like:

sed 's/[\t ][\t ]*/ /g' < datafile.txt | cut -d' ' -f3,5

where the (gnu) sed will replace any tab or space characters with a single space.

For a variant - here is a perl solution too:

perl -lanE 'say "$F[2] $F[4]"' < datafile.txt

Answer 3

For the sake of completeness:

while read -r _ _ one _ two _; do
    echo "$one $two"
done < file.txt

Instead of _ an arbitrary variable (such as junk) can be used as well. The point is just to extract the columns.

Demo:

$ while read -r _ _ one _ two _; do echo "$one $two"; done < /tmp/file.txt
1657 19.6117
1410 18.8302
3078 18.6695
2434 14.0508
3129 13.5495