I want to print column 1 of this file. I used this command: awk '{print $1}' but it just printed the first word of the 1st column.
DATA
ABC transporters ABC transporters Alanine, aspartate and glutamate metabolism Alanine, aspartate alpha-Linolenic acid metabolism alpha-Linolenic acid metabolism Aminoacyl-tRNA biosynthesis Aminoacyl-tRNA biosynthesis Amino sugar and nucleotide sugar metabolism Amino sugar and nucleotide Arachidonic acid metabolism Arachidonic Output:
ABC Alanine, alpha-Linolenic Aminoacyl-tRNA Amino Arachidonic Desired Output:
ABC transporters Alanine, aspartate and glutamate metabolism alpha-Linolenic acid metabolism Aminoacyl-tRNA biosynthesis Amino sugar and nucleotide sugar metabolism Arachidonic acid metabolism 4 Answers
What I can see is that your columns are delimited by two space.
so with awk:
awk -F '\\s\\s' '{print $1}' Since this seems to be a fixed-width column, you can just cut the corresponding characters. The widest column Alanine, aspartate and glutamate metabolism seems to be 44 characters wide, so:
$ cut -c1-44 foo ABC transporters Alanine, aspartate and glutamate metabolism alpha-Linolenic acid metabolism Aminoacyl-tRNA biosynthesis Amino sugar and nucleotide sugar metabolism Arachidonic acid metabolism As the second column obviously repeats the beginning of the first column, I take this as criterion for the cut with sed, thus it does not depend on the column width:
sed 's/^\(.*\)\(.*\) \1$/\1\2/' First pattern is the repeated part, backreferenced as \1 at the end of the line. You could add ;s/ *$// to remove the trailing spaces if they bother you.
Building upon muru's answer that the column is specified with fixed width, using egrep command with option -o will allow you to print just the matched (non-empty) parts of a matching line specified by the search pattern. By default, however, entire line will be printed.
$ egrep -o "^.{44}" foo 
