Print the first column

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