I know this should be an easy one by googling but was not successful. Sorry for that.
I would like to print the first line of groups defined the value in the first column. Delimiter is tab.
Input:
A 5
A 3
B 2
B 1
B 77
C 4
C 10000
D 99
Output:
A 5
B 2
C 4
D 99
The shortest one:
awk -F'\t' '!a[$1]++' file
The output:
A 5
B 2
C 4
D 99
!a[$1]++ - ensures line printing on encountering the first unique value of the 1st column
{print $0} see https://unix.stackexchange.com/questions/159695/how-does-awk-a0-work/159697#159697
– Archemar
Jun 21 '17 at 11:41
Something like can do the work:
awk -F\t 'BEGIN {A=""} {if ($1!=A) { print $0; A=$1}}' input_file
When you initialize variable A select initial value which is not in to the list of existing in column 1
-F'\t', with -F\t it outputs all lines for me. https://ibb.co/bZaMNk
– RomanPerekhrest
Jun 21 '17 at 12:58
Here are two non-awk options:
sort u foo -k 1,1
Use sort purely for the -u (--unique) capability. Use only the first character (-k 1,1) for comparison.
rev foo | uniq -f 1 | rev
Use uniq. The -f option only allows specifying a start field for comparison, so we use some legerdemain to first reverse (rev) the input before uniq, and then rev that output again.
