Print first colum is the condition is true using awk

Question

I have a binary file which looks like this,

Orthogroup  A   B   C   D   E   F
OG0000000   1   1   1   0   0   0
OG0000001   1   1   1   0   0   0
OG0000002   1   1   1   0   0   0
OG0000003   1   1   1   0   0   0
OG0000004   0   0   0   1   1   0
OG0000005   0   0   0   1   1   0
OG0000006   0   0   0   1   1   0
OG0000007   0   0   0   1   1   0
OG0000008   1   0   0   0   0   0
OG0000009   1   0   0   0   0   0
OG0000010   1   0   0   0   0   0

I want to use awk to print the first column if column D and E is 1 and rest is 0.

OG0000004
OG0000005
OG0000006
OG0000007

I tried to do this with individual files of each column using the awk command

awk 'NR==FNR{arr[$0];next} $0 in arr' file1 file2 

but doing this I had to generate multiple files and I thought using a binary file will be a more clean way.

Please Help,

Answer 1

I would do something like this:

$ cat test.tsv 
groupA  1       1       1       0       0       0
groupB  1       1       0       0       0       0
groupC  1       0       1       0       0       0
groupD  1       1       1       0       0       0
groupE  0       0       0       1       1       1
groupF  0       0       0       1       1       0

$ awk '($5==1 && $6==1 && $2+$3+$4+$7==0){print $1}'
groupF

Answer 2

Straightforwardly

awk '$2==0 && $3==0 && $4==0 && $5==1 && $6==1 && $7==0 {print $1}' bin.file

A little spiffier

awk '$5 && $6 && !($2 || $3 || $4 || $7) {print $1}' bin.file