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Perhaps also:
awk '$5!~/^[01]$/' file
or
awk '$5>1' file

Hey Amit,

grep doesn't really work on delimited row data, rather it works as individual character in the file.

in your case awk is always a better approach. You may try the below: (assuming you...

I think you need to specify a field separator as a comma.


Owner@Owner-PC ~
$ awk -F, '!a[$1]++' filename
ENSG00000003137,ENST00000001146
ENSG00000003402,ENST00000309955


Owner@Owner-PC ~...

Of course with -F flag:
awk -F, '!a[$1]++' file

Please give an example of what your output should look like. Because the file as listed has 0s and 1s on all lines, and therefore grep as you propose to use it will delete all lines,


Andrew

Try

$ echo chr6 27262994 27263391 194 | awk '$3=$2+$4' OFS=\\t
chr6 27262994 27263188 194OR

if you want to mask column 4 then
$ echo chr6 27262994 27263391 194 | awk...

If you want first column for each entry then just use..

awk '{for(i=2;i<=NF;i++){if(X[$i,i]++){$i=""}}}1' OFS="\t" file

Please use code tags to preserve formatting in the data samples. Your input is barely readable.
That one line? or multiple lines?
What's the expected output?

can you see the below post and use CODE tag

https://www.unix.com/how-post-unix-linux-forums/167686-forum-video-tutorial-how-use-code-tags.html