bash - Print out lines have value great than 1 with awk -


i trying pull out data in file of value in line great 1. guys know easy way this?

awk -f"\t" '{if(($3>1)||($4>1)||($5>1)||($6>1)||($7>1)||($8>1)||($9>1)||($10>1)||($11>1)||($12>1)||($13>1)||($14>1)||($15>1)||($16>1)||($17>1)||($18>1))print$0}' file1.txt > file2.txt

you can use for loop in awk instead of writing many many conditions:

awk -f"\t" '{for(i=3; i<=18; i++){if($i>1){print $0; next;}}}' filename

if wish comapare column 3 last column, use nf instead of 18 in loop condition.


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