awk: how to filter out a row that all entries (16 columns) are 0

Question

My CSV file is delimited by tab, and I want to filter out those rows that have all entries (in 16 columns) of 0 value. I am now doing this

awk '$1 != 0 && $2 != 0 && ....omitted && $16 != 0 {print $0}' file.csv > newFile.csv

As you can see this is so tiring for inputing the same conditions for all 16 columns. Is there any easier way?

Answer 1

What about something like

grep -Ev '^0+([[:space:]]+0+){15}$' file.csv

or

awk --posix '!/^0+([[:space:]]+0+){15}/'

Answer 2

How about something like this (assuming you have 16 fields throughout)

awk '{for (i=1; i<=16; ++i) if($i != 0) {print;next}}' file.csv > newFile.csv