Remove line with awk in vcf.gz

Question

[Sillpositive]

Hello everyone.

I have a vcf.gz file and I want to filter the columns that contain "." .

I did this zcat file.vcf.gz | grep -v "#" | awk ' $4=="." || $5=="." ' .

however I don't know how to delete them and save the new file in vcf.gz format .

Thank you for your help.

[Sillpositive]

Thank you very much for your answer Yes my objective was to delete the lines containing ".'' So I will use instead ($4 !="." && $5 !=".").

Answer 1

[finswimmer]

Don't use grep/awk/... to filter vcf files. Instead use programs that are specialized on doing this, like bcftools.

Column 4 and 5 are the REF and ALT column. So you like to exclude all rows that have no value there:

$ bcftools view -e "REF=='.'||ALT=='.'" -o output.vcf.gz input.vcf.gz

[Sillpositive]

Thank you finswimmer for you response ! But with bcftool it's possible to keep the vcf.gz format ? thank you !

[finswimmer]

Yes, of course. I've edited my answer.

[cpad0112]

is output type not necessary? -O ?

[Sillpositive]

I also used -Oz option to have vcf.gz format !

Answer 2

[ATpoint]

zcat file.vcf.gz | awk '$1 ~ /^#/ {print $0;next} {if ($4 == "." || $5 == "." ) print }' | bgzip > new.vcf.gz

Will print all entries where $4 or $5 is . to a new compressed VCF file (bgzip for compression) preserving the header lines starting with #.

$1 ~ /^#/ {print $0;next} essentially means that if the line starts with # then print it (to preserve header lines).

{if ($4 == "." || $5 == "." ) print } tests if $4 or $5 is . and prints the entire row if true.

If you wanted entries with no . in either of the columns, that would be ($4 != "." && $5 != "." )

---

Edit: Agree with finswimmer that specialized tools such as bcftools are preferred to avoid any possible file corruption.

[Sillpositive]

Thank you very much for your answer Yes my objective was to delete the lines containing ".'' So I will use instead ($4 !="." && $5 !=".").