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.
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written
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Sillpositive •
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
Thank you finswimmer for you response ! But with bcftool it's possible to keep the vcf.gz format ? thank you !
Yes, of course. I've edited my answer.
I also used -Oz option to have vcf.gz format !
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.
Thank you very much for your answer Yes my objective was to delete the lines containing ".''
So I will use instead ($4 !="." && $5 !=".").
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Thank you very much for your answer Yes my objective was to delete the lines containing ".'' So I will use instead
($4 !="." && $5 !=".").