I went through the manuals and can't find the meaning of the below flag:

-filt_rules_l 'filter==0'

Here is the link to manual for this flag: filt_rules_l

Can anyone explain?

IMPUTE version:


Edit: The full IMPUTE2 code is irrelevant, I am not using the command to impute, I received an imputed data and the script used to impute. This one flag was a mystery to me. I will update the post with impute software version when I find out.


I have never used this program, but reading the manual in the link it has to mean the following.

The manual says:

Writing -filt_rules_l 'eur.maf<0.05' on the command line would tell the program to remove any variants with eur.maf values less than 0.05 from the reference panel.

So, -filt_rules_l 'filter==0' would mean that the program removes any variants with filter values of 0 from the reference panel. In the legend file there should be a column named filter, which are filtered out when the value is 0.

Here is the example legend file, in your case there must be a custom additional column named "filter":

$ head ALL_1000G_phase1integrated_v3_chr1_impute.legend

id position a0 a1 afr.aaf amr.aaf asn.aaf eur.aaf afr.maf amr.maf asn.maf eur.maf
rs58108140 10583 G A 0.0407 0.1713 0.1311 0.2071 0.0407 0.1713 0.1311 0.2071
rs189107123 10611 C G 0.0142 0.0276 0.0140 0.0211 0.0142 0.0276 0.0140 0.0211
  • $\begingroup$ Thank you, are you suggesting that reference panel has column named "filter"? I will have to check. But I think "eur.maf" column is calculated on the fly from the reference file, need to check. $\endgroup$
    – zx8754
    Nov 17 '17 at 10:48
  • 1
    $\begingroup$ If you want to use your command it should. If not it won't work, and you need to find the column on which you want to filter. $\endgroup$
    – benn
    Nov 17 '17 at 10:55
  • 1
    $\begingroup$ Yes, sorry I have missed the _l bit, meaning this filter works from legend file, which in my case apparently had a modified version of Phase3 legend file with extra "filter" column. All makes sense now. Thank you. $\endgroup$
    – zx8754
    Nov 20 '17 at 9:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.