Thank you, actually, there are two species. The 3rd genome is the combination of them (1st: h. 2nd: m and 3rd: hm). I want to make sure that reads mapped to all the three genomes were removed in the 3rd bam file. Below is my code:

''' p1 <- ScanBamParam(what=c('qname','mapq'),tagFilter =list(xf=c(25)) ,tag=c('xf'))

in 10X's bam file, xf=25 means that read is uniquely mapped to a genome, and was used for counting UMI. So I only look at reads with xf=25 among the 3 bam files.

filter <- FilterRules(list(fun = function(hm_reads) {

anab_m <- scanBam(BAM_m, param=p1)

anab_h <- scanBam(BAM_h, param=p1)

df_m <- do.call(DataFrame, anab_m[[1]])

df_h <- do.call(DataFrame, anab_h[[1]])

Find qnames: which are mapped to the 3 genome at the same time

internames<-Reduce(intersect,list(hm_reads$qname,df_h$qname,df_m$qname))

pickl<-rep(TRUE,nrow(hm_reads))

names(pickl)<-hm_reads$qname

If there are multiple mapped reads, then remove it in hm bam file

if(length(internames)>0){ pickl[internames]<-FALSE }

return(pickl) }))

print('now, subset BAM files')

outbam<-paste(OUTPATH,Sample,'-',IND_SECTION,'.bam',sep='')

open(BAM_h)

open(BAM_m)

filterBam(BAM_hm, destination=outbam, filter = filter, param = p1)

close(BAM_h)

close(BAM_m) '''

May I ask if this is correct? Many thanks!