Compare and filter three BAM files
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wt215 • 0
@wt215-12405
Last seen 6 weeks ago
United Kingdom

Hi,

I mapped fastq to 3 different reference genome. Now I want to find out reads which were mapped to the first two genome, then filter them out in the 3rd bam file.

I will then process the 3rd bam file for obtaining UMI count.

Is there any simple way to achieve this?

Many thanks!

RNAseq 10X • 143 views
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swbarnes2 ★ 1.2k
@swbarnes2-14086
Last seen 15 minutes ago
San Diego

If you think you have a mix of three different organisms in the ix, the better way to do this is to make one reference with all three genomes. Any reads that align best to one genome will align there.

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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!

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