Involve tag count information in coverage function (IRanges, chipseq package)?
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Xiaohui Wu ▴ 280
@xiaohui-wu-4141
Last seen 10.3 years ago
Hi all, I'm using chipseq package to find peak in my count data. The R code is like this: ranges = IRanges(start=tag$start, end=tag$end,width=tag$end- tag$start+1, names = tag$tag_id) gr <- GRanges(seqnames = seqnames,ranges = ranges,strand =strand) gr.ext <- resize(gr, width = 200) cov <- coverage(gr.ext) islands <- slice(cov, lower = 1) thd=peakCutoff(cov, fdr = 0.05) peaks <- slice(cov, lower = thd) I found that the processing is not I expected, where tag was expanded to fragment and then calculate coverage. For my data, the data format is like: <chr,strand,tag_start,tag_end,count>, where count is the count for that tag. How to use the "count" information to calculate the coverage? Otherwise I think I need to repeat the tag for 'count' times and the do coverage, this is too trouble. Thank you! Regards, Xiaohui [[alternative HTML version deleted]]
ChIPSeq Coverage chipseq ChIPSeq Coverage chipseq • 1.2k views
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Entering edit mode
Xiaohui Wu ▴ 280
@xiaohui-wu-4141
Last seen 10.3 years ago
Sorry to bother you, problem solved. There is 'weight' parameter for cov <- coverage(gr.ext,weight=..) Xiaohui ---------------------- Hi all, I'm using chipseq package to find peak in my count data. The R code is like this: ranges = IRanges(start=tag$start, end=tag$end,width=tag$end- tag$start+1, names = tag$tag_id) gr <- GRanges(seqnames = seqnames,ranges = ranges,strand =strand) gr.ext <- resize(gr, width = 200) cov <- coverage(gr.ext) islands <- slice(cov, lower = 1) thd=peakCutoff(cov, fdr = 0.05) peaks <- slice(cov, lower = thd) I found that the processing is not I expected, where tag was expanded to fragment and then calculate coverage. For my data, the data format is like: <chr,strand,tag_start,tag_end,count>, where count is the count for that tag. How to use the "count" information to calculate the coverage? Otherwise I think I need to repeat the tag for 'count' times and the do coverage, this is too trouble. Thank you! Regards, Xiaohui [[alternative HTML version deleted]]
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