Question

summing integer RLE

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gen ▴ 30

@gen-7383

Last seen 9.7 years ago

United States

Hi,

Say you have something like this RleListexample

RleList of length 4
$seq1
integer-Rle of length 1575 with 42 runs
  Lengths: 982   1   1   2   3   2   2   1   1   1 ...   1   1   1   2   1   1   3   1 531
  Values :   0   2   3   6   8  10  12  13  15  17 ...  12  11  10   8   7   5   4   2   0

$seq2
integer-Rle of length 1584 with 1 run
  Lengths: 1584
  Values :    0

$seq1
integer-Rle of length 1575 with 42 runs
  Lengths: 982   1   1   2   3   2   2   1   1   1 ...   1   1   1   2   1   1   3   1 531
  Values :   0   2   3   6   8  10  12  13  15  17 ...  12  11  10   8   7   5   4   2   0

$seq2
integer-Rle of length 1584 with 1 run
  Lengths: 1584
  Values :    0

How would you add or average the values of only the seq1 RLEs into one single RLE object?

When I try sum(RleListexample) it collapses everything into one number. When I try RleListexample$seq1 it doesn't provide me with the two RLEs in the list it just provides me with one.

rle list irange grangeslist • 2.1k views

ADD COMMENT • link updated 9.7 years ago by Chris Seidel ▴ 80 • written 9.7 years ago by gen ▴ 30

score 1 · Answer 1 · 2015-04-01

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Chris Seidel ▴ 80

@chris-seidel-5840

Last seen 3.6 years ago

United States

Your Rle object has the same name for more than one element. Since this doesn't provide specificity, if you want to sum a particular element, you might refer to it specifically (i.e. not by name):

sum(RleListexample[[1]])

Or if you want to sum all the elements named "seq1" you might create a boolean index:

sum(RleListexample[names(RleListexample) == "seq1"])

ADD COMMENT • link 9.7 years ago Chris Seidel ▴ 80

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Hi, thanks but the answer collapses the RLE into one number. I need to find the average RLE in an RLElist. IE for two sets of coverage data. Combine them into one set and then average it out for mean coverage.

thanks

ADD REPLY • link 9.7 years ago gen ▴ 30

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Maybe you are looking for something like (RleListexample[1:2] + RleListexample[3:4]) / 2 ? As Chris suggested, it's probably more usual to think of averaging two separate objects

library(GenomicAlignments)
cvg1 = coverage(BamFile("first.bam"))
cvg2 = coverage(BamFile("second.bam"))
avecvg = (cvg1 + cvg2) / 2

ADD REPLY • link 9.7 years ago Martin Morgan 25k

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cool it works for individual RLEs. I might need to do something with lists though.

ADD REPLY • link 9.7 years ago gen ▴ 30