Closed skchronicles closed 5 years ago
We are currently transforming cpm filter, and by doing so, we are making making the filtering much less stringent. val1 = 0.5 val1=(val1/max(tot))*1e6 filter <- apply(cpm(mydata), 1, function(x) length(x[x>val1])>=val2)
val1 [1] 0.008896055
Issue Resolved: https://github.com/CCBR/Pipeliner/commit/576bb9a442ea36374beeaec0872011ea1dd441dd
skchronicles
commented
5 years ago
We are currently transforming cpm filter, and by doing so, we are making making the filtering much less stringent. val1 = 0.5 val1=(val1/max(tot))*1e6 filter <- apply(cpm(mydata), 1, function(x) length(x[x>val1])>=val2)