add count comparison functionality to clonalCompare()

[Qile0317]

To address #428, this PR adds the argument proportion = TRUE to clonalCompare(). The default behvaiour of the function is identical to previous versions, but when prop = FALSE it uses counts instead of frequecies.

A minor change is that when exportTable=TRUE the output now is sorted first if the argument was specified. Technically a breaking change but I doubt people would be affected.

The functionality works but when I test the output, I divide the exported counts by the total number of clones and try to see if it would be identical to the proportions but its not. I think I am mis-understanding the exact behaviour here @ncborcherding. Is it instead supposed to be divided only by potentially filtered counts?

[ncborcherding]

For proportion:

tbl[, 2] <- tbl[, 2] / sum(tbl[, 2])

This will take the individual clone count and divide by the size of the entire repertoire (number of clones * counts). If we were going to do by clones, it would be:

tbl[, 2] <- tbl[, 2] / length(tbl[, 2])

Sorry I went ahead and made a commit before I saw your question, which I think was probably you working through things. My bad.

Nick

[Qile0317]

@ncborcherding All good - I synced up my stuff and updated the test. Could you possible edit the added test file to reflect the intended behaviour? A bit confused here.

I was (perhaps mistakenly) just thinking that the feature requester in #428 wanted the Y axis to be raw totaled counts of all clones. But from your comment above the definition for when proportion=FALSE, it is normalized by repertoire diversity (number of clonotypes) - right?

[Qile0317]

@ncborcherding All good - I synced up my stuff and updated the test. Could you possible edit the added test file to reflect the intended behaviour? A bit confused here.

I was (perhaps mistakenly) just thinking that the feature requester in #428 wanted the Y axis to be raw totaled counts of all clones. But from your comment above the definition for when proportion=FALSE, it is normalized by repertoire diversity (number of clonotypes) - right?

ok, so I did this now everything is passing. I checked that for each sample group the scale factor from proportion to counts are identical. @ncborcherding Should I change the description of the function to highlight this "diversity normalization"? Otherwise this is ready for merge.