Hi All

I'm strugglin with the analysis of DEP in a protein dataset i have. I have managed to analyse to this point in R using variants of MSGF etc and quantified via SI. Having run the samples via DAPAR and Prostar, i have knowldege of the number (or at least the region) of DEP to expect in this particular comparison (eg. CTRL to 0.01 mg/L). Using msmsTests, i create the hypothesis tests (H0 and H1), subset the data set to only these two groups and then run msms.EdgeR() with div = to the colSums().

This all runs fine but when i get to the portion or adding FDR to the data via test.results, something weird happens where instead of getting a dataset (so to speak) of protein id, raw spectral counts for control condition, raw spectral conts for comparitive condition, logfold change (LFC.AV) (logfc, LR, p.value etc), the second column (which is the raw spectral counts for the ctrl condition is always NA irrespective of which concentration i choose. I''m unsure why this is the case, when the msms.EdgeR runs ok with the comparisons. Has this happened to anyone else? Is there a particular reason this might occur that i can check to see if my data has been entered incorrectly?

I'm scratching my head over what would cause this. I did wonder whether it had anything to do with my estimates of dispersion used in the msms.EdgeR model build but i'm unsure how to proceed now.

Any help would be much appreciated. I can of course provide the code but i am hoping someone else has stumbled across this.

Many thanks :) 

I'll attempt an answer here. I would suggest you reformulate your logical operator that checks whether you have any missing or negative values. The double operators && and || can be very confusing. Here's an example showing that they don't necessarily work in cases such as yours.

> x
   F1  F2   F3
P1  1 1.2  1.3
P2  1 1.0  1.0
P3  1 1.0  1.0
P4  1 1.0 -1.0
>  is.na(x) || x < 0
[1] FALSE
> anyNA(x)
[1] FALSE
> any(x < 0)
[1] TRUE

Also, care should be taken in the presence of NAs

> x[4, 3] <- 1
> x[2, 2] <- NA
> x
   F1  F2  F3
P1  1 1.2 1.3
P2  1  NA 1.0
P3  1 1.0 1.0
P4  1 1.0 1.0
>  is.na(x) || x < 0
[1] FALSE
> anyNA(x)
[1] TRUE
> any(x < 0)
[1] NA

You can also use any(..., na.rm = TRUE).

I would suggest to use the following syntax

> x
   F1  F2  F3
P1  1 1.2 1.3
P2  1  NA 1.0
P3  1 1.0 1.0
P4  1 1.0 1.0
> any(is.na(x) | x < 0)
[1] TRUE
> x[2, 2] <- 1
> x[4, 3] <- -1
> x
   F1  F2   F3
P1  1 1.2  1.3
P2  1 1.0  1.0
P3  1 1.0  1.0
P4  1 1.0 -1.0
> any(is.na(x) | x < 0)
[1] TRUE

Re the double form of the logical operators, have a look at the manual page (?& will get you there), which says:

‘&’ and ‘&&’ indicate logical AND and ‘|’ and ‘||’ indicate logical OR. The shorter form performs elementwise comparisons in much the same way as arithmetic operators. The longer form evaluates left to right examining only the first element of each vector. Evaluation proceeds only until the result is determined. The longer form is appropriate for programming control-flow and typically preferred in ‘if’ clauses.

Hope this help!