Hi !

I need to compare the transcriptome of 5 cases and 5 controls using one-color Agilent microarrays. Weights were computed by the following function:

#My weight function
myFlagFun <- function(x) {
#Weight only strongly positive spots 1, everything else 0
present <- x$gIsPosAndSignif == 1
probe <- x$ControlType == 0
manual <- x$IsManualFlag == 0
strong <- x$gIsWellAboveBG == 1
y <- as.numeric(present & probe & manual & strong)

#Weight weak spots 0.5

weak <- strong == FALSE
weak <- (present & probe & manual & weak)
weak <- grep(TRUE,weak)
y[weak] <- 0.5

#Weight flagged spots 0.5

sat <- x$gIsSaturated == 0
xdr <- x$gIsLowPMTScaledUp == 0
featureOL1 <- x$gIsFeatNonUnifOL == 0
featureOL2 <- x$gIsFeatPopnOL == 0
flagged <- (sat & xdr & featureOL1 & featureOL2)
flagged <- grep(FALSE, flagged)
good <- grep(TRUE, y==1)
flagged <- intersect(flagged, good)
y[flagged] <- 0.5
y
}

Then I used the following script:

RG <- read.maimages(targets, source="agilent",green.only=TRUE, wt.fun=myFlagFun)

RGb <- backgroundCorrect(RG, method="none")

MA <- normalizeBetweenArrays(RGb, method="quantile")

MA.ave <- avereps(MA, ID=MA$genes$ProbeName)

f <- factor(targets$Condition, levels = unique(targets$Condition))
design <- model.matrix(~0 + f)
colnames(design) <- levels(f)

fit <- lmFit(MA.ave, design)

If I understood correctly, the weights are used in gene-wise weighted least squares regression to estimate the coefficients in the linear model. So they will affect the logFC values. My problem is that I need to get back to normalized individual measures so that their means are in line with coefficients calculated by the lmFit function. 

Here is an example of what I get on a single gene:

In controls:

The corresponding coefficient is 

If I understand you correctly, you want to compute some "normalised" value that accounts for the weight of each observation. In general, this is not possible, as the weights scale the contribution of each observation to the sum of squares and thus only make sense in the context of least-squares regression. However, you can get something for group means. Consider the weighted mean for your first example:

y <- c(5.61618123133188, 5.0746766862945, 4.8972404255748, 4.90207357931074, 5.68229237143083)
w <- c(1, 0.5, 0.5, 0, 1)
weighted.mean(y, w) # Gives 5.428144

The weighted mean of each group is calculated as sum(y*w)/sum(w), so you could define a modified expression value as:

z <- y*w/mean(w)
mean(z) # Gives 5.428144

Of course, var(z) will not be the sample variance, and z itself is largely gibberish when you look at its values - for example, a weight of zero gets a modified expression of zero, which is misleading as the corresponding value in y is decidedly non-zero. But, as requested, the (weighted) mean is the same as what you would get from performing weighted linear regression via lmFit.

I don't know what you intend to use z for, but it is far better to feed both the original observations and their associated weights into downstream procedures that can handle weights. If a procedure cannot handle weights, then supplying y without modification is almost certainly better than using z. The values of z are only "correct" in the sense that they give the same weighted mean as you would get from weighted linear regression - trying to compute any other statistic will probably give (wildly!) wrong results.

Thank you very much Aaron,

Actually, if I don't take into account the attributed weights the resulting logFC may be quite different from what I get with lmFit. When using the above example, I get -0.0807573981440699 if the weights are taken into account and 0.0813813767914962 if they are not. Even though the logFC is not significant we get opposite results. I told you that I would like to compare 5 cases and 5 controls but this is la little more complicated. I would like to compare two conditions over time using the TTCA R package. I would need to use normalized individual data to run the analysis. For each time point, I'm afraid that if I don't take into account these weighted corrections, discrepancies between limma and TTCA may lead to different results. If you have a better approach to handle this issue, please let me know what to do.

Thank you again for your time.