Hi Gordon Thanks for the response. I have another question. I am reading Imagene output files using read.maimages (Limma) or ImaGeneData$read (Aroma). The former can read both files simultaneously whereas the latter reads each file separately. I was using read.maimages until I found that I could not get the flag information from the data. At some point of pre-processing I need to exclude the spots with certain flag values associated with it (the flags are attached during image quantification). Suppose I would like to exclude all the spots with a flag value of more than 0. When I do this: myfun<-function(x) as.numeric(x$flags > 0) RG<-read.maimages(files,source="imagene",wt.fun=myfun) I get the message that it reads the images and then: Error in "[<-"(`*tmp*`,,I,value=numeric(0)) : Nothing to replace with I know that the files specified in the variable 'files' does have flags with higher values than zero. Was there a problem during the reading? Is there any other way to find the flag information? Also, I found that ImaGeneData$read (Aroma) does include flag information in the returned object. But I would have to read the flags manually and conditionally insert NAs for the corresponding R,G values. Thanks and regards, Saroj -----Original Message----- From: Gordon K Smyth [mailto:smyth@wehi.EDU.AU] Sent: Wednesday, July 14, 2004 6:34 PM To: saroj@wayne.edu Cc: bioconductor@stat.math.ethz.ch Subject: Re: [BioC] Difference between normalizeWithinArrays and stat.ma Print-tip loess normalization in limma is identical to that in sma (deliberately). However the limma command accommodates weights while the stat.ma() does not. I like sma but it is a no longer under development. Better to use one of the BioC packages under active development and support such as marray or limma. Gordon

Hi all, I have a basic question (perhaps too basic!) regarding RG to MA transformation. I understand the logic of the MA transformation as described by Terry Speed and other documents. My situation is a bit different from cDNA arrays. In this case, the red intensity refers to variable reactivity of a sample (hopefully, containing antibodies) against known antigens on the chip (each spot has different antigens). The green channel refers to reactivity against a constant protein (each spot has the same one) that is arrayed for the purpose of checking against variable protein deposit because of print-tip variation, day-to-day variation of the way in which the proteins are prepared, etc. The green intensity across the spots is never constant within a chip, indicating the variations as mentioned above. Therefore I think that the ratio of even a very reactive spot in the chip might or might not achieve a red:green ratio of 1. I wonder, if it has any impact on the issue of RG-to-MA transformation? M = log2(R/G) makes perfect sense to me, but I am not able to understand the significance of A = 1/2(log2(R.G)) in this case. Any pointers would be helpful. Thanks and regards Saroj Mohapatra --------------------- Saroj K Mohapatra, MD Research Associate Karmanos Cancer Institute Wayne State University School of Medicine 110 E. Warren, Room 311 Detroit MI 48201 313-833-0715 x2424 saroj@wayne.edu