How to make contrast table to combine two conditions/treatments against another one (in a multi groups experiments
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xyliu00 ▴ 20
@xyliu00-8370
Last seen 6.8 years ago
United States

Hi,

Novice users here, with both R in general and bioconductor/limma in particular.

I can perform Affy array data analysis following the tutorials. But I have something too complex for me.

The data set I am working on has multiple groups or treatments. The lab wants to compare between those treatments, which is easy to do. It is also requested to combine groups to a larger one and comparing to another group. 

For example, I have A, B, C, D four groups. A is the control, B, C, D are different treatments or treated for different period time. It is easy for me to make contrast table to compare B, C, D to A separately. 

Now, if B and C are samples treated with the same reagent, but for different period of time, and I need to compare B+C vs A; also C and D have some thing in common, and I need to make a comparison C+D vs B.

How do make the design and contrast matrix to do all of those. Can I include all of these in one matrix, or I need to make multiple matrix?

Sorry for lacking of better description. I am not familiar with all the right terminology. R is quite different for me, coming from more formal programming languages.

Thanks a lot! 

limma design and contrast matrix • 7.7k views
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Aaron Lun ★ 27k
@alun
Last seen 38 minutes ago
The city by the bay

Let's say we have a grouping vector for your samples, like below:

grouping <- c("A", "A", "B", "B", "C", "C", "D", "D")

We use this to construct a design matrix for our experiment:

grouping <- factor(grouping)
design <- model.matrix(~0 + grouping)
colnames(design) <- levels(grouping)

Now, to recapitulate your easy comparisons - if we want to compare between any two groups, say, A and B, we can do something like this (and feed it to contrasts.fit as described in the user's guide):

con <- makeContrasts(A - B, levels=design)

For each gene, the above contrast will test whether the average log-expression for samples of group A is equal to that of group B. For a more complex comparison between, say, B + C against A, we can do something like this:

con.2 <- makeContrasts(A - (B+C)/2, levels=design)

This will test for whether the log-expression of group A is equal to the average of groups B and C. As you can see, the same design matrix is used, so construction of a separate matrix is not necessary. Keep in mind, though, that DE genes from con.2 are not guaranteed to be DE in both A - B and A - C comparisons. Consider a gene that is strongly DE between A and B, yet not DE between A and C. This gene will still be able reject the null in con.2, as the log-expression in A will be different from the average of B and C.

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hi @ Aaron Lun, actually I have done the same thing you suggest here but I am getting an error.

My codes are:

 

> samples <- c(eset1$characteristics, eset$characteristics)
 

> samples <- as.factor(samples)

>samples

>design <- model.matrix(~0 + samples)

>colnames(design) <- c( "TUMOR", "NORMAL")

>design

>levels(samples)

>library(limma)

>fit <- lmFit(filteredEset, design)

>contrast.matrix <- makeContrasts("TUMOR-NORMAL", levels = "samples")

 

Error:  Error in eval(expr, envir, enclos) : object 'TUMOR' not found

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A cursory inspection of the user's guide would reveal the correct call:

makeContrasts(TUMOR - NORMAL, levels=design)
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als not working

same error.

 

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Are you sure you're running the code correctly? Here's an example that works on my machine:

samples <- factor(rep(c("TUMOR", "NORMAL"), each=3))
design <- model.matrix(~0 + samples)
colnames(design) <- levels(samples)
con <- makeContrasts(TUMOR - NORMAL, levels=design)
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