5
$\begingroup$

I know that Voom function from limma package from Bioconductor converts raw counts into log-CPM values and then Normalization is applied on that, with normalize.method argument.

I would like to know clearly

  1. how this normalization is done (statistically) ?

  2. normalization across samples and normalization across genes. But don't the know how it is done statistically.

$\endgroup$
  • $\begingroup$ Hi! Did you read the papers in the reference section of voom help? What kind of normalization across samples and normalization across genes are you interested in? There are many normalization methods and for which kind of experiment do you mean: RNA-seq or others? $\endgroup$ – llrs Sep 13 '17 at 12:19
  • $\begingroup$ 1) Ya I have seen that: normalization method to be applied to the logCPM values. Choices are as for the method argument of normalizeBetweenArrays when the data is single-channel. So when I'm using counts data does it mean normalization between samples? 2) for RNA-Seq $\endgroup$ – stack_learner Sep 13 '17 at 14:04
  • 3
    $\begingroup$ Please edit your question to add extra information. Comments are easy to miss, hard to read and can be deleted without warning. Also, it would really help if you explain what kind of data you are dealing with, what experiment generated it, and mention whether you've looked at the relevant R docs and the papers they link to, explaining which parts are not clear to you. $\endgroup$ – terdon Sep 13 '17 at 14:14
  • 2
    $\begingroup$ Harold Pimentel has useful blog posts on normalisation across genes and across samples $\endgroup$ – heathobrien Sep 13 '17 at 15:54
0
$\begingroup$

I'd like to know answer for this myself. It seems the normalize.method argument is set to none by default. You can edit by looking at the methods in normalizeBetweenArrays. But how do I know if and when I have to use normalize.method in my analysis?
The output from voom is log2CPM of count values (adding 0.5 to them).

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.