Hi,
I fit a GLM with edgeR like this:
>colnames(fit)
A B C D
For contrast = c(-1,0,1,0), I guess logFC = log2 (average CPM in C/
average
CPM in A).
But for contrast = c(1,-1,-1,1), how is logFC calculated?
Thanks,
Xinwei
[[alternative HTML version deleted]]
Dear Xinwei,
Well, in the first case you get C-A (C minus A). In the second case
you
get A-B-C+D.
Best wishes
Gordon
> Date: Mon, 1 Apr 2013 23:10:42 -0400
> From: Xinwei Han <xinwei.han at="" duke.edu="">
> To: <bioconductor at="" stat.math.ethz.ch="">
> Subject: [BioC] logFC in GLM of egdeR
>
> Hi,
>
> I fit a GLM with edgeR like this:
>
>> colnames(fit)
> A B C D
>
> For contrast = c(-1,0,1,0), I guess logFC = log2 (average CPM in C/
average
> CPM in A).
>
> But for contrast = c(1,-1,-1,1), how is logFC calculated?
>
> Thanks,
> Xinwei
>
______________________________________________________________________
The information in this email is confidential and
intend...{{dropped:4}}
HI Gordon,
Thanks. Does that mean in the second case, basically logFC = log2 (
(average CPM in D+ average CPM in A) / (average
CPM in B +average CPM in C) ), except for other adjustments from
negative
binomial?
Thanks,
Xinwei
On Tue, Apr 2, 2013 at 7:03 PM, Gordon K Smyth <smyth@wehi.edu.au>
wrote:
> Dear Xinwei,
>
> Well, in the first case you get C-A (C minus A). In the second case
you
> get A-B-C+D.
>
> Best wishes
> Gordon
>
> Date: Mon, 1 Apr 2013 23:10:42 -0400
>> From: Xinwei Han <xinwei.han@duke.edu>
>> To: <bioconductor@stat.math.ethz.**ch <bioconductor@stat.math.ethz.ch="">>
>> Subject: [BioC] logFC in GLM of egdeR
>>
>> Hi,
>>
>> I fit a GLM with edgeR like this:
>>
>> colnames(fit)
>>>
>> A B C D
>>
>> For contrast = c(-1,0,1,0), I guess logFC = log2 (average CPM in C/
>> average
>> CPM in A).
>>
>> But for contrast = c(1,-1,-1,1), how is logFC calculated?
>>
>> Thanks,
>> Xinwei
>>
>>
> ______________________________**______________________________**____
______
> The information in this email is confidential and
inte...{{dropped:10}}
Dear Xinwei,
No, not quite.
It would be best to try to think in terms of linear comparisons on the
log-scale, because that is how glms works. A-B-C+D means
(average log CPM in A)
minus
(average log CPM in B)
minus
(average log CPM in C)
add
(average log CPM in D)
You can think of it as twice
(average log CPM in A and D)
minus
(average log CPM in B and C).
or you could think of it as
(log fold change from A to B)
plus
(log fold change from D to C)
but none of these are quite the same as what you wrote. Your
expression
interchanges averages and logs in a way that is not correct.
Note that the logFC you get from the contrast c(1,-1,-1,1) is exactly
equal to the logFC you get from c(1,-1,0,0) plus what you get from
(0,0,-1,1). It just adds up in the natural way.
Best wishes
Gordon
On Tue, 2 Apr 2013, Xinwei Han wrote:
> HI Gordon,
>
> Thanks. Does that mean in the second case, basically logFC = log2 (
> (average CPM in D+ average CPM in A) / (average CPM in B +average
CPM in
> C) ), except for other adjustments from negative binomial?
>
> Thanks,
> Xinwei
>
>
> On Tue, Apr 2, 2013 at 7:03 PM, Gordon K Smyth <smyth at="" wehi.edu.au=""> wrote:
>
>> Dear Xinwei,
>>
>> Well, in the first case you get C-A (C minus A). In the second
case you
>> get A-B-C+D.
>>
>> Best wishes
>> Gordon
>>
>> Date: Mon, 1 Apr 2013 23:10:42 -0400
>>> From: Xinwei Han <xinwei.han at="" duke.edu="">
>>> To: <bioconductor at="" stat.math.ethz.**ch="" <bioconductor="" at="" stat.math.ethz.ch="">>
>>> Subject: [BioC] logFC in GLM of egdeR
>>>
>>> Hi,
>>>
>>> I fit a GLM with edgeR like this:
>>>
>>> colnames(fit)
>>>>
>>> A B C D
>>>
>>> For contrast = c(-1,0,1,0), I guess logFC = log2 (average CPM in
C/
>>> average
>>> CPM in A).
>>>
>>> But for contrast = c(1,-1,-1,1), how is logFC calculated?
>>>
>>> Thanks,
>>> Xinwei
>>>
>>>
>> ______________________________**______________________________**___
_______
>> The information in this email is confidential and intended solely
for the
>> addressee.
>> You must not disclose, forward, print or use it without the
permission of
>> the sender.
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_______
>>
>
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