Hi Petr,

Thank you. Yes, normalized.matrix" and "normalized" are the same, my mistakes.

Operators: like Cyclic loess normalization and quantile Normalization,
from limma or affy packages.

Would cv= sd/mean be a better replacement for sd?

On 4/5/12, Petr Savicky [via R]
<ml-node+s789695n4535407h40@n4.nabble.com> wrote:
>
>
> On Thu, Apr 05, 2012 at 07:46:52AM -0700, ali_protocol wrote:
>> Hi all,
>>
>> I have a matrix  (n*2), I want to compare 2 operators (2 normalization for
>> array results) on these matrix.
>> The 2 columns should ideally become the same after operations
>> (normalization). So to compare operations,
>> I do this for each normalization:
>>
>> s= sum (apply (normalized.matrix, 2,sd))
>> c= cor (normalized[,1],normalized [,2], method='pearson')
>>
>>
>> I expect that if normalization 1 is superior, s should be less and c
>> greater
>> than normalization2, but both s and c change in 1 direstion. Is this
>> possible or am I doing something wrong?
>
> Hi.
>
> Is "normalized.matrix" and "normalized" the same matrix?
>
> Can you specify, which operators you use for normalization?
>
> Without having this information, i guess that comparing the
> correlations alone can be used, since it does not depend on
> the scaling the numbers. By an appropriate scaling factor,
> the sd may be changed to any value, but this does not say
> much about the amount of information in the data. On the other
> hand, the correlation does not change by scaling, so it may
> be a more reliable measure.
>
> Note that
>
>   apply(normalized.matrix, 2, sd)
>
> is the same as
>
>   sd(normalized.matrix)
>
> Hope this helps.
>
> Petr Savicky.
>
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