[R] Inverting a scale(X)

[Godfrey van der Linden]

On 2010-07-03, at 17:43 , Peter Ehlers wrote:

> 
> On 2010-07-03 0:05, Godfrey van der Linden wrote:
>> G'day, All.
>> 
>> I have been trying to trackdown a problem in my R analysis script. I perform a scale() operation on a matrix then do further work.
>> 
>> Is there any way of inverting the scale() such that
>>     sX<- scale(X)
>>     Xprime<- inv.scale(x); 	# does inv.scale exist?
>> 
>> resulting in Xprime_{ij} == X_{ij} where Xprime_{ij} \in R
>> 
>> There must be some way of doing it but I'm such a newb that I haven't been able to find it.
>> 
>> Thanks
>> 
>> Godfrey
>> 
> 
> If your sX hasn't lost the "scaled:center" and
> "scaled:scale" attributes that it got from the
> scale() operation, then you can just reverse
> the scaling procedure using those. Multiply
> columns by the "scale" attribute, then add the
> "center" attribute. Something like:
> 
> MN <- attr(sx, "scaled:center")
> SD <- attr(sx, "scaled:scale")
> Xprime <- t(apply(sx, 1, function(x){x * SD + MN}))
> 
> If the attributes have been lost by your further
> work, then I'm afraid you're out of luck.
> 
>  -Peter Ehlers

Thanks for this, I had forgotten the transpose function existed. I did maintain the attributes, though I was surprised how many times I had to move them manually in my script.

Anyway I also tried a functional programming solution and I'm sort of curious what the differences are? I used rep to build out the SD and MN vectors. Something like this (though I have lost the precise code now and would have to regenerate it)

nr = nrow(sx)
Xprime = sx * rep(SD, each=nr) + rep(MN, each=nr);

Is there any way of determining which approach is more efficient?

Thanks again.

Godfrey