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
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