R help - Jan 2013 - centering matrix

? stato filtrato un testo allegato il cui set di caratteri non era
indicato...
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Hello,

The question is a bit confusing.
If you nedd to compute B = X'X, all you have to do is

B <- t(X) %*% X

If you want to compute the other formula, the following avoids loops.

n <- nrow(d)
B <- -d^2/2 - rowSums(d^2)/n - colSums(d^2)/n + sum(d^2)/n^2


Hope this helps,

Rui Barradas

Em 28-01-2013 22:35, Eleonora Schiano escreveu:
> I have a matrix d representing distances.
> I need to find B=X'X
> X=matrix that generated these distances.+
>
> B[i,j]=-(1/2)*{[d[i,j]^2-[(1/n)*sum for i=1...n(d[i,j])^2]-[(1/n)sum for
> j=1...n(d[i,j])^2]+[(1/n^2)sum for i=1...n sum for j=1...n (d[i,j])^2]}
>
> it's right if i do
> B=matrix(,nrow=dim(d)[1],ncol=dim(d)[2])
> for (i in 1:dim(d)[1])
> for (j in 1:dim(d)[2]){
> B=(-(1/2))*(d[i,j]^2-d[i,]^2-d[,j]^2+d[,]^2)
> }
> or i have to put
> B[i,j]=(-(1/2))*(d[i,j]^2-d[i,]^2-d[,j]^2+d[,]^2)
> ??
>
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
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> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

Hello,

Please keep the discussion on the R-Help list, there's no reason not to.
As for your question, can you be more specific?
Also, take a look at function ?scale.

Rui Barradas
Em 29-01-2013 23:31, Eleonora Schiano escreveu:
> i have to do multidimensional scaling.
> Can you help me?
>
>
> 2013/1/29 Rui Barradas <ruipbarradas at sapo.pt>
>
>> Hello,
>>
>> The question is a bit confusing.
>> If you nedd to compute B = X'X, all you have to do is
>>
>> B <- t(X) %*% X
>>
>> If you want to compute the other formula, the following avoids loops.
>>
>> n <- nrow(d)
>> B <- -d^2/2 - rowSums(d^2)/n - colSums(d^2)/n + sum(d^2)/n^2
>>
>>
>> Hope this helps,
>>
>> Rui Barradas
>>
>> Em 28-01-2013 22:35, Eleonora Schiano escreveu:
>>
>>> I have a matrix d representing distances.
>>> I need to find B=X'X
>>> X=matrix that generated these distances.+
>>>
>>> B[i,j]=-(1/2)*{[d[i,j]^2-[(1/**n)*sum for
i=1...n(d[i,j])^2]-[(1/n)sum
>>> for
>>> j=1...n(d[i,j])^2]+[(1/n^2)sum for i=1...n sum for j=1...n
(d[i,j])^2]}
>>>
>>> it's right if i do
>>> B=matrix(,nrow=dim(d)[1],ncol=**dim(d)[2])
>>> for (i in 1:dim(d)[1])
>>> for (j in 1:dim(d)[2]){
>>> B=(-(1/2))*(d[i,j]^2-d[i,]^2-**d[,j]^2+d[,]^2)
>>> }
>>> or i have to put
>>> B[i,j]=(-(1/2))*(d[i,j]^2-d[i,**]^2-d[,j]^2+d[,]^2)
>>> ??
>>>
>>>          [[alternative HTML version deleted]]
>>>
>>> ______________________________**________________
>>> R-help at r-project.org mailing list
>>>
https://stat.ethz.ch/mailman/**listinfo/r-help<https://stat.ethz.ch/mailman/listinfo/r-help>
>>> PLEASE do read the posting guide http://www.R-project.org/**
>>> posting-guide.html
<http://www.R-project.org/posting-guide.html>
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>>
>