whoops, backwards
new.x <- x[!remove.set,]
On Fri, Jul 20, 2012 at 2:51 PM, Richard M. Heiberger <r...@temple.edu>wrote:
> This works to multiply the ith row of a by the ith value of b.
> It might be what you can use
>
> a <- matrix(1:30, 6, 5)
> b <- 1:6
>
> a
> a*b
>
>
>
> To simplify your code, I think you can do this---one multiplication
>
> xA <- x %*% A
>
> Now you can do the tests on xA and not have any matrix multiplications
> inside the loop.
> The next line is not tested but I hope gives you the idea
>
> remove.set <- (xA[,1] < .5) || (xA[,2] < 52) || (xA[,3] > 150)
>
> new.x <- x[remove.set,]
>
> You now have new.x with no explicit loops at all, because this takes
> advantage of vector operations.
>
> Rich
>
> On Fri, Jul 20, 2012 at 2:34 PM, wwreith <reith_will...@bah.com> wrote:
>
>> That is faster than what I was doing and reducing 15% of my iterations it
>> still very helpful.
>>
>> Next question.
>>
>> I need to multiply each row x[i,] of the matrix x by another matrix A.
>> Specifically
>>
>> for(i in 1:n)
>> {
>> If (x[i,]%*%A[,1]<.5 || x[i,]%*%A[,2]<42 || x[i,]%*%A[,3]>150)
>> {
>> x<-x[-i,]
>> n<-n-1
>> }. #In other words remove row i from x if it does not meet criteria (>=.5,
>> >=42, <=150). When multiplied to A
>> }
>> Is there a better way than using a for loop for this or x<-x[-i,] for that
>> matter? I assume building a new matrix would be worse.
>>
>> Ideally I want to also exclude some x[,i] as well example if x[1,] is
>> better
>> than x[2,] in all three categories i.e. bigger, bigger, and smaller than
>> x[2,] when multiplied to A then I want to exclude x[2,] as well. Any
>> suggestions on whether it is better to do this all at once or in stages?
>>
>> Thanks for helping!
>>
>>
>>
>> --
>> View this message in context:
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>> Sent from the R help mailing list archive at Nabble.com.
>>
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>
>
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