[swift-users] hello, first time poster and accelerate question

[Jacob Bandes-Storch]

I must have forgotten how matrices work—I think it's actually this
direction. And I also lost an m :-)

[image: Inline image 1]


On Fri, Nov 18, 2016 at 7:55 PM, Jacob Bandes-Storch <jtbandes at gmail.com>
wrote:

> Can't you achieve this with a single vector*matrix multiplication?
>
> [1/n,  1/n,  ...,  1/n] * A  =  [mean(col 1); mean(col 2); ...; mean(col
> n)]
>
> Or in more legible form:
>
> [image: Inline image 1]
>
> Jacob
>
> On Fri, Nov 18, 2016 at 5:36 AM, Yuma Decaux via swift-users <
> swift-users at swift.org> wrote:
>
>> Hello everyone,
>>
>> Nice to meet y'all, and here's my first question:
>>
>> I am currently setting a semantically easy to use accelerate operation
>> extension set for my research.
>>
>> I have done all the basic operations, vector to scalar, vector to vector,
>> matrix mult matrix, transpose etc, but would like to know what the best
>> approach might be for getting the mean for a matrix, as I am not sure if
>> the result is what I think it is. In fact, I would like an output of m by 1
>> from an n by m matrix with means of each column vector.
>>
>> The function is:
>> func vDSP_meanvD(UnsafePointer<Double>, vDSP_Stride,
>> UnsafeMutablePointer<Double>, vDSP_Length)
>>
>> I assume for a vector this is fine since it is 1 dimensional, but for
>> matrices here is my question:
>>
>> What is the approach to take?
>> 1-Do I slice my matrix into m copies of size n where M_i(n by m) and pass
>> them all through vDSP_meanvD?
>> 2- Does the stride argument take care of this, in that if I have stride
>> of n, then the ranges [0:n], [n+1: 2n], [2n+1: 3n] will have their mean and
>> std computed?
>>
>>
>> Thanks for any pointer
>>
>> Best regards,
>>
>>
>>
>>
>> _______________________________________________
>> swift-users mailing list
>> swift-users at swift.org
>> https://lists.swift.org/mailman/listinfo/swift-users
>>
>
>
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