[R] matrix column division by vector

[Bert Gunter]

I'm wrong. Jeff's approach DOES use the matrix API and so I would say
is better than mine.

Cheers,
Bert

Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374

"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
H. Gilbert Welch




On Wed, May 14, 2014 at 9:22 AM, Bert Gunter <bgunter at gene.com> wrote:
> Yes, but Jeff and I advocated essentially the same approach and the
> difference between our versions is unimportant.
>
> All the other approaches are at least aesthetically less desirable.
>
> One "criticism" of our solution: it relies on the underlying
> implementation of matrices rather than the matrix() API. In principle,
> the implementation could change while the API remained constant.
> However, the reality is that this would never happen (it would break
> thousands of lines of code that use this approach because R was not in
> the past and really still isn't entirely OO). But full disclosure
> demands ...
>
> Cheers,
> Bert
>
> -- Bert
>
> Bert Gunter
> Genentech Nonclinical Biostatistics
> (650) 467-7374
>
> "Data is not information. Information is not knowledge. And knowledge
> is certainly not wisdom."
> H. Gilbert Welch
>
>
>
>
> On Wed, May 14, 2014 at 9:07 AM, David L Carlson <dcarlson at tamu.edu> wrote:
>> Bert wins the race:
>>
>>> system.time(replicate(1e5, m/rep(v,e=2)))
>>    user  system elapsed
>>    0.25    0.00    0.25
>>> system.time(replicate(1e5, m/matrix( v, ncol=ncol(m), nrow=nrow(m), byrow=TRUE)))
>>    user  system elapsed
>>    0.42    0.00    0.42
>>> system.time(replicate(1e5, t(t(m)/v)))
>>    user  system elapsed
>>    1.31    0.00    1.33
>>> system.time(replicate(1e5, sweep(m, 2, v, "/")))
>>    user  system elapsed
>>    3.39    0.00    3.40
>>> system.time(replicate(1e5, t(apply(m, 1, function(x) x/v))))
>>    user  system elapsed
>>    5.04    0.01    5.06
>>
>> David C
>>
>> -----Original Message-----
>> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Jeff Newmiller
>> Sent: Wednesday, May 14, 2014 10:28 AM
>> To: carol white; carol white; r-help at stat.math.ethz.ch
>> Subject: Re: [R] matrix column division by vector
>>
>> Please post in plain text... your email is getting distorted and hard to read by the HTML.
>>
>> I don't know how to use do.call for this, but when you understand how vectors recycle and matrices and arrays are laid out in memory (read the Introduction to R document if not) then the following comes to mind:
>>
>> mat2 <- m / matrix( v, ncol=ncol(m), nrow=nrow(m), byrow=TRUE )
>>
>> ---------------------------------------------------------------------------
>> Jeff Newmiller                        The     .....       .....  Go Live...
>> DCN:<jdnewmil at dcn.davis.ca.us>        Basics: ##.#.       ##.#.  Live Go...
>>                                       Live:   OO#.. Dead: OO#..  Playing
>> Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
>> /Software/Embedded Controllers)               .OO#.       .OO#.  rocks...1k
>> ---------------------------------------------------------------------------
>> Sent from my phone. Please excuse my brevity.
>>
>> On May 14, 2014 7:51:36 AM PDT, carol white <wht_crl at yahoo.com> wrote:
>>>Hi,
>>>What is the elegant script to divide the columns of a matrix by the
>>>respective position of a vector elements?
>>>
>>>m=rbind(c(6,4,2),c(3,2,1))
>>>
>>>v= c(3,2,1)
>>>
>>>res= 6/3�� 4/2� 2/1
>>>������� 3/3�� 2/2 �� 1/1
>>>
>>>
>>>this is correct�
>>>mat2 = NULL
>>>
>>>for (i in 1: ncol(m))
>>>
>>>��� mat2 = cbind(mat2, m[,i]/ v[i])
>>>
>>>
>>>but how to do more compact and elegant with for ex do.call?
>>>
>>>Many thanks
>>>
>>>Carol
>>>       [[alternative HTML version deleted]]
>>>
>>>
>>>
>>>------------------------------------------------------------------------
>>>
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>>
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