[R] effective matrix subset

[Patrick Burns]

That may be a better solution, but I don't think
it is clearly a better solution.

I presume you mean that your computation is the
most time efficient.  That seems believable to me.
It is not the most human efficient -- it will take
some one reading the code non-trivial effort to
understand it.

Whether time or code clarity are more important
depends on the particular application.

Patrick Burns
patrick at burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide for the Unwilling S User")

jgarcia at ija.csic.es wrote:
> It seems that this solution provided by Dan (and also available in
> SPoetry; I'm sorry I didn't notice it) is the fastest and simplest. I was
> using a more standard approach:
>
> V <- t(A)[(0:(nrow(A)-1))*ncol(A)+X],
>
> That wasn't bad, but I was confident that you, R gurus, could outperform
> this. This is clearly a much better solution.
>
> Thanks all, and best wishes,
> Javier
> ------
>
>
>
>
>   
>> on 08/09/2008 06:52 AM Dan Davison wrote:
>>     
>>> On Sat, Aug 09, 2008 at 06:29:59AM -0500, Marc Schwartz wrote:
>>>       
>>>> on 08/09/2008 06:01 AM jgarcia at ija.csic.es wrote:
>>>>         
>>>>> Hi;
>>>>> If we have a matrix A, and a vector X, where length(X)=nrow(A), and X
>>>>> contains a wanted column for each row in A, in row ascending order.
>>>>> How
>>>>> would be the most effective way to extract the desired vector V (with
>>>>> length(V)=nrow(A))?
>>>>>           
>>>> A <- matrix(1:20, 4, 5)
>>>>
>>>>         
>>>>> A
>>>>>           
>>>>      [,1] [,2] [,3] [,4] [,5]
>>>> [1,]    1    5    9   13   17
>>>> [2,]    2    6   10   14   18
>>>> [3,]    3    7   11   15   19
>>>> [4,]    4    8   12   16   20
>>>>
>>>>
>>>> # Create an arbitrary set of indices, one for each row in A
>>>> X <- c(2, 5, 1, 4)
>>>>
>>>>         
>>>>> X
>>>>>           
>>>> [1] 2 5 1 4
>>>>
>>>>
>>>> Presumably you want:
>>>>
>>>> V <- c(A[1, 2], A[2, 5], A[3, 1], A[4, 4])
>>>>
>>>>         
>>>>> V
>>>>>           
>>>> [1]  5 18  3 16
>>>>
>>>>
>>>> If so, then:
>>>>
>>>>         
>>>>> sapply(seq(nrow(A)), function(i) A[i, X[i]])
>>>>>           
>>>> [1]  5 18  3 16
>>>>         
>>> Or
>>>
>>>       
>>>> A[cbind(seq(nrow(A)), X)]
>>>>         
>>> [1]  5 18  3 16
>>>
>>> Dan
>>>       
>> Better (and faster) solution Dan.
>>
>> I can't blame the lack of coffee on missing that one this morning. I
>> have had a full pot already over the past 6 hours, working on shifting
>> my internal clock and getting ready to begin my journey to Dortmund
>> later tonight...
>>
>> Safe travels to all who are going.
>>
>> Marc
>>
>>
>>     
>
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