Thanks a lot to all of you!

On Tue, Apr 26, 2011 at 10:34 AM, Peter Ehlers <ehlers@ucalgary.ca> wrote:

> On 2011-04-26 05:26, Dimitris Rizopoulos wrote:
>
>> one way is the following:
>>
>> a<- matrix(rnorm(9), 3 ,3)
>>
>> aa<- a[order(row(a), -a)]
>> matrix(aa, nrow(a), byrow = TRUE)[, 2]
>>
>
> That's clever, Dmitris. And very fast!
>
> Lars: my apologies; I didn't read your request
> carefully. You asked for more _efficient_ and
> I just thought 'shorter code'. I should know
> that whenver I think 'apply', I should think
> 'matrix'.
>
> Peter Ehlers
>
>
>
>>
>> I hope it helps.
>>
>> Best,
>> Dimitris
>>
>>
>> On 4/26/2011 2:01 PM, Lars Bishop wrote:
>>
>>> Hi,
>>>
>>> I need to extract the second largest element from each row of a
>>> matrix. Below is my solution, but I think there should be a more
>>> efficient
>>> way to accomplish the same, or not?
>>>
>>>
>>>   set.seed(1)
>>>   a<- matrix(rnorm(9), 3 ,3)
>>>   sec.large<- as.vector(apply(a, 1, order, decreasing=T)[2,])
>>>   ans<- sapply(1:length(sec.large), function(i) a[i, sec.large[i]])
>>>   ans
>>>
>>> Thanks in advance for your help,
>>> Lars.
>>>
>>>        [[alternative HTML version deleted]]
>>>
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>>>
>>>
>>
>

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