I think Enrico's solution is probably better overall and doesn't
require as much ugly behind-the-scenes trickery, but here's another
fun way that seems to run ever-so-marginally faster on my machine.
The vapply call is messy, but it seems to get the job done -- if it's
not clear, the point is to break matRandom into a list where each
element was previously one column in preparation for the do.call();
I'd welcome any insight into a slicker way to do so.
t0 <- system.time(matRandom <- matrix(runif(6000*3000),ncol=3000))
# I have to bump up columns to see any meaningful difference
## Enrico's
t1 <- system.time({ test1 <- matRandom[ ,1L];
for (i in seq.int(2L, ncol(matRandom)))
test1 <- pmax(test1, matRandom[ ,i])
})
## Mine
t2 <- system.time({
temp <- vapply(seq.int(ncol(matRandom)), function(i,x) list(x[,i]),
vector("list",1) , matRandom)
test2 <- do.call(pmax, temp)
})
identical(test1, test2)
TRUE
t0
user system elapsed
2.58 0.10 2.69
t1
user system elapsed
1.63 0.00 1.63
t2
user system elapsed
1.25 0.00 1.25
Michael
PS -- It makes me very happy that building matRandom is the slowest
step. All hail the mighty vectorization of R!
On Wed, Oct 12, 2011 at 9:10 AM, Enrico Schumann
<enricoschum...@yahoo.de> wrote:
>
> Hi Wolfgang,
>
> how about a loop?
>
> matRandom <- matrix(runif(n=600000), ncol=6)
>
> ## variant 1
> system.time(test1 <- pmax(matRandom[,1], matRandom[,2], matRandom[,3],
> matRandom[,4], matRandom[,5], matRandom[,6]))
>
> User System verstrichen
> 0.01 0.00 0.01
>
>
> ## variant 2
> system.time(test2 <- apply(matRandom, 1, max))
>
> User System verstrichen
> 0.56 0.00 0.56
>
>
> ## variant 3
> system.time({
> test3 <- matRandom[ ,1L]
> ## add a check that ncol(matrix) > 1L
> for (i in 2:ncol(matRandom))
> test3 <- pmax(test3, matRandom[ ,i])
>
> })
> User System verstrichen
> 0.01 0.00 0.01
>
>
>
>> all.equal(test1,test2)
> [1] TRUE
>
>> all.equal(test1,test3)
> [1] TRUE
>
>
> Regards,
> Enrico
>
> Am 12.10.2011 13:06, schrieb Wolfgang Wu:
>>
>> I am having the following problem. I want to calculate the maximum of each
>> row in a matrix. If I pass in the matrix split up by each column then this
>> is no problem and works great. However I don't know how many columns I have
>> in advance. In the example below I have 3 columns, but the number of columns
>> is not fix. So how do I do this?
>>
>>
>> matRandom<- matrix(runif(n=30), ncol=3);
>> #Does not work
>> pmax(matRandom)
>> #Does work
>> pmax(matRandom[,1], matRandom[,2], matRandom[,3])
>>
>>
>> I am aware that I can do it with the apply function, but the calculation
>> is time sensitive so fast execution is important.
>>
>>
>> #Apply might be too slow
>>
>> matRandom<- matrix(runif(n=300000), ncol=3);
>> system.time(test<- pmax(matRandom[,1], matRandom[,2], matRandom[,3]))
>> system.time(test<- apply(matRandom, 1, max))
>>
>>
>>> matRandom<- matrix(runif(n=300000), ncol=3);
>>> system.time(test<- pmax(matRandom[,1], matRandom[,2], matRandom[,3]))
>>
>> user system elapsed
>> 0.02 0.00 0.02
>>>
>>> system.time(test<- apply(matRandom, 1, max))
>>> user system elapsed
>>
>> 2.37 0.00 2.38
>>
>>
>>
>>
>> Thanks for your help.
>>
>> Regards.
>>
>>
>> Wolfgang Wu
>>
>> ______________________________________________
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> --
> Enrico Schumann
> Lucerne, Switzerland
> http://nmof.net/
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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