[R] Sum efficiently from large matrix according to re-occuring levels of factor?
hadley wickham
h.wickham at gmail.com
Tue Jul 22 06:08:19 CEST 2008
Using Jim's index with my method gives you the best of both worlds:
x <- matrix(sample(20, 1e6 * 3, replace = T), ncol = 3)
system.time({
dataBreaks <- cumsum(c(0, (diff(x[, 2] + x[, 1] * max(x[, 2])) != 0)))
# sum up column 3 and output the first two columns with the indices
result <- lapply(split(seq(nrow(x)), dataBreaks), function(.sect){
c(x[.sect[1], 1:2], sum(x[.sect, 3]))
})
a <- do.call(rbind, result)
})
system.time({
index <- cumsum(c(0, (diff(x[, 2] + x[, 1] * max(x[, 2])) != 0)))
b <- cbind(x[!duplicated(index), 1:2], tapply(x[, 3], index, sum))
})
all.equal(a, b)
On my computer, Jim's method took 60 seconds and mine took 16.
Hadley
On Sun, Jul 20, 2008 at 8:41 PM, Ralph S. <ruffel1 at hotmail.com> wrote:
>
> yes - thank you very much! slowly getting to the full power of R . . .
>
> ----------------------------------------
>> Date: Sun, 20 Jul 2008 21:21:35 -0400
>> From: jholtman at gmail.com
>> To: ruffel1 at hotmail.com
>> Subject: Re: [R] Sum efficiently from large matrix according to re-occuring levels of factor?
>> CC: h.wickham at gmail.com; r-help at r-project.org
>>
>> Does this do what you want:
>>
>>> # following up on another idea that was presented
>>> # where are the breaks
>>> dataBreaks <- cumsum(c(0, (diff(x[, 2] + x[, 1] * max(x[, 2])) != 0)))
>>> # sum up column 3 and output the first two columns with the indices
>>> result <- lapply(split(seq(nrow(x)), dataBreaks), function(.sect){
>> + c(x[.sect[1], 1:2], sum(x[.sect, 3]))
>> + })
>>> do.call(rbind, result)
>> [,1] [,2] [,3]
>> 0 1 7 3
>> 1 2 4 2
>> 2 3 2 3
>> 3 1 7 10
>>
>>
>> On Sun, Jul 20, 2008 at 7:57 PM, Ralph S. wrote:
>>>
>>> The first and second column are actually indices of another matrix (my example may make this not sufficiently clear). I want to compare the sum with that corresponding entry, and then record the result of that.
>>>
>>> Any idea?
>>>
>>> Best,
>>>
>>> Ralph
>>>
>>>
>>>
>>> ----------------------------------------
>>>> Date: Sun, 20 Jul 2008 16:50:41 -0700
>>>> From: h.wickham at gmail.com
>>>> To: ruffel1 at hotmail.com
>>>> Subject: Re: [R] Sum efficiently from large matrix according to re-occuring levels of factor?
>>>> CC: r-help at r-project.org
>>>>
>>>> On Sun, Jul 20, 2008 at 4:47 PM, hadley wickham wrote:
>>>>> On Sun, Jul 20, 2008 at 4:16 PM, Ralph S. wrote:
>>>>>>
>>>>>> Hi,
>>>>>>
>>>>>> I am trying to calculate the sum for each occurrence of the level of a factor in a very large matrix. In addition, I want to save that sum together with the information of the level of the factor and the level of a second factor.
>>>>>>
>>>>>> My matrix looks like this:
>>>>>>
>>>>>> x<-matrix(c(1,1,1,2,2,3,3,1,1,7,7,7,4,4,2,2,7,7,1,1,1,1,1,1,2,5,5),9,3)
>>>>>>
>>>>>> I want to sum according to the levels in the first column and save the sum with the information of the level in the first and the second column in a new matrix.
>>>>>>
>>>>>> That is, I want output in the matrix of form:
>>>>>>
>>>>>> 1 7 3
>>>>>> 2 4 2
>>>>>> 3 2 3
>>>>>> 1 7 10
>>>>>>
>>>>>
>>>>> Why that and not:
>>>>>
>>>>> 1 7 13
>>>>> 2 4 2
>>>>> 3 2 3
>>>>>
>>>>> ?
>>>>
>>>> Here's a solution for that case:
>>>>
>>>> index <- x[, 2] + x[, 1] * max(x[, 2])
>>>> cbind(x[!duplicated(index), 1:2], tapply(x[, 3], index, sum))
>>>>
>>>> It takes about half a second for a million row matrix.
>>>>
>>>> Hadley
>>>>
>>>>
>>>>
>>>> --
>>>> http://had.co.nz/
>>>
>>> _________________________________________________________________
>>> With Windows Live for mobile, your contacts travel with you.
>>>
>>> 072008
>>> ______________________________________________
>>> R-help at 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.
>>>
>>
>>
>>
>> --
>> Jim Holtman
>> Cincinnati, OH
>> +1 513 646 9390
>>
>> What is the problem you are trying to solve?
>
> _________________________________________________________________
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--
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