Here is one way of doing it:
> compMat2 <- function(A, B) { # rows of B present in A
+ B0 <- B[!duplicated(B), ]
+ na <- nrow(A); nb <- nrow(B0)
+ AB <- rbind(A, B0)
+ ab <- duplicated(AB)[(na+1):(na+nb)]
+ return(sum(ab))
+ }
>
>
> set.seed(8237)
> A <- matrix(sample(1:1000, 2*67420, replace=TRUE), 67420, 2)
> B <- matrix(sample(1:1000, 2*59199, replace=TRUE), 59199, 2)
>
> system.time({
+ # convert for comparison
+ A.1 <- apply(A, 1, function(x) paste(x, collapse = ' '))
+ B.1 <- apply(B, 1, function(x) paste(x, collapse = ' '))
+ count <- sum(B.1 %in% A.1)
+ })
user system elapsed
1.77 0.00 1.79
>
>
> count
[1] 3905
>
On Fri, Dec 2, 2011 at 2:46 PM, Hans W Borchers
<hwborch...@googlemail.com> wrote:
> Michael Kao <mkao006rmail <at> gmail.com> writes:
>
>>
> Well, taking a second look, I'd say it depends on the exact formulation.
>
> In the applications I have in mind, I would like to count each occurrence
> in B only once. Perhaps the OP never thought about duplicates in B
>
> Hans Werner
>
>>
>> Here is an example based on the duplicated function
>>
>> test.mat1 <- matrix(1:20, nc = 5)
>>
>> test.mat2 <- rbind(test.mat1[sample(1:5, 2), ], matrix(101:120, nc = 5))
>>
>> compMat <- function(mat1, mat2){
>> nr1 <- nrow(mat1)
>> nr2 <- nrow(mat2)
>> mat2[duplicated(rbind(mat1, mat2))[(nr1 + 1):(nr1 + nr2)], ]
>> }
>>
>> compMat(test.mat1, test.mat2)
>>
>
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--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
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