[R] comparing two matrices

[Dimitris Rizopoulos]

I think the following should work in your case:

mat1 <- data.matrix(expand.grid(0:2, 0:2, 0:2))
mat2 <- mat1[c(19, 16, 13, 24, 8), ]
############
ind1 <- apply(mat1, 1, paste, collapse = "/")
ind2 <- apply(mat2, 1, paste, collapse = "/")
match(ind2, ind1)


I hope it helps.

Best,
Dimitris


Quoting Adrian Dusa <dusa.adrian at gmail.com>:

> Hello Marc and Dimitris,
>
> There was an error in my first example (therefore not reproducible), so
> mat1 <- expand.grid(0:2, 0:2, 0:2)
> mat2 <- mat1[c(19, 16, 13, 24, 8), ]
>
> Your solution works if and only if the elements in both matrices are unique.
> Unfortunately, it does not apply for my matrices where elements do repeat
> (only the rows are unique).
>
>> which(apply(matrix(mat1 %in% mat2, dim(mat1)), 1, all))
> integer(0)
>
>> which((mat1 %in% mat2)[1:nrow(mat1)])
> integer(0)
>
>
> Another solution would be using base 3 operations:
> mat1 <- expand.grid(0:2, 0:2, 0:2)[, 3:1]
> mat2 <- mat1[c(19, 16, 13, 24, 8), ]
>
> mylines <- mat2[, 1]
> for (i in 2:ncol(mat2)) {mylines <- 3*mylines + mat2[, i]}
> mylines + 1
> [1] 19 16 13 24  8
>
>
> I was still hoping for a direct matrix function to avoid the for() loop.
> Thanks,
> Adrian
>
>
> On Sunday 21 January 2007 01:06, Marc Schwartz wrote:
>> On Sun, 2007-01-21 at 00:14 +0200, Adrian Dusa wrote:
>> > Dear helpeRs,
>> >
>> > I have two matrices:
>> > mat1 <- expand.grid(0:2, 0:2, 0:2)
>> > mat2 <- aa[c(19, 16, 13, 24, 8), ]
>> >
>> > where mat2 is always a subset of mat1
>> >
>> > I need to find the corersponding row numbers in mat1 for each row in
>> > mat2. For this I have the following code:
>> >
>> > apply(mat2, 1, function(x) {
>> >     which(apply(mat1, 1, function(y) {
>> >         sum(x == y)
>> >         }) == ncol(mat1))
>> >     })
>> >
>> > The code is vectorized, but I wonder if there is a simpler (hence faster)
>> > matrix computation that I miss.
>> >
>> > Thank you,
>> > Adrian
>>
>> I have not fully tested this, but how about:
>>
>> mat1 <- matrix(1:20, ncol = 4, byrow = TRUE)
>> mat2 <- matrix(1:60, ncol = 4, byrow = TRUE)
>> mat2 <- mat2[sample(15), ]
>>
>> > mat1
>>
>>      [,1] [,2] [,3] [,4]
>> [1,]    1    2    3    4
>> [2,]    5    6    7    8
>> [3,]    9   10   11   12
>> [4,]   13   14   15   16
>> [5,]   17   18   19   20
>>
>> > mat2
>>
>>       [,1] [,2] [,3] [,4]
>>  [1,]   13   14   15   16
>>  [2,]    5    6    7    8
>>  [3,]   41   42   43   44
>>  [4,]   17   18   19   20
>>  [5,]   21   22   23   24
>>  [6,]   25   26   27   28
>>  [7,]   53   54   55   56
>>  [8,]    9   10   11   12
>>  [9,]   57   58   59   60
>> [10,]   33   34   35   36
>> [11,]   49   50   51   52
>> [12,]   45   46   47   48
>> [13,]    1    2    3    4
>> [14,]   29   30   31   32
>> [15,]   37   38   39   40
>>
>> > which(apply(matrix(mat2 %in% mat1, dim(mat2)), 1, all))
>>
>> [1]  1  2  4  8 13
>>
>>
>> HTH,
>>
>> Marc Schwartz
>
> --
> Adrian Dusa
> Romanian Social Data Archive
> 1, Schitu Magureanu Bd
> 050025 Bucharest sector 5
> Romania
> Tel./Fax: +40 21 3126618 \
>           +40 21 3120210 / int.101
>
>



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