快速生成矩阵的笛卡尔积 [英] Quickly generate the cartesian product of a matrix
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问题描述
假设我有一个矩阵x
,其中包含10行和2列.我想生成一个新矩阵M
,其中包含x
中每对唯一的行-即一个具有55行4列的新矩阵.
例如,
x <- matrix (nrow=10, ncol=2, 1:20)
M <- data.frame(matrix(ncol=4, nrow=55))
k <- 1
for (i in 1:nrow(x))
for (j in i:nrow(x))
{
M[k,] <- unlist(cbind (x[i,], x[j,]))
k <- k + 1
}
所以x
是:
[,1] [,2]
[1,] 1 11
[2,] 2 12
[3,] 3 13
[4,] 4 14
[5,] 5 15
[6,] 6 16
[7,] 7 17
[8,] 8 18
[9,] 9 19
[10,] 10 20
然后M
有4列,前两行是x
的一行,接下来的2行是x
的另一行:
> head(M,10)
X1 X2 X3 X4
1 1 11 1 11
2 1 11 2 12
3 1 11 3 13
4 1 11 4 14
5 1 11 5 15
6 1 11 6 16
7 1 11 7 17
8 1 11 8 18
9 1 11 9 19
10 1 11 10 20
在R中是否有更快或更简单的方法(或同时有这两种方法)?
解决方案
expand.grid()
函数对此有用:
R> GG <- expand.grid(1:10,1:10)
R> GG <- GG[GG[,1]>=GG[,2],] # trim it to your 55 pairs
R> dim(GG)
[1] 55 2
R> head(GG)
Var1 Var2
1 1 1
2 2 1
3 3 1
4 4 1
5 5 1
6 6 1
R>
现在您有了'n *(n + 1)/2'个子集,您可以简单地索引原始矩阵.
Let's say I have a matrix x
which contains 10 rows and 2 columns. I want to generate a new matrix M
that contains each unique pair of rows from x
- that is, a new matrix with 55 rows and 4 columns.
E.g.,
x <- matrix (nrow=10, ncol=2, 1:20)
M <- data.frame(matrix(ncol=4, nrow=55))
k <- 1
for (i in 1:nrow(x))
for (j in i:nrow(x))
{
M[k,] <- unlist(cbind (x[i,], x[j,]))
k <- k + 1
}
So, x
is:
[,1] [,2]
[1,] 1 11
[2,] 2 12
[3,] 3 13
[4,] 4 14
[5,] 5 15
[6,] 6 16
[7,] 7 17
[8,] 8 18
[9,] 9 19
[10,] 10 20
And then M
has 4 columns, the first two are one row from x
and the next 2 are another row from x
:
> head(M,10)
X1 X2 X3 X4
1 1 11 1 11
2 1 11 2 12
3 1 11 3 13
4 1 11 4 14
5 1 11 5 15
6 1 11 6 16
7 1 11 7 17
8 1 11 8 18
9 1 11 9 19
10 1 11 10 20
Is there either a faster or simpler (or both) way of doing this in R?
解决方案
The expand.grid()
function useful for this:
R> GG <- expand.grid(1:10,1:10)
R> GG <- GG[GG[,1]>=GG[,2],] # trim it to your 55 pairs
R> dim(GG)
[1] 55 2
R> head(GG)
Var1 Var2
1 1 1
2 2 1
3 3 1
4 4 1
5 5 1
6 6 1
R>
Now you have the 'n*(n+1)/2' subsets and you can simple index your original matrix.
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