计算R中矩阵中有序对的数量

Question

给定矩阵m如下(行方式排列为1-5):

    # [,1] [,2] [,3] [,4] [,5]
 # [1,]    1    5    2    4    3
 # [2,]    2    1    4    3    5
 # [3,]    3    4    1    2    5
 # [4,]    4    1    3    2    5
 # [5,]    4    3    1    2    5
 # [6,]    1    4    2    3    5
 # [7,]    4    3    2    5    1
 # [8,]    4    1    3    5    2
 # [9,]    1    2    3    4    5
# [10,]    4    3    2    1    5

我想知道每个元素1-5在每行的另一个元素之前的次数(即考虑所有可能的对)

例如,对于所有行中的对(1,5),1先行5,9次.另一个例子,对于对(3,1),3在1所有行之前是4次.我希望所有行中所有可能的对都有相同的结果.那是,

# (1, 2), (1, 3), (1, 4), (1, 5)
# (2, 1), (2, 3), (2, 4), (2, 5)
# (3, 1), (3, 2), (3, 4), (3, 5)
# (4, 1), (4, 2), (4, 3), (4, 5)
# (5, 1), (5, 2), (5, 3), (5, 4)

m <- structure(c(1L, 2L, 3L, 4L, 4L, 1L, 4L, 4L, 1L, 4L, 5L, 1L, 4L, 
1L, 3L, 4L, 3L, 1L, 2L, 3L, 2L, 4L, 1L, 3L, 1L, 2L, 2L, 3L, 3L, 
2L, 4L, 3L, 2L, 2L, 2L, 3L, 5L, 5L, 4L, 1L, 3L, 5L, 5L, 5L, 5L, 
5L, 1L, 2L, 5L, 5L), .Dim = c(10L, 5L))

如何在R中有效地做到这一点？

编辑

你会如何为这个矩阵做同样的事情？

      # [,1] [,2] [,3] [,4] [,5]
 # [1,]    3    4    1    5    0
 # [2,]    1    2    5    3    0
 # [3,]    3    5    0    0    0
 # [4,]    4    5    0    0    0
 # [5,]    3    4    1    5    2
 # [6,]    3    1    2    0    0
 # [7,]    4    1    5    2    0
 # [8,]    4    3    5    2    0
 # [9,]    5    2    0    0    0
# [10,]    5    4    2    0    0

m <- structure(c(3, 1, 3, 4, 3, 3, 4, 4, 5, 5, 4, 2, 5, 5, 4, 1, 1, 
3, 2, 4, 1, 5, 0, 0, 1, 2, 5, 5, 0, 2, 5, 3, 0, 0, 5, 0, 2, 2, 
0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0), .Dim = c(10L, 5L))

Answer 1

这是一个矢量化解决方案,没有apply:

func <- function(a,b) sum((which(!t(m-b)) - which(!t(m-a)))>0)

#> func(1,5)
#[1] 9
#> func(5,1)
#[1] 1

要生成所有想要的组合,您可以简单地执行:

N = combn(1:5, 2)
cbind(N, N[nrow(N):1,])

然后,您只需要一个循环来迭代列并应用该函数.