每个变量的平均距离

Question

可以使用下面给出的公式计算每个变量的平均距离。这里d表示感兴趣的变量与其父变量的平均距离。 p和q代表该变量对其父节点的不同状态的条件概率，i代表子节点的不同状态，n代表父节点集的状态数。

以下是两个州父母的例子。 我想要计算的是：

   Average {[(0.8286-0.6308)^2],[(0.1364-0.2347)^2],...,[(0.0017-0.0049)^2]}
   =0.0107

当我有超过3个州时，我需要找到：

Average {[(a-b)^2+(a-c)^2+(b-c)^2)],....

我试过了：

      x1<-c(0.8286,0.1364,0.0300,0.0033,0.0017)

      x2<-c(0.6308,0.2347,0.0807,0.0489,0.0049)

      dist(rbind(x1,x2))

但它只是给我欧几里得距离。

Answer 1

抱歉，起初我有一个误会。现在这就是你真正能做的事情：

d <- function(mat) {
  ind <- as.numeric(combn(nrow(mat), 2))
  n <- length(ind) / 2
  mean(apply(mat, 2, function(x) {y <- x[ind]; sum((y[seq(from = 1, length = n, by = 2)] - y[seq(from = 2, length = n, by = 2)])^2)}))/n
  }

示例，假设您有概率表：

set.seed(0); mat <- matrix(runif(20), 4, 5)

#           [,1]      [,2]       [,3]      [,4]      [,5]
# [1,] 0.8966972 0.9082078 0.66079779 0.1765568 0.4976992
# [2,] 0.2655087 0.2016819 0.62911404 0.6870228 0.7176185
# [3,] 0.3721239 0.8983897 0.06178627 0.3841037 0.9919061
# [4,] 0.5728534 0.9446753 0.20597457 0.7698414 0.3800352

d(mat) # 0.1775407

有关2个州的示例数据：

x1<-c(0.8286,0.1364,0.0300,0.0033,0.0017)
x2<-c(0.6308,0.2347,0.0807,0.0489,0.0049)
d(rbind(x1,x2))  # 0.01068956

Answer 2

除非我误解了这个问题，否则答案就是

mean((x1-x2)^2)

证明：

> (x1-x2)^2
[1] 0.03912484 0.00966289 0.00257049 0.00207936 0.00001024

> 0.8286-0.6308
[1] 0.1978
> 0.1978^2
[1] 0.03912484