图节点之间的乘法距离

Question

我想找出图中所有节点之间的距离，而不是将边权重相加，我想将它们相乘。

举个例子：

library(igraph)

# create a weighted adjacency matrix
mx <- structure(c(0, 0.5, 0, 0, 0, 0.5, 0, 0.5, 0.5, 0, 0, 0.5, 0, 0, 0.5, 0, 0.5, 
    0, 0, 0, 0, 0, 0.5, 0, 0), .Dim = c(5L, 5L))

## convert to igraph object
mx2 <- graph.adjacency(mx, weighted = TRUE)

我可以得到所有节点之间的距离如下：

shortest.paths(mx2)

 [,1] [,2] [,3] [,4] [,5]
[1,]  0.0  0.5  1.0  1.0  1.5
[2,]  0.5  0.0  0.5  0.5  1.0
[3,]  1.0  0.5  0.0  1.0  0.5
[4,]  1.0  0.5  1.0  0.0  1.5
[5,]  1.5  1.0  0.5  1.5  0.0

但这通过将我想乘以它们的相关权重相加来计算所有节点之间的距离，这将导致以下结果：

      [,1] [,2] [,3]  [,4]  [,5]
[1,] 0.000 0.50 0.25 0.250 0.125
[2,] 0.500 0.00 0.50 0.500 0.250
[3,] 0.250 0.50 0.00 0.250 0.500
[4,] 0.250 0.50 0.25 0.000 0.125
[5,] 0.125 0.25 0.50 0.125 0.000

据我所知，使用 igraph 中的 "out of the box" 选项无法做到这一点，我正在努力自己弄清楚（在实际数据中，矩阵要大得多，并且各种尺寸）。任何建议将不胜感激。

Answer 1

这是一个提案。可能有很大的改进空间，但它给出了预期的输出。这个想法是为每对节点提取最短路径，然后乘以与每条路径关联的权重（我使用了这个中的一些代码）。

shortest.paths.multi <- function(mx) {
  output <- mx
  mx2 <- graph.adjacency(mx, weighted = TRUE)
  for (r in 1:nrow(mx)){
    for (c in 1:nrow(mx)){
      SP <- shortest_paths(mx2, from = r, to = c)
      VP <- SP$vpath[[1]]
      EP <- rep(VP, each=2)[-1]
      EP <- EP[-length(EP)]
      output[r, c] <- prod(E(mx2)$weight[get.edge.ids(mx2, EP)])
    }
  }
  diag(output) <- 0
  output
}

shortest.paths.multi(mx)

      [,1] [,2] [,3]  [,4]  [,5]
[1,] 0.000 0.50 0.25 0.250 0.125
[2,] 0.500 0.00 0.50 0.500 0.250
[3,] 0.250 0.50 0.00 0.250 0.500
[4,] 0.250 0.50 0.25 0.000 0.125
[5,] 0.125 0.25 0.50 0.125 0.000

编辑

这里可能是编写此函数的更好方法：

shortest.paths.multi <- function(r, c){ 
  SP <- shortest_paths(mx2, from = r, to = c)
  VP <- SP$vpath[[1]]
  EP <- rep(VP, each=2)[-1]
  EP <- EP[-length(EP)]
  prod(E(mx2)$weight[get.edge.ids(mx2, EP)])
}

VecFun <- Vectorize(shortest.paths.multi)

output <- outer(1:nrow(mx), 1:ncol(mx), FUN = VecFun)
diag(output) <- 0
output

Answer 2

按照上面的答案编写函数肯定是更好的方法。但是考虑这个问题的另一种方法是，如果你想要权重的乘积，并且 shortest.paths() 给你权重的总和，那么如果你将权重的对数输入 shortest.paths() 函数，则结果将等于权重的乘积。

这在实践中比我想象的要复杂一些，因为您的权重介于 0 和 1 之间并且 shortest.paths() 算法不接受负权重，但您可以通过乘以-1 计算权重前后。

library(igraph)

## Cheat and log it
ln.mx <- structure(c(0, 0.5, 0, 0, 0, 0.5, 0, 0.5, 0.5, 0, 0, 0.5, 0, 0, 0.5, 0, 0.5, 
                  0, 0, 0, 0, 0, 0.5, 0, 0), .Dim = c(5L, 5L))
ln.mx <- ifelse(ln.mx!=0, log(ln.mx), 0) 

## convert to igraph object
ln.mx2 <- graph.adjacency(ln.mx, weighted = TRUE)

# The issue with the approach is that the shortest.path algorithm doesn't like
# negative weights. Since your weights fall in (0,1) their log is negative.
# We multiply edge weights by -1 to swap the sign, and then will 
# multiply again by -1 to get
# the result
E(ln.mx2)$weight <- -1*E(ln.mx2)$weight

# The result is just the regular shortest paths algorithm,
# times -1 (to undo the step above) and exponentiated to undue the logging
res <- exp(shortest.paths(ln.mx2)* -1)

# its still not perfect since the diagonal distance defaults to
# zero and exp(0) is 1, not 0. So we manually reset the diagonal
diag(res) <- 0

# The result is as hoped
res
#>       [,1] [,2] [,3]  [,4]  [,5]
#> [1,] 0.000 0.50 0.25 0.250 0.125
#> [2,] 0.500 0.00 0.50 0.500 0.250
#> [3,] 0.250 0.50 0.00 0.250 0.500
#> [4,] 0.250 0.50 0.25 0.000 0.125
#> [5,] 0.125 0.25 0.50 0.125 0.000

图节点之间的乘法距离

multiplicative distance between graph nodes

r

igraph

adjacency-matrix