Question

我想找到图形中所有节点之间的距离，而不是总和我想要乘以它们的边缘权重。

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)

我可以按如下方式获取所有节点之间的距离：

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

但是这会通过将我想要乘以它们的相关权重相加来计算所有节点之间的距离，这将产生以下结果：

      [,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“开箱即用”选项无法做到这一点，我正在努力自己解决这个问题（在真实数据中，矩阵要大得多，而且各种尺寸）。任何建议将不胜感激。

Answer 1

这是一个提案。可能还有很大的改进空间，但它可以提供预期的产量。我们的想法是为每对节点提取最短路径，然后乘以与每条路径相关的权重（我使用了这个answer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

修改

这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

在上面的答案中写一个函数肯定是更好的方法。但另一种思考问题的方法是，如果你想要权重的产品，而shortest.paths()给你权重的总和，那么如果你给shortest.paths()函数提供权重的对数，那么权重的指数就是结果将等于权重的乘积。

这在练习中比我想象的要好一些，因为你的权重在0和1之间，shortest.paths()算法不接受负权重，但你可以通过乘以-1来解决在计算权重之前和之后。

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

2 个答案: