我有三种不同的矩阵，他们的Krackhardt效率对我来说似乎不对。前两个矩阵在拓扑上是等价的，但它们的效率是不同的。任何人都有不一致的解释？
对于第一个矩阵，效率为1：
A <- matrix(c(0,1,0,0,0,0,0,0,0,0,
0,0,1,0,0,0,0,0,0,0,
0,0,0,1,0,0,0,0,0,0,
0,0,0,0,1,0,0,0,0,0,
0,0,0,0,0,1,0,0,0,1,
0,0,0,0,0,0,1,0,0,0,
0,0,0,0,0,0,0,1,0,0,
0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,1,0),ncol=10)
A_net <- network(A,directed=TRUE)
g_eff <- efficiency(A_net)
plot.network(A_net, vertex.col = "white", vertex.border = col_ama,
usearrows=FALSE, edge.col=col_gri, vertex.lwd = 3.5,
vertex.cex = 3.5)
title(paste("Efficiency =",round(g_eff,3)))
效率不同的等效矩阵：
A <- matrix(c(0,0,0,0,1,0,0,0,0,0,
1,0,1,0,0,0,0,0,0,0,
0,0,0,0,0,0,1,0,0,0,
0,0,1,0,0,0,0,0,0,0,
0,0,0,0,0,1,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,1,0,
0,0,0,1,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,1,
0,0,0,0,0,0,0,0,0,0),ncol=10)
A_net <- network(A,directed=FALSE)
g_eff <- efficiency(A_net)
plot.network(A_net, vertex.col = "white", vertex.border = col_ama,
usearrows=FALSE, edge.col=col_gri, vertex.lwd = 3.5,
vertex.cex = 3.5)
title(paste("Efficiency =",round(g_eff,3)))
这个第三个矩阵有两个具有最小边数（n_i-1）的分量，但它们的效率不是一个。它既不匹配帮助中的公式：
(1 - [ |E(G)| - Sum(N_i-1,i=1,..,n) ]/[ Sum((N_i-1)^2,i=1,..,n) ] = 1-[8-(2+4)]/[4+16] = .9)
第三个矩阵：
A <- matrix(c(0,0,0,0,1,0,0,0,0,0,
1,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,1,0,0,0,
0,0,1,0,0,0,0,0,0,0,
0,0,0,0,0,1,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,1,0,
0,0,0,1,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,1,
0,0,0,0,0,0,0,0,0,0),ncol=10)
A_net <- network(A,directed=FALSE)
g_eff <- efficiency(A_net)
g_eff
plot.network(A_net, vertex.col = "white", vertex.border = col_ama,
usearrows=FALSE, edge.col=col_gri, vertex.lwd = 3.5,
vertex.cex = 3.5)
title(paste("Efficiency =",round(g_eff,3)))
答案 0 :(得分：0)
看来你的例子是错误的：在第一种情况下，你正在使用定向网络，而在第二种情况下是一个无向网络（检查你的网络强制声明 - 顺便说一句，你可以只使用矩阵）。这是一个示例，前两个实际上是等价的：
> A <- matrix(c(0,1,0,0,0,0,0,0,0,0,
+ 0,0,1,0,0,0,0,0,0,0,
+ 0,0,0,1,0,0,0,0,0,0,
+ 0,0,0,0,1,0,0,0,0,0,
+ 0,0,0,0,0,1,0,0,0,1,
+ 0,0,0,0,0,0,1,0,0,0,
+ 0,0,0,0,0,0,0,1,0,0,
+ 0,0,0,0,0,0,0,0,0,0,
+ 0,0,0,0,0,0,0,0,0,0,
+ 0,0,0,0,0,0,0,0,1,0),ncol=10)
> A
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 0 0 0 0 0 0 0 0
[2,] 1 0 0 0 0 0 0 0 0 0
[3,] 0 1 0 0 0 0 0 0 0 0
[4,] 0 0 1 0 0 0 0 0 0 0
[5,] 0 0 0 1 0 0 0 0 0 0
[6,] 0 0 0 0 1 0 0 0 0 0
[7,] 0 0 0 0 0 1 0 0 0 0
[8,] 0 0 0 0 0 0 1 0 0 0
[9,] 0 0 0 0 0 0 0 0 0 1
[10,] 0 0 0 0 1 0 0 0 0 0
> B<-matrix(c(0,0,0,0,1,0,0,0,0,0,
+ 1,0,1,0,0,0,0,0,0,0,
+ 0,0,0,0,0,0,1,0,0,0,
+ 0,0,1,0,0,0,0,0,0,0,
+ 0,0,0,0,0,1,0,0,0,0,
+ 0,0,0,0,0,0,0,0,0,0,
+ 0,0,0,0,0,0,0,0,1,0,
+ 0,0,0,1,0,0,0,0,0,0,
+ 0,0,0,0,0,0,0,0,0,1,
+ 0,0,0,0,0,0,0,0,0,0),ncol=10)
> A
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 0 0 0 0 0 0 0 0
[2,] 1 0 0 0 0 0 0 0 0 0
[3,] 0 1 0 0 0 0 0 0 0 0
[4,] 0 0 1 0 0 0 0 0 0 0
[5,] 0 0 0 1 0 0 0 0 0 0
[6,] 0 0 0 0 1 0 0 0 0 0
[7,] 0 0 0 0 0 1 0 0 0 0
[8,] 0 0 0 0 0 0 1 0 0 0
[9,] 0 0 0 0 0 0 0 0 0 1
[10,] 0 0 0 0 1 0 0 0 0 0
> B
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 1 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 0
[3,] 0 1 0 1 0 0 0 0 0 0
[4,] 0 0 0 0 0 0 0 1 0 0
[5,] 1 0 0 0 0 0 0 0 0 0
[6,] 0 0 0 0 1 0 0 0 0 0
[7,] 0 0 1 0 0 0 0 0 0 0
[8,] 0 0 0 0 0 0 0 0 0 0
[9,] 0 0 0 0 0 0 1 0 0 0
[10,] 0 0 0 0 0 0 0 0 1 0
> efficiency(A)
[1] 1
> efficiency(B)
[1] 1
如果您通过设置“directed = FALSE”来对称 - 这是您在示例中对其中一个网络所做的 - 那么您将获得0.889的效率。请注意，这与我们应该得到的相符：
1 - （18 - 9）/（选择（10,2）* 2-9）= 0.889
（请记住，Krackhardt效率会将所有网络视为有向图，因此相互边缘会被视为两条边。此外，我注意到您似乎错误地使用了手册中的公式，这可能是您混淆的一部分。）
你的第三个矩阵再次是效率1，因为它没有多余的边缘：
> C<-matrix(c(0,0,0,0,1,0,0,0,0,0,
+ 1,0,0,0,0,0,0,0,0,0,
+ 0,0,0,0,0,0,1,0,0,0,
+ 0,0,1,0,0,0,0,0,0,0,
+ 0,0,0,0,0,1,0,0,0,0,
+ 0,0,0,0,0,0,0,0,0,0,
+ 0,0,0,0,0,0,0,0,1,0,
+ 0,0,0,1,0,0,0,0,0,0,
+ 0,0,0,0,0,0,0,0,0,1,
+ 0,0,0,0,0,0,0,0,0,0),ncol=10)
> C
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 1 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 0
[3,] 0 0 0 1 0 0 0 0 0 0
[4,] 0 0 0 0 0 0 0 1 0 0
[5,] 1 0 0 0 0 0 0 0 0 0
[6,] 0 0 0 0 1 0 0 0 0 0
[7,] 0 0 1 0 0 0 0 0 0 0
[8,] 0 0 0 0 0 0 0 0 0 0
[9,] 0 0 0 0 0 0 1 0 0 0
[10,] 0 0 0 0 0 0 0 0 1 0
> efficiency(C)
[1] 1
您的问题源于对称化（使用directed = FALSE将C强制转换为网络对象）。这会增加额外的边缘，导致
1-（16 - 3 - 5）/（选择（4,2）* 2 - 3 +选择（6,2）* 2 - 5）= 0.7647059
这相当于sna给你的东西：
> efficiency(symmetrize(C))
[1] 0.7647059
希望能够解决问题！