我有三个不同的矩阵，它们的Krackhardt效率对我来说似乎是错误的。前两个矩阵在拓扑上是等价的，但它们的效率不同。有人对不一致有解释吗？
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)))
不同效率的等效矩阵：
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)))
第三个矩阵有两个边数最少的分量（n_i-1），但它们的效率不是一个。它与帮助中的公式也不匹配：
(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)))
您的示例似乎有误：在第一种情况下，您使用的是有向网络，而在第二种情况下，您使用的是无向网络（检查您的网络强制声明 - 顺便说一句，您可以只使用矩阵）。 下面是一个演示，证明前两个实际上是等效的：
> 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
如果你对称化 - 这就是你在示例中通过设置"directed=FALSE"对其中一个网络所做的 - 那么你得到的效率为0.889。 请注意，这与我们应该得到的相匹配：
1 - （18 - 9）/（选择（10,2）*2-9） = 0.889
（请记住，Krackhardt 效率将所有网络视为二合字母，因此相互边算作两条边。 另外，我注意到您似乎从手册页中错误地复制了公式，这可能是您困惑的一部分。
你的第三个矩阵又是效率 1，因为它没有多余的边：
> 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）。 这增加了额外的边缘，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 为您提供的：
> efficiency(symmetrize(C))
[1] 0.7647059
希望能把事情弄清楚！