提取從``從包igraph`功能all_shortest_paths`信息中的R

Question

對於此示例曲線圖：

set.seed(1) 
g <- make_chordal_ring(15, matrix(c(3, 12, 4, 7, 8, 11), nr = 2)) 
k <- Vectorize(all_shortest_paths, "from", F)(g, V(g), 7) 

我們開始在圖中任何給定的節點，並使用節點7（參考結尾的所有最短路徑節點）。我想要的是計算節點Y在從節點X到節點7的最短路徑中存在的次數。

如果我表示從節點1到節點7通過節點2的最短路徑數N（1,2,7），並用n（1,7）總數的從節點1到節點7的最短路徑我想辦法，以產生一個表，看起來像：

我真的堅持這樣做，例如，如果我們看一下k的輸出：

> k[[1]][1] 
$res 
$res[[1]] 
+ 3/15 vertices: 
[1] 1 4 7 

我不知道該怎麼路徑1,4,7隔離和算上這對N（1,4,7）

Answer 1

我會去一個簡單的循環：

# initialize your matrix with all zeros 
m <- matrix(0,nrow=vcount(g),ncol=vcount(g)+1) 

# iterate over elements of k 
for(fromVertex in 1:length(k)){ 

    # iterate over res entry of each element of k 
    for(path in k[[fromVertex]]$res){ 

    # path is a vertex sequence, same type as V(g), 
    # calling as.integer we get the vertices indexes inside the sequence 
    verticesOfPath <- as.integer(path) 

    # we exclude the first and the last vertex (from,to) 
    innerVertices <- verticesOfPath[c(-1,-length(verticesOfPath))] 

    if(length(innerVertices) > 0){ 
     # this is not a direct path 
     m[verticesOfPath[1],innerVertices] <- m[verticesOfPath[1],innerVertices] + 1 
    } 
    # add 1 to the last column 
    m[verticesOfPath[1],ncol(m)] <- m[verticesOfPath[1],ncol(m)] + 1 
    } 
} 

結果：

> m 
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] 
[1,] 0 0 0 1 0 0 0 0 0  0  0  0  0  0  0  1 
[2,] 0 0 0 0 0 1 0 0 0  0  0  0  0  0  0  1 
[3,] 0 0 0 1 0 0 0 0 0  0  0  0  0  0  0  1 
[4,] 0 0 0 0 0 0 0 0 0  0  0  0  0  0  0  1 
[5,] 0 0 0 1 0 1 0 0 0  0  0  0  0  0  0  2 
[6,] 0 0 0 0 0 0 0 0 0  0  0  0  0  0  0  1 
[7,] 0 0 0 0 0 0 0 0 0  0  0  0  0  0  0  1 
[8,] 0 0 0 0 0 0 0 0 0  0  0  0  0  0  0  1 
[9,] 0 0 0 0 0 0 0 1 0  1  0  0  0  0  0  2 
[10,] 0 0 0 0 0 0 0 0 0  0  0  0  0  0  0  1 
[11,] 0 0 0 0 0 0 0 0 0  1  0  0  0  0  0  1 
[12,] 0 0 0 0 0 0 0 1 0  0  0  0  0  0  0  1 
[13,] 0 0 0 0 0 0 0 0 0  1  0  0  0  0  0  1 
[14,] 0 0 0 0 0 1 0 0 0  0  0  0  0  0  0  1 
[15,] 0 0 0 0 0 0 0 1 0  0  0  0  0  0  0  1 

如果頂點具有名稱屬性，你可以將它們設置爲矩陣的行和col名稱：

rownames(m) <- V(g)$name 
colnames(m) <- c(V(g)$name,'TOT') 

> m 
    A B C D E F G H I J K L M N O TOT 
A 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 
B 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 
C 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 
D 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 
E 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 2 
F 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 
G 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 
H 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 
I 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 2 
J 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 
K 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 
L 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 
M 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 
N 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 
O 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1