我的嘗試是：
library(igraph)
######################################################################
set.seed(5)
g <- sample_gnm(10, 4)
xy <- cbind(runif(10), runif(10))
par(mfrow=c(1,2))
plot(g, vertex.size=5, layout=xy)
num_point <- length(V(g)[degree(g)==0])
for(k in 1:num_point){
points = V(g)[degree(g)==0]
for(i in 1:length(E(g))) { # loop over all edges
head <- get.edgelist(g)[i,][1]; h <- c(V(g)[head]$x, V(g)[head]$y)
tail <- get.edgelist(g)[i,][2]; t <- c(V(g)[tail]$x, V(g)[tail]$y)
d <- NULL
# loop over all points
for(j in points) d <- c(d, runif(1))
E(g)[i]$d <- min(d) # local min
E(g)[i]$p <- points[which(d == min(d))]
} # i
ei = which.min(E(g)$d) # edge with the global min
vi = E(g)[ei]$p
# head and tail of edge with global min
head <- get.edgelist(g)[E(g)[ei],][1]; tail <- get.edgelist(g)[E(g)[ei],][2]
g <- add_edges(g, c(head, V(g)[vi],
V(g)[vi],
tail));
g <- delete_edges(g, get.edge.ids(g, c(head, tail) ))
}
plot(g, vertex.size=5, layout=xy)
問題。當邊數增加 1 且點數減少 1 evety 步長時，如何組織所有邊上的回圈？可以看出，我沒有以顯式形式使用 k 變數。
uj5u.com熱心網友回復：
我認為您可以使用加上終止條件而不是for回圈，repeat即不再有孤立的頂點
repeat {
points <- V(g)[degree(g) == 0]
for (i in 1:length(E(g))) { # loop over all edges
head <- get.edgelist(g)[i, ][1]
h <- c(V(g)[head]$x, V(g)[head]$y)
tail <- get.edgelist(g)[i, ][2]
t <- c(V(g)[tail]$x, V(g)[tail]$y)
d <- NULL
# loop over all points
for (j in points) d <- c(d, runif(1))
E(g)[i]$d <- min(d) # local min
E(g)[i]$p <- points[which(d == min(d))]
} # i
ei <- which.min(E(g)$d) # edge with the global min
vi <- E(g)[ei]$p
# head and tail of edge with global min
head <- get.edgelist(g)[E(g)[ei], ][1]
tail <- get.edgelist(g)[E(g)[ei], ][2]
g <- add_edges(g, c(
head, V(g)[vi],
V(g)[vi],
tail
))
g <- delete_edges(g, get.edge.ids(g, c(head, tail)))
if (sum(degree(g) == 0) == 0) {
break
}
}
uj5u.com熱心網友回復：
我會建議你為此使用遞回并放棄回圈——對樹和圖形結構使用遞回肯定會讓你的生活更輕松。
回答：
- 維護所有葉節點的堆疊
- 每次通過匹配葉節點值來迭代清空堆疊時
- 如果堆疊有新值和計數！= 到舊計數。
- 現在再次迭代。
轉載請註明出處，本文鏈接：https://www.uj5u.com/qiye/526278.html
標籤：rfor循环递归图形
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