我有一个节点列表，我需要随机分配＆＃39; p＆＃39;枢纽到＆＃39; n＆＃39;客户端。
我得到了以下数据，第一行显示：
以下行显示：
下面我将展示原始数据，添加colnames()它看起来像这样：
total_nodes hubs_required total_capacity
50 5 120
node number x_coordinate y_coordinate node_demand
1 2 62 3
2 80 25 14
3 36 88 1
4 57 23 14
. . . .
. . . .
. . . .
50 1 58 2
提供了x和y值，因此我们可以计算欧几里德距离。
nodes:
50 5 120
1 2 62 3
2 80 25 14
3 36 88 1
4 57 23 14
5 33 17 19
6 76 43 2
7 77 85 14
8 94 6 6
9 89 11 7
10 59 72 6
11 39 82 10
12 87 24 18
13 44 76 3
14 2 83 6
15 19 43 20
16 5 27 4
17 58 72 14
18 14 50 11
19 43 18 19
20 87 7 15
21 11 56 15
22 31 16 4
23 51 94 13
24 55 13 13
25 84 57 5
26 12 2 16
27 53 33 3
28 53 10 7
29 33 32 14
30 69 67 17
31 43 5 3
32 10 75 3
33 8 26 12
34 3 1 14
35 96 22 20
36 6 48 13
37 59 22 10
38 66 69 9
39 22 50 6
40 75 21 18
41 4 81 7
42 41 97 20
43 92 34 9
44 12 64 1
45 60 84 8
46 35 100 5
47 38 2 1
48 9 9 7
49 54 59 9
50 1 58 2
我从第一行提取了信息。
nodes <- as.matrix(read.table(data))
header<-colnames(nodes)
clean_header <-gsub('X','',header)
requested_hubs <- as.numeric(clean_header[2])
max_supply_capacity <- as.numeric(clean_header[3])
我需要随机选择5个节点，这些节点将充当集线器
set.seed(37)
node_to_hub <-nodes[sample(nrow(nodes),requested_hubs,replace = FALSE),]
然后随机我需要为每个集线器分配节点，计算集线器与每个节点之间的距离，当超出max_supply_capacity（120）时，选择以下集线器和重复这个过程。
在最后一次迭代之后，我需要返回所有集线器的累积距离总和。
我需要重复此过程100次并返回累积距离总和的min()值。
这是我完全陷入困境的地方，因为当我不得不随机选择元素时，我不确定如何循环使用矩阵。
我有以下要素：
capacity <- c(numeric()) # needs to be <= to 120
distance_sum <- c(numeric())
global_hub_distance <- c(numeric())
欧氏距离（圆形）的公式如下所示，但我不确定在分配节点时如何反映随机选择。
distance <-round(sqrt(((node_to_hub[i,2]-nodes[i,2]))^2+(node_to_hub[random,3]-nodes[random,3])^2))
我认为我需要的循环的想法如下，但正如我之前提到的，我不知道如何处理样本客户端选择以及随机客户端的距离计算。
for(i in 1:100){
node_to_hub
for(i in 1:nrow(node_to_hub){
#Should I randomly sample the clients here???
while(capacity < 120){
node_demand <- nodes[**random**,3]
distance <-round(sqrt(((node_to_hub[i,2]-nodes[i,2]))^2+(node_to_hub[**random**,3]-nodes[**random**,3])^2))
capacity <-c(capacity, node_demand)
distance_sum <- c(distance_sum,distance)
}
global_hub_distance <- c(global_hub_distance,distance_sum)
capacity <- 0
distance_sum <- 0
}
min(global_hub_distance)
}
答案 0 :(得分：1)
不完全确定您要找的是什么，但此代码可能能够为您提供帮助。它并不是非常快，因为在完成整个节点列表后，只需在整个节点列表上找到一个字符串，找到超过120的位置，而不是在达到total_capacity之后停止使用一段时间。
nodes <- structure(list(node_number = 1:50,
x = c(2L, 80L, 36L, 57L, 33L, 76L, 77L, 94L,
89L, 59L, 39L, 87L, 44L, 2L, 19L, 5L,
58L, 14L, 43L, 87L, 11L, 31L, 51L, 55L,
84L, 12L, 53L, 53L, 33L, 69L, 43L, 10L,
8L, 3L, 96L, 6L, 59L, 66L, 22L, 75L, 4L,
41L, 92L, 12L, 60L, 35L, 38L, 9L, 54L, 1L),
y = c(62L, 25L, 88L, 23L, 17L, 43L, 85L, 6L, 11L,
72L, 82L, 24L, 76L, 83L, 43L, 27L, 72L, 50L,
18L, 7L, 56L, 16L, 94L, 13L, 57L, 2L, 33L, 10L,
32L, 67L, 5L, 75L, 26L, 1L, 22L, 48L, 22L, 69L,
50L, 21L, 81L, 97L, 34L, 64L, 84L, 100L, 2L, 9L, 59L, 58L),
node_demand = c(3L, 14L, 1L, 14L, 19L, 2L, 14L, 6L,
7L, 6L, 10L, 18L, 3L, 6L, 20L, 4L,
14L, 11L, 19L, 15L, 15L, 4L, 13L,
13L, 5L, 16L, 3L, 7L, 14L, 17L,
3L, 3L, 12L, 14L, 20L, 13L, 10L,
9L, 6L, 18L, 7L, 20L, 9L, 1L, 8L,
5L, 1L, 7L, 9L, 2L)),
.Names = c("node_number", "x", "y", "node_demand"),
class = "data.frame", row.names = c(NA, -50L))
total_nodes = nrow(nodes)
hubs_required = 5
total_capacity = 120
iterations <- 100
track_sums <- matrix(NA, nrow = iterations, ncol = hubs_required)
colnames(track_sums) <- paste0("demand_at_hub",1:hubs_required)
然后我更喜欢使用距离函数，在这种情况下，A和B是2个单独的向量，c（x，y）和c（x，y）。
euc.dist <- function(A, B) round(sqrt(sum((A - B) ^ 2))) # distances
The Loop：
for(i in 1:iterations){
# random hub selection
hubs <- nodes[sample(1:total_nodes, hubs_required, replace = FALSE),]
for(h in 1:hubs_required){
# sample the nodes into a random order
random_nodes <- nodes[sample(1:nrow(nodes), size = nrow(nodes), replace = FALSE),]
# cumulative sum their demand, and get which number passes 120,
# and subtract 1 to get the node before that
last <- which(cumsum(random_nodes$node_demand) > total_capacity) [1] - 1
# get sum of all distances to those nodes (1 though the last)
all_distances <- apply(random_nodes[1:last,], 1, function(rn) {
euc.dist(A = hubs[h,c("x","y")],
B = rn[c("x","y")])
})
track_sums[i,h] <- sum(all_distances)
}
}
min(rowSums(track_sums))
修改
作为一个功能：
hubnode <- function(nodes, hubs_required = 5, total_capacity = 120, iterations = 10){
# initialize results matrices
track_sums <- node_count <- matrix(NA, nrow = iterations, ncol = hubs_required)
colnames(track_sums) <- paste0("demand_at_hub",1:hubs_required)
colnames(node_count) <- paste0("nodes_at_hub",1:hubs_required)
# user defined distance function (only exists wihtin hubnode() function)
euc.dist <- function(A, B) round(sqrt(sum((A - B) ^ 2)))
for(i in 1:iterations){
# random hub selection
assigned_hubs <- sample(1:nrow(nodes), hubs_required, replace = FALSE)
hubs <- nodes[assigned_hubs,]
assigned_nodes <- NULL
for(h in 1:hubs_required){
# sample the nodes into a random order
assigned_nodes <- sample((1:nrow(nodes))[-assigned_hubs], replace = FALSE)
random_nodes <- nodes[assigned_nodes,]
# cumulative sum their demand, and get which number passes 120,
# and subtract 1 to get the node before that
last <- which(cumsum(random_nodes$node_demand) > total_capacity) [1] - 1
# if there are none
if(is.na(last)) last = nrow(random_nodes)
node_count[i,h] <- last
# get sum of all distances to those nodes (1 though the last)
all_distances <- apply(random_nodes[1:last,], 1, function(rn) {
euc.dist(A = hubs[h,c("x","y")],
B = rn[c("x","y")])
})
track_sums[i,h] <- sum(all_distances)
}
}
return(list(track_sums = track_sums, node_count = node_count))
}
output <- hubnode(nodes, iterations = 100)
node_count <- output$node_count
track_sums <- output$track_sums
plot(rowSums(node_count),
rowSums(track_sums), xlab = "Node Count", ylab = "Total Demand", main = paste("Result of", 100, "iterations"))
min(rowSums(track_sums))