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问题
我正在寻找一种方法来有效地对固定对象周围发生的随机选择的采样横断面进行排序。这些横断面一旦生成,就需要以一种在空间上有意义的方式进行排序,以使行进的距离最小化。这将通过确保当前横断面的终点尽可能靠近下一个横断面的起点来实现。此外,没有一个横断面可以重复。
因为有数千条断面需要订购,这是手动完成的一项非常繁琐的任务,我正在尝试使用 R 来自动化此过程。我已经生成了横断面,每个横断面都有一个起点和终点,其位置使用 360 度系统指示(例如,0 是北,90 是东,180 是南,270 是西)。我还生成了一些似乎指示下一个横断面的起点和 ID 的代码,但此代码存在一些问题:(1) 根据所考虑的起点和终点,它可能会产生错误,(2 ) 它没有实现我最终需要它实现的目标,并且 (3) 照原样,代码本身似乎过于复杂,我不禁想知道是否有更直接的方法来做到这一点。
理想情况下,代码会导致横断面被重新排序,以便它们匹配它们应该飞行的顺序,而不是它们最初输入的顺序。
数据
为简单起见,我们假设只有 10 个横断面需要排序。
# Transect ID for the start point
StID <- c(seq(1, 10, 1))
# Location of transect start point, based on a 360-degree circle
StPt <- c(342.1, 189.3, 116.5, 67.9, 72, 208.4, 173.2, 97.8, 168.7, 138.2)
# Transect ID for the end point
EndID <- c(seq(1, 10, 1))
# Location of transect start point, based on a 360-degree circle
EndPt <- c(122.3, 313.9, 198.7, 160.4, 166, 26.7, 312.7, 273.7, 288.8, 287.5)
# Dataframe
df <- cbind.data.frame(StPt, StID, EndPt, EndID)
# Create two new columns that will be populated using a loop
df$StPt_Next <- NA
df$ID_Next <- NA
# Also create a list to be populated as end and start points are matched
used <- c(df$StPt[1]) #puts the start point of transect #1 into the used vector since we will start with 1 and do not want to have it used again
# Then, for every row in the dataframe...
for (i in seq(1,length(df$EndPt)-1, 1)){ # Selects all rows except the last one as the last transect should have no "next" transect
# generate some print statements to indicate that the script is indeed running while you wait....
print(paste("######## ENDPOINT", i, ":", df$EndPt[i], " ########"))
print(paste("searching for a start point that fits criteria to follow this endpoint",sep=""))
# sequentially select each end point
valueEndPt <- df[i,1]
# and order the index by taking the absolute difference of end and start points and, if this value is greater than 180, also subtract from 360 so all differences are less than 180, then order differences from smallest to largest
orderx <- order(ifelse(360-abs(df$StPt-valueEndPt) > 180,
abs(df$StPt-valueEndPt),
360-abs(df$StPt-valueEndPt)))
tmp <- as.data.frame(orderx)
# specify index value
index=1
# for as long as there is an "NA" present in the StPt_Next created before for loop...
while (is.na(df$StPt_Next[i])) {
#select the value of the ordered index in sequential order
j=orderx[index]
# if the start point associated with a given index is present in the list of used values...
if (df$StPt[j] %in% used){
# then have R print a statement indicate this is the case...
print(paste("passing ",df$StPt[j], " as it has already been used",sep=""))
# and move onto the next index
index=index+1
# break statement intended to skip the remainder of the code for values that have already been used
next
# if the start point associated with a given index is not present in the list of used values...
} else {
# then identify the start point value associated with that index ID...
valueStPt <- df$StPt[j]
# and have R print a statement indicating an attempt is being made to use the next value
print(paste("trying ",df$StPt[j],sep=""))
# if the end transect number is different from the start end transect number...
if (df$EndID[i] != df$StID[j]) {
# then put the start point in the new column...
df$StPt_Next[i] <- df$StPt[j]
# note which record this start point came from for ease of reference/troubleshooting...
df$ID_Next[i] <- j
# have R print a statement that indicates a value for the new column has beed selected...
print(paste("using ",df$StPt[j],sep=""))
# and add that start point to the list of used ones
used <- c(used,df$StPt[j])
# otherwise, if the end transect number matches the start end transect number...
} else {
# keep NA in this column and try again
df$StPt_Next[i] <- NA
# and indicate that this particular matched pair can not be used
print(paste("cant use ",valueStPt," as the column EndID (related to index in EndPt) and StID (related to index in StPt) values are matching",sep=""))
}# end if else statement to ensure that start and end points come from different transects
# and move onto the next index
index=index+1
}# end if else statement to determine if a given start point still needs to be used
}# end while loop to identify if there are still NA's in the new column
}# end for loop
StPt StID EndPt EndID StPt_Next ID_Next
1 342.1 1 122.3 1 67.9 4
2 189.3 2 313.9 2 173.2 7
3 116.5 3 198.7 3 97.8 8
4 67.9 4 160.4 4 72.0 5
5 72.0 5 166.0 5 116.5 3
6 208.4 6 26.7 6 189.3 2
7 173.2 7 312.7 7 168.7 9
8 97.8 8 273.7 8 138.2 10
9 168.7 9 288.8 9 208.4 6
10 138.2 10 287.5 10 NA NA
StPt StID EndPt EndID StPt_Next ID_Next TransNum
1 342.1 1 122.3 1 67.9 4 1
4 67.9 4 160.4 4 72.0 5 2
5 72.0 5 166.0 5 116.5 3 3
3 116.5 3 198.7 3 97.8 8 4
8 97.8 8 273.7 8 138.2 10 5
10 138.2 10 287.5 10 NA NA 6
EndPt <- c(158.7,245.1,187.1,298.2,346.8,317.2,74.5,274.2,153.4,246.7,193.6,302.3,6.8,359.1,235.4,134.5,111.2,240.5,359.2,121.3,224.5,212.6,155.1,353.1,181.7,334,249.3,43.9,38.5,75.7,344.3,45.1,285.7,155.5,183.8,60.6,301,132.1,75.9,112,342.1,302.1,288.1,47.4,331.3,3.4,185.3,62,323.7,188,313.1,171.6,187.6,291.4,19.2,210.3,93.3,24.8,83.1,193.8,112.7,204.3,223.3,210.7,201.2,41.3,79.7,175.4,260.7,279.5,82.4,200.2,254.2,228.9,1.4,299.9,102.7,123.7,172.9,23.2,207.3,320.1,344.6,39.9,223.8,106.6,156.6,45.7,236.3,98.1,337.2,296.1,194,307.1,86.6,65.5,86.6,296.4,94.7,279.9)
StPt <- c(56.3,158.1,82.4,185.5,243.9,195.6,335,167,39.4,151.7,99.8,177.2,246.8,266.1,118.2,358.6,357.9,99.6,209.9,342.8,106.5,86.4,35.7,200.6,65.6,212.5,159.1,297,285.9,300.9,177,245.2,153.1,8.1,76.5,322.4,190.8,35.2,342.6,8.8,244.6,202,176.2,308.3,184.2,267.2,26.6,293.8,167.3,30.5,176,74.3,96.9,186.7,288.2,62.6,331.4,254.7,324.1,73.4,16.4,64,110.9,74.4,69.8,298.8,336.6,58.8,170.1,173.2,330.8,92.6,129.2,124.7,262.3,140.4,321.2,34,79.5,263,66.4,172.8,205.5,288,98.5,335.2,38.7,289.7,112.7,350.7,243.2,185.4,63.9,170.3,326.3,322.9,320.6,199.2,287.1,158.1)
EndID <- c(seq(1, 100, 1))
StID <- c(seq(1, 100, 1))
df <- cbind.data.frame(StPt, StID, EndPt, EndID)
最佳答案
正如@chinsoon12 指出隐藏在您的问题中,您有一个(不对称的)旅行商问题。不对称的出现是因为您的 transec 的起点和终点不同。
ATSP 是一个著名的 NP 完全问题。因此,即使对于中等规模的问题,精确的解决方案也非常困难(有关更多信息,请参阅 wikipedia)。因此,在大多数情况下,我们能做的最好的事情是近似或启发式。正如您提到的,有数千条断面,这至少是一个中等规模的问题。
不是从一开始就编写 ATSP 近似算法,而是有一个现有的 R 的 TSP 库。这包括几个近似算法。引用文档是 here .
以下是我对您的问题使用的 TSP 包。从设置开始(假设我在您的问题中运行了 StPt
、 StID
、 EndPt
和 EndID
。
install.packages("TSP")
library(TSP)
library(dplyr)
# Dataframe
df <- cbind.data.frame(StPt, StID, EndPt, EndID)
# filter to 6 example nodes for requested comparison
df = df %>% filter(StID %in% c(1,3,4,5,8,10))
[row,col]
矩阵中是从(结束)横断面的成本/距离
row
到(起点)横断面
col
.此代码创建整个距离矩阵。
# distance calculation
transec_distance = function(end,start){
abs_dist = abs(start-end)
ifelse(360-abs_dist > 180, abs_dist, 360-abs_dist)
}
# distance matrix
matrix_distance = matrix(data = NA, nrow = nrow(df), ncol = nrow(df))
for(start_id in 1:nrow(df)){
start_point = df[start_id,'StPt']
for(end_id in 1:nrow(df)){
end_point = df[end_id,'EndPt']
matrix_distance[end_id,start_id] = transec_distance(end_point, start_point)
}
}
answer = solve_TSP(as.ATSP(matrix_distance))
# get length of cycle
print(answer)
# sort df to same order as solution
df_w_answer = df[as.numeric(answer),]
# add info about next transect to each transect
df_w_answer = df_w_answer %>%
mutate(visit_order = 1:nrow(df_w_answer)) %>%
mutate(next_StID = lead(StID, order_by = visit_order),
next_StPt = lead(StPt, order_by = visit_order))
# add info about next transect to each transect (for final transect)
df_w_answer[df_w_answer$visit_order == nrow(df_w_answer),'next_StID'] =
df_w_answer[df_w_answer$visit_order == 1,'StID']
df_w_answer[df_w_answer$visit_order == nrow(df_w_answer),'next_StPt'] =
df_w_answer[df_w_answer$visit_order == 1,'StPt']
# compute distance between end of each transect and start of next
df_w_answer = df_w_answer %>% mutate(dist_between = transec_distance(EndPt, next_StPt))
EndID
至
next_StID
,并且您将覆盖(一个很好的近似值)最小距离内的每个断面。
# as path (without returning to start)
min_distance = sum(df_w_answer$dist_between) - max(df_w_answer$dist_between)
path_start = df_w_answer[df_w_answer$dist_between == max(df_w_answer$dist_between), 'next_StID']
path_end = df_w_answer[df_w_answer$dist_between == max(df_w_answer$dist_between), 'EndID']
print(sprintf("minimum cost path = %.2f, starting at node %d, ending at node %d",
min_distance, path_start, path_end))
1 --> 5 --> 8 --> 4 --> 3 --> 10 --> 1
.
关于r - 使用 R 对随机生成的横断面进行有效排序,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/62124593/
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