Question

我有一个具有30,000条记录的数据框，格式如下：

ID | Name | Latitude | Longitude | Country |
1  | Hull | 53.744   | -0.3456   | GB      |

我想选择一条记录作为起始位置，并选择一条记录作为目的地，并返回最短路径的路径（列表）。

我正在使用Geopy查找两点之间的距离，以公里为单位

import geopy.distance

coords_1 = (52.2296756, 21.0122287)
coords_2 = (52.406374, 16.9251681)

print (geopy.distance.vincenty(coords_1, coords_2).km)

我已经从以下教程中阅读了如何在python中执行A *： https://www.redblobgames.com/pathfinding/a-star/implementation.html

但是他们创建了一个网格系统来浏览。

这是数据框中记录的直观表示：

这是我到目前为止的代码，但是找不到路径：

def calcH(start, end):
    coords_1 = (df['latitude'][start], df['longitude'][start])
    coords_2 = (df['latitude'][end], df['longitude'][end])
    distance = (geopy.distance.vincenty(coords_1, coords_2)).km
    return distance

^计算点之间的距离

def getneighbors(startlocation):
    neighborDF = pd.DataFrame(columns=['ID', 'Distance'])
    coords_1 = (df['latitude'][startlocation], df['longitude'][startlocation])
    for index, row in df.iterrows():
        coords_2 = (df['latitude'][index], df['longitude'][index])
        distance = round((geopy.distance.vincenty(coords_1, coords_2)).km,2)
        neighborDF.loc[len(neighborDF)] = [index, distance]
    neighborDF = neighborDF.sort_values(by=['Distance'])
    neighborDF = neighborDF.reset_index(drop=True)

    return neighborDF[1:5]

^返回4个最接近的位置（忽略自身）

openlist = pd.DataFrame(columns=['ID', 'F', 'G', 'H', 'parentID'])
closedlist = pd.DataFrame(columns=['ID', 'F', 'G', 'H', 'parentID'])

startIndex = 25479 # Hessle
endIndex = 8262 # Leeds

h = calcH(startIndex, endIndex)
openlist.loc[len(openlist)] = [startIndex,h, 0, h, startIndex]

while True:

#sort the open list by F score
openlist = openlist.sort_values(by=['F'])
openlist = openlist.reset_index(drop=True)

currentLocation = openlist.loc[0]
closedlist.loc[len(closedlist)] = currentLocation
openlist = openlist[openlist.ID != currentLocation.ID]

if currentLocation.ID == endIndex:
    print("Complete")
    break

adjacentLocations = getneighbors(currentLocation.ID)

if(len(adjacentLocations) < 1):
    print("No Neighbors: " + str(currentLocation.ID))
else:
    print(str(len(adjacentLocations)))

for index, row in adjacentLocations.iterrows():
    if adjacentLocations['ID'][index] in closedlist.values:
        continue

    if (adjacentLocations['ID'][index] in openlist.values) == False:

        g = currentLocation.G + calcH(currentLocation.ID, adjacentLocations['ID'][index])
        h = calcH(adjacentLocations['ID'][index], endIndex)
        f = g + h
        openlist.loc[len(openlist)] = [adjacentLocations['ID'][index], f, g, h, currentLocation.ID]
    else:
        adjacentLocationInDF = openlist.loc[openlist['ID'] == adjacentLocations['ID'][index]] #Get location from openlist
        g = currentLocation.G + calcH(currentLocation.ID, adjacentLocations['ID'][index])
        f = g + adjacentLocationInDF.H
        if float(f) < float(adjacentLocationInDF.F):
            openlist = openlist[openlist.ID != currentLocation.ID]
            openlist.loc[len(openlist)] = [adjacentLocations['ID'][index], f, g, adjacentLocationInDF.H, currentLocation.ID]

if (len(openlist)< 1):
    print("No Path")
    break

从关闭列表中查找路径：

# return the path
pathdf = pd.DataFrame(columns=['name', 'latitude', 'longitude', 'country'])
def getParent(index):

    parentDF = closedlist.loc[closedlist['ID'] == index]
    pathdf.loc[len(pathdf)] = [df['name'][parentDF.ID.values[0]],df['latitude'][parentDF.ID.values[0]],df['longitude'][parentDF.ID.values[0]],df['country'][parentDF.ID.values[0]]]
    if index != startIndex:
        getParent(parentDF.parentID.values[0])

getParent(closedlist['ID'][len(closedlist)-1])

当前，此A *的实现找不到完整的路径。有什么建议吗？

编辑：我尝试将考虑的邻居数量从4增加到10，但得到了一条路径，但没有一条最佳路径：

我们正试图从Hessle到利兹。

^个可用节点

原始数据： Link

Answer 1

我仍然不确定您的方式有什么问题，尽管肯定有一些问题，如评论中已经提到的那样。

您以x in dataframe.values格式进行的检查将检查x是否是values返回的numpy数组中值的 any ，不一定是ID字段
在打开列表中使用数据框而不是适当的堆，而在关闭列表中使用散列集会不必要地降低搜索速度，因为您必须一直搜索并排序整个列表（不确定熊猫是否可以通过索引来加快查找速度，但是排序肯定会花费时间）

无论如何，我发现这是一个有趣的问题，并尝试了一下。但是事实证明，将数据帧用作某种伪堆确实非常慢，并且我还发现数据帧索引非常混乱（并且可能容易出错？），因此我将代码更改为使用{{ 3}}的数据和openlist的适当namedtuple堆，以及dict的{{1}}映射节点到其父节点。另外，检查的次数少于代码中的检查次数（例如，某个节点是否已经在打开列表中），而这些检查实际上并不重要。

closedlist

这为我提供了一条从Hessle到利兹的路线，这似乎更合理：

import csv, geopy.distance, collections, heapq

Location = collections.namedtuple("Location", "ID name latitude longitude country".split())
data = {}
with open("stations.csv") as f:
    r = csv.DictReader(f)
    for d in r:
        i, n, x, y, c = int(d["id"]), d["name"], d["latitude"], d["longitude"], d["country"]
        if c == "GB":
            data[i] = Location(i,n,x,y,c)

def calcH(start, end):
    coords_1 = (data[start].latitude, data[start].longitude)
    coords_2 = (data[end].latitude, data[end].longitude)
    distance = (geopy.distance.vincenty(coords_1, coords_2)).km
    return distance

def getneighbors(startlocation, n=10):
    return sorted(data.values(), key=lambda x: calcH(startlocation, x.ID))[1:n+1]

def getParent(closedlist, index):
    path = []
    while index is not None:
        path.append(index)
        index = closedlist.get(index, None)
    return [data[i] for i in path[::-1]]


startIndex = 25479 # Hessle
endIndex = 8262 # Leeds

Node = collections.namedtuple("Node", "ID F G H parentID".split())

h = calcH(startIndex, endIndex)
openlist = [(h, Node(startIndex, h, 0, h, None))] # heap
closedlist = {} # map visited nodes to parent

while len(openlist) >= 1:
    _, currentLocation = heapq.heappop(openlist)
    print(currentLocation)

    if currentLocation.ID in closedlist:
        continue
    closedlist[currentLocation.ID] = currentLocation.parentID

    if currentLocation.ID == endIndex:
        print("Complete")
        for p in getParent(closedlist, currentLocation.ID):
            print(p)
        break

    for other in getneighbors(currentLocation.ID):
        g = currentLocation.G + calcH(currentLocation.ID, other.ID)
        h = calcH(other.ID, endIndex)
        f = g + h
        heapq.heappush(openlist, (f, Node(other.ID, f, g, h, currentLocation.ID)))

即使您因必须使用熊猫（？）而不能使用它，也许也可以帮助您最终发现实际的错误。

Python：A *从具有经度和纬度的数据帧路由

1 个答案: