Python:来自数据帧的A*路由经度和纬度

Question

我有一个包含以下格式的30,000条记录的数据框:

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,将一条记录作为目标,并返回最短路径的路径(列表).

我使用Geopy来找到以km为单位的点之间的距离

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了解了如何在python中执行A*:https: //www.redblobgames.com/pathfinding/a-star/implementation.html

然而,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,但它找不到路径:

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]

^返回最近的4个位置(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

从封闭列表中查找路径:

# 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])

目前,A*的这种实现并没有找到完整的路径.有什么建议？

编辑:我已经尝试将所考虑的邻居数量从4增加到10,并且我得到了一条路径但不是最佳路径:

我们正试图从赫斯勒到利兹.

^可用节点

原始数据: 链接

Answer 1

我仍然不确定你的方法有什么问题，尽管肯定有一些问题，正如评论中已经提到的。

• 仅考虑最近的四个（或就此而言，任何固定数量的）邻居可能会导致死胡同或图表的某些部分被完全切断，例如，不在“最近的 X”范围内的孤立城市他们的邻居
• 您在表单中的检查x in dataframe.values将检查numpy 数组中是否x有任何values值返回，而不一定是 ID 字段
• 使用数据帧而不是开放列表的适当堆和封闭列表的哈希集会使搜索不必要地变慢，因为您必须始终搜索和排序整个列表（不确定 Pandas 是否可以加快搜索速度）使用索引查找，但排序肯定需要时间）

无论如何，我发现这是一个有趣的问题并尝试了一下。但事实证明，使用数据帧作为某种伪堆确实非常慢，而且我发现数据帧索引非常混乱（并且可能容易出错？），所以我更改了用于namedtuple数据的代码以及 的适当heapq堆openlist以及dict到其父节点的映射节点closedlist。此外，检查次数比代码中少（例如，节点是否已在 openlist 中），而这些检查实际上并不重要。

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)))

这给了我这条从赫塞尔到利兹的路径，这似乎更合理：

Location(ID=25479, name='Hessle', latitude='53.717567', longitude='-0.442169', country='GB')
Location(ID=8166, name='Brough', latitude='53.726452', longitude='-0.578255', country='GB')
Location(ID=25208, name='Eastrington', latitude='53.75481', longitude='-0.786612', country='GB')
Location(ID=25525, name='Howden', latitude='53.764526', longitude='-0.86068', country='GB')
Location(ID=7780, name='Selby', latitude='53.78336', longitude='-1.06355', country='GB')
Location(ID=26157, name='Sherburn-In-Elmet', latitude='53.797142', longitude='-1.23176', country='GB')
Location(ID=25308, name='Garforth Station', latitude='53.796211', longitude='-1.382083', country='GB')
Location(ID=8262, name='Leeds', latitude='53.795158', longitude='-1.549089', country='GB')

即使你不能使用它，因为你必须使用 Pandas（？），也许这可以帮助你最终发现你的实际错误。