Question

我有一个元组列表和python中的单个点，例如[（1,2），（2,5），（6,7），（9,3）]和（2,1），我想弄清楚各个点的所有组合可能创建的最快路径到点列表。（基本上我想找到从（2,1）开始到达所有点的最有效方法）。我有一个manhattanDistance函数，它可以取2点并输出距离。但是，我的算法给出了不一致的答案（启发式因某种原因而关闭）

实现这一目标的正确方法是什么？

这是我以前的算法：

def bestPath(currentPoint,goalList):
    sum = 0
    bestList = []
    while len(goallist) > 0:
        for point in list:
            bestList.append((manhattanD(point,currentPoint),point))
        bestTup = min(bestList)
        bestList = []
        dist = bestTup[0]
        newP = bestTup[1]
        currentPoint = newP
        sum += dist
return sum

Answer 1

由于您没有那么多要点，您可以轻松使用尝试各种可能性的解决方案。

以下是您可以做的事情：

首先获得所有组合：

>>> list_of_points = [(1,2) , (2,5), (6,7), (9,3)]
>>> list(itertools.permutations(list_of_points))
[((1, 2), (2, 5), (6, 7), (9, 3)),
((1, 2), (2, 5), (9, 3), (6, 7)),
((1, 2), (6, 7), (2, 5), (9, 3)),
((1, 2), (6, 7), (9, 3), (2, 5)),
((1, 2), (9, 3), (2, 5), (6, 7)),
((1, 2), (9, 3), (6, 7), (2, 5)),
((2, 5), (1, 2), (6, 7), (9, 3)),
((2, 5), (1, 2), (9, 3), (6, 7)),
((2, 5), (6, 7), (1, 2), (9, 3)),
((2, 5), (6, 7), (9, 3), (1, 2)),
((2, 5), (9, 3), (1, 2), (6, 7)),
((2, 5), (9, 3), (6, 7), (1, 2)),
((6, 7), (1, 2), (2, 5), (9, 3)),
((6, 7), (1, 2), (9, 3), (2, 5)),
((6, 7), (2, 5), (1, 2), (9, 3)),
((6, 7), (2, 5), (9, 3), (1, 2)),
((6, 7), (9, 3), (1, 2), (2, 5)),
((6, 7), (9, 3), (2, 5), (1, 2)),
((9, 3), (1, 2), (2, 5), (6, 7)),
((9, 3), (1, 2), (6, 7), (2, 5)),
((9, 3), (2, 5), (1, 2), (6, 7)),
((9, 3), (2, 5), (6, 7), (1, 2)),
((9, 3), (6, 7), (1, 2), (2, 5)),
((9, 3), (6, 7), (2, 5), (1, 2))]

然后创建一个函数，为您提供组合的长度：

def combination_length(start_point, combination):
    lenght = 0
    previous = start_point  
    for elem in combination:
        lenght += manhattanDistance(previous, elem)

    return length

最后一个测试每种可能性的函数：

def get_shortest_path(start_point, list_of_point):
    min = sys.maxint
    combination_min = None
    list_of_combinations = list(itertools.permutations(list_of_points))
    for combination in list_of_combination:
        length = combination_length(start_point, combination)
        if length < min:
            min = length
            combination_min = combination

    return combination_min

然后最后你可以：

import sys, itertools

def combination_length(start_point, combination):
    lenght = 0
    previous = start_point  
    for elem in combination:
        lenght += manhattanDistance(previous, elem)

    return length

def get_shortest_path(start_point, list_of_point):
    min = sys.maxint
    combination_min = None
    list_of_combinations = list(itertools.permutations(list_of_points))
    for combination in list_of_combination:
        length = combination_length(start_point, combination)
        if length < min:
            min = length
            combination_min = combination

    return combination_min

list_of_points = [(1,2) , (2,5), (6,7), (9,3)]
print get_shortest_path((2,1), list_of_points)

Answer 2

如果这与旅行商问题类似，那么您想查看NetworkX python模块。

Answer 3

尝试找到所有组合，然后检查最短距离。

找到python列表中点之间最短距离的清洁方法？

3 个答案: