从Python中的元组列表中查找路径

Question

我有以下形式的元组列表：

data = [('Abe', 'Bob', '3'), 
        ('Abe', 'Frank', '5'),
        ('Abe', 'George', '4'),
        ('Carl', 'Bob', '1'),
        ('Dan', 'Carl', '2')]

此数据模拟无向图，其中从'Abe'到大小为3的'Bob'之间存在联系。由于该图是无向的，因此也意味着从'Bob'到大小为3的'Abe'之间存在联系。好。

我需要显示两个输入之间是否存在连接及其权重。 例如，如果输入为“ Abe”，“ Dan”，则结果将返回从“ Abe”到“ Dan”的最短（最小节点跳，不是最不重要的权重）路径，即Abe，Bob，Carl，Dan和权重3 + 1 + 2 = 6。

我有这段代码显示“ Abe”是否会到达“ Dan”，但是我不知道如何返回路径。

def get_path_and_weight(data, start, end):

    reachable = [start]
    added = -1

    while added != 0:
        added = 0
        for first, second, weight in data:
            if(first in reachable) and (second not in reachable):
                reachable.append(second)
                added += 1
            if(first not in reachable) and (second in reachable): 
                reachable.append(first)
                added += 1

        if (end in reachable):
            print("YES")
            #return path
        else:
            print("NO")

Answer 1

有许多已经开发和调试过的软件包可以做到这一点，例如， networkx

import networkx as nx

data = [('Abe', 'Bob', '3'), 
        ('Abe', 'Frank', '5'),
        ('Abe', 'George', '4'),
        ('Carl', 'Bob', '1'),
        ('Dan', 'Carl', '2')]

g = nx.Graph()
for e in data:
    g.add_edge(e[0], e[1], distance=int(e[2]))
>>> nx.shortest_path(g, 'Abe', 'Bob', 'distance'), nx.shortest_path_length(g, 'Abe', 'Bob', 'distance')
(['Abe', 'Bob'], 3)

Answer 2

您可以生成所有可能的路径，然后按权重对它们进行排序。 注意我已经将数据中的权重更改为数字，而不是字符串：

data = [
    ('Abe', 'Bob', 3), 
    ('Abe', 'Frank', 5),
    ('Abe', 'George', 4),
    ('Carl', 'Bob', 1),
    ('Dan', 'Carl', 2),
]

WEIGHT = 0
NODES = slice(1, None)

def get_path_and_weight(data, start, end):

    paths = [[0, start]]
    added = True

    while added:
        added = False

        for first, second, weight in data:
            for path in paths:
                candidate = None

                if (first in path[NODES]) and (second not in path[NODES]):
                    candidate = second
                elif (first not in path[NODES]) and (second in path[NODES]):
                    candidate = first

                if candidate:
                    new_path = list(path)
                    new_path.append(candidate)
                    new_path[WEIGHT] += weight

                    if new_path not in paths:
                        paths.append(new_path)
                        added = True

    for path in sorted(paths):
        if end in path[NODES]:
            return path

    return None

然后，您可以将其称为：

weight, *path = get_path_and_weight(data, "Abe", "Dan")

print(path, "with weight", weight)

给出结果：

['Abe', 'Bob', 'Carl', 'Dan'] with weight 6

并且由于它返回路径或None ，因此您仍然可以将其用作谓词函数：

if get_path_and_weight(data, "Abe", "Dan"):
    print("connected")

Answer 3

def get_rpath_with_weight(data,start,end):
    rpath = []
    reachable=False
    nxt_dst = start
    weight_ = 0
    rpath.append(nxt_dst)
    for datum in data:
        if nxt_dst in datum:
            #print datum
            fm_ = datum[0]
            to_ = datum[1]
            weight_ = weight_ + int(datum[2])
            if fm_ == nxt_dst:
               nxt_dst = to_
            else:
               nxt_dst = fm_
            if nxt_dst == end:
               reachable=True
            rpath.append(nxt_dst)
    print rpath,weight_,reachable


get_rpath_with_weight(data,'Abe','Dan')

get_rpath_with_weight(data,'Dan','Frank')

样本输出

['Abe', 'Bob', 'Carl', 'Dan'] 6 True
['Dan', 'Carl'] 2 False

上面的示例可以获取路径并确定是否可访问，但是我认为您也需要进一步增强它以处理多个路径。