我试图两次运行Dijkstra函数,以便找出3个点之间的最短距离。当我运行它一次时,它工作正常,但是两次在第38和44行返回一个关键错误。
代码:
graph = {'c1': {'c2':4, 'L1':3}, 'c2':{'c1':4, 'c3':3, 'L1':2.5}, 'c3':{'c2':3, 'L1':2}, 'L1':{'c1':3, 'c2':2.5, 'c3':2}}
def dijkstra(graph, start, goal):
shortest_distance = {}
predecessor = {}
unseenNodes = graph
infinity = float('inf')
path = []
for node in unseenNodes:
shortest_distance[node] = infinity
shortest_distance[start] = 0
#print(shortest_distance)
while unseenNodes:
minNode = None
for node in unseenNodes:
if minNode is None:
minNode = node
elif shortest_distance[node] < shortest_distance[minNode]:
minNode = node
for childNode, weight in graph[minNode].items():
if weight + shortest_distance[minNode] < shortest_distance[childNode]:
shortest_distance[childNode] = weight + shortest_distance[minNode]
predecessor[childNode] = minNode
unseenNodes.pop(minNode)
currentNode = goal
while currentNode != start:
try:
path.insert(0, currentNode)
currentNode = predecessor[currentNode]
except KeyError:
print('Path not reachable')
break
path.insert(0, start)
if shortest_distance[goal] != infinity:
print('Shortest distance: ' + str(shortest_distance[goal]))
print('Path:' + str(path))
dijkstra(graph, 'L1', 'c2')
dijkstra(graph, 'c2', 'c3')
这是我得到的错误:
Path not reachable
dijkstra(graph, 'c2', 'c3')
File "E:\Work\Delivery_API\Delivery.py", line 38, in dijkstra
if shortest_distance[goal] != infinity:
KeyError: 'c3'
答案 0 :(得分:3)
执行unseenNodes = graph
时,将变量unseenNodes
绑定到由graph
引用的相同字典。
这意味着在unseenNodes
上所做的任何更改也会在<{> {1}} 上反映。
因此,在第一个graph
中一切顺利。
但,在此运行期间,您执行了一些dijkstra(graph, 'L1', 'c2')
,这再次相当于执行unseenNodes.pop(minNode)
。因此,到您进行graph.pop(minNode)
时,您的图表已经改变,看起来并不像您想的那样。这就是为什么仅运行一次即可使用的原因。
简单的补救方法是处理dijkstra(graph, 'c2', 'c3')
的副本,以确保其完好无损。您需要做的就是将分配更改为:graph
。这样会创建unseenNodes = dict(graph)
的副本({{1}现在引用了另一个单独的字典,而不是graph
引用的字典)。