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我已经为 Dijkstra 算法复制并粘贴了这个答案复制并粘贴到我的项目中。经过几次简单的测试后似乎还可以。

在我的具体实现中，我需要算法返回一个节点列表。所以我必须修改原始代码，使其始终返回一个列表。更具体地说，我删除了return "string"那里的所有行。我修改的代码如下：

## using Dijkstra Algorithm ##
def choosePath(s, t):
    net = {'0':{'1':138, '9':150},
       '1':{'0':138, '2':178, '8':194},
       '2':{'1':178, '3':47.5},
       '3':{'2':47.5, '4':70},
       '4':{'3':70, '5':70},
       '5':{'4':70, '6':36},
       '6':{'5':36, '7':50},
       '7':{'6':50, '8':81},
       '8':{'7':81, '9':138, '1':194},
       '9':{'8':138, '0':150}}
    # sanity check
    if s == t:
        return []
    # create a labels dictionary
    labels={}
    # record whether a label was updated
    order={}
    # populate an initial labels dictionary
    for i in net.keys():
        if i == s: labels[i] = 0 # shortest distance form s to s is 0
        else: labels[i] = float("inf") # initial labels are infinity
    from copy import copy
    drop1 = copy(labels) # used for looping
    ## begin algorithm
    while len(drop1) > 0:
        # find the key with the lowest label
        minNode = min(drop1, key = drop1.get) #minNode is the node with the smallest label
        # update labels for nodes that are connected to minNode
        for i in net[minNode]:
            if labels[i] > (labels[minNode] + net[minNode][i]):
                labels[i] = labels[minNode] + net[minNode][i]
                drop1[i] = labels[minNode] + net[minNode][i]
                order[i] = minNode
        del drop1[minNode] # once a node has been visited, it's excluded from drop1
    ## end algorithm
    # print shortest path
    temp = copy(t)
    rpath = []
    path = []
    while 1:
        rpath.append(temp)
        if order.has_key(temp):
            temp = order[temp]
        if temp == s:
            rpath.append(temp)
            break
    for j in range(len(rpath)-1,-1,-1):
        path.append(rpath[j])
    return [junctions[int(elem)] for elem in path]

然后当我运行它时，我最终得到以下错误：

>>> Traceback (most recent call last):
  File "C:\Users\...\simulation.py", line 162, in choosePath
    rpath.append(temp)
MemoryError

显然，这是因为我删除了返回“字符串”行。但是，我没有找出哪个删除使它死亡。为什么会这样？

我怎样才能让它再次工作并且总是返回一个列表而不是我希望的字符串？

4

1 回答 1

2

我怀疑您的问题是您将错误的参数传递给函数。你想打电话choosePath('0', '9')。字符串。不是整数。

可笑的是，如果您删除的程序的任何部分仍然存在，它会抓住这一点并停止程序。使用这部分，它会捕获您的输入是否错误。

if net.has_key(s)==False:
    return "There is no start node called " + str(s) + "."
if net.has_key(t)==False:
    return "There is no terminal node called " + str(t) + "."

有了这部分，它会捕捉到它是否永远无法找到解决方案。

else: return "There is no path from " + str(s) + " to " + str(t) + "."

完整性检查并不是绝对必要的，因为正如您所提到的，在您的网络中保证了一条路径。检查仍然很好，因为如果你确实选择改变周围的东西，你就会知道计算机会因为明显的错误而叫你出来。一种选择是用异常替换它们，因为除非出现严重错误，否则这些消息都不应该真正出现。这就是我在下面的代码中选择的。

class NoPathException(Exception):
    pass

def choosePath(s, t):
    net = {'0':{'1':138, '9':150},
       '1':{'0':138, '2':178, '8':194},
       '2':{'1':178, '3':47.5},
       '3':{'2':47.5, '4':70},
       '4':{'3':70, '5':70},
       '5':{'4':70, '6':36},
       '6':{'5':36, '7':50},
       '7':{'6':50, '8':81},
       '8':{'7':81, '9':138, '1':194},
       '9':{'8':138, '0':150}}
    # sanity check
    if s == t:
        return []
    if not net.has_key(s):
        raise ValueError("start node argument not in net")
    if not net.has_key(t):
        raise ValueError("end node argument not in net")
    # create a labels dictionary
    labels={}
    # record whether a label was updated
    order={}
    # populate an initial labels dictionary
    for i in net.keys():
        if i == s: labels[i] = 0 # shortest distance form s to s is 0
        else: labels[i] = float("inf") # initial labels are infinity
    from copy import copy
    drop1 = copy(labels) # used for looping
    ## begin algorithm
    while len(drop1) > 0:
        # find the key with the lowest label
        minNode = min(drop1, key = drop1.get) #minNode is the nod2 with the smallest label
        # update labels for nodes that are connected to minNode
        for i in net[minNode]:
            if labels[i] > (labels[minNode] + net[minNode][i]):
                labels[i] = labels[minNode] + net[minNode][i]
                drop1[i] = labels[minNode] + net[minNode][i]
                order[i] = minNode
        del drop1[minNode] # once a node has been visited, it's excluded from drop1
    ## end algorithm
    # print shortest path
    temp = copy(t)
    rpath = []
    path = []
    while 1:
        rpath.append(temp)
        if order.has_key(temp):
            temp = order[temp]
        else:
            raise NoPathException("no path to solution")
        if temp == s:
            rpath.append(temp)
            break
    for j in range(len(rpath)-1,-1,-1):
        path.append(rpath[j])

    return path

测试

a = choosePath('3', '9')
print(a)
['3', '4', '5', '6', '7', '8', '9']

这是你要找的输出吗？

于 2013-10-04T10:13:25.817 回答