Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
467 views
in Technique[技术] by (71.8m points)

python - How can I correct the error ' AttributeError: 'dict_keys' object has no attribute 'remove' '?

I was trying shortest path finder using dijkstra algorithm but It seems not working. Can't figure out what the problem is. Here are the code and the error message. (I'm working on Python 3.5. https://www.youtube.com/watch?v=LHCVNtxb4ss)

graph = {
    'A': {'B': 10, 'D': 4, 'F': 10},
    'B': {'E': 5, 'J': 10, 'I': 17},
    'C': {'A': 4, 'D': 10, 'E': 16},
    'D': {'F': 12, 'G': 21},
    'E': {'G': 4},
    'F': {'E': 3},
    'G': {'J': 3},
    'H': {'G': 3, 'J': 3},
    'I': {},
    'J': {'I': 8},
}

def dijkstra(graph, start, end):
    D = {}
    P = {}
    for node in graph.keys():
        D[node]= -1
        P[node]=""
    D[start]=0
    unseen_nodes=graph.keys()
    while len(unseen_nodes) > 0:
        shortest=None
        node=' '
        for temp_node in unseen_nodes:
            if shortest==None:
                shortest = D[temp_node]
                node = temp_node
            elif D[temp_node]<shortest:
                    shortest=D[temp_node]
                    node=temp_node
        unseen_nodes.remove(node)
        for child_node, child_value in graph[node].items():
            if D[child_node] < D[node] + child_value:
                D[child_node] = D[node] + child_value
                P[child_node]=node
    path = []
    node = end
    while not (node==start):
        if path.count(node)==0:
            path.insert(0, node)
            node=P[node]
        else:
            break
    path.insert(0, start)
    return path

AttributeError: 'dict_keys' object has no attribute 'remove'

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

In Python 3, dict.keys() returns a dict_keys object (a view of the dictionary) which does not have remove method; unlike Python 2, where dict.keys() returns a list object.

>>> graph = {'a': []}
>>> keys = graph.keys()
>>> keys
dict_keys(['a'])
>>> keys.remove('a')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'dict_keys' object has no attribute 'remove'

You can use list(..) to get a keys list:

>>> keys = list(graph)
>>> keys
['a']
>>> keys.remove('a')
>>> keys
[]

unseen_nodes = graph.keys()

to

unseen_nodes = list(graph)

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...