iter，值，字典中的项不起作用 [英] iter, values, item in dictionary does not work

问题描述

有这个python代码

Having this python code

edges = [(0, [3]), (1, [0]), (2, [1, 6]), (3, [2]), (4, [2]), (5, [4]), (6, [5, 8]), (7, [9]), (8, [7]), (9, [6])]
graph = {0: [3], 1: [0], 2: [1, 6], 3: [2], 4: [2], 5: [4], 6: [5, 8], 7: [9], 8: [7], 9: [6]}
cycles = {}
while graph:
    current = graph.iteritems().next()
    cycle = [current]
    cycles[current] = cycle
    while current in graph:
        next = graph[current][0]
        del graph[current][0]
        if len(graph[current]) == 0:
            del graph[current]
        current = next
        cycle.append(next)


def traverse(tree, root):
    out = []
    for r in tree[root]:
        if r != root and r in tree:
            out += traverse(tree, r)
        else:
            out.append(r)
    return out

print ('->'.join([str(i) for i in traverse(cycles, 0)]))



Traceback (most recent call last):
  File "C:\Users\E\Desktop\c.py", line 20, in <module>
    current = graph.iteritems().next()
AttributeError: 'dict' object has no attribute 'iteritems'

我还尝试了itervalues，iterkeys ...但是不起作用
如何修改代码？

I also tried itervalues, iterkeys... but that does not work How to modify code?

推荐答案

您正在使用Python 3;使用 dict.items（）

You are using Python 3; use dict.items() instead.

Python 2 dict.iter * / code>方法已经在Python 3中重命名，其中 dict.items（）默认情况下返回字典视图而不是列表。字典视图以与Python 2相同的方式 dict.iteritems（）做为迭代。

The Python 2 dict.iter* methods have been renamed in Python 3, where dict.items() returns a dictionary view instead of a list by default now. Dictionary views act as iterables in the same way dict.iteritems() do in Python 2.

从<一个href =http://docs.python.org/3.1/whatsnew/3.0.html#views-and-iterators-instead-of-lists> Python 3最新的文档：




• dict 方法 dict.keys（） ， dict.items（）和 dict.values（）返回views的列表。例如，这不再适用： k = d.keys（）; k.sort（）。使用 k = sorted（d）（这也适用于Python 2.5，同样有效）。


• 另外， code> dict.iterkeys（）， dict.iteritems（）和 dict.itervalues（） code>方法不再受支持。



• dict methods dict.keys(), dict.items() and dict.values() return "views" instead of lists. For example, this no longer works: k = d.keys(); k.sort(). Use k = sorted(d) instead (this works in Python 2.5 too and is just as efficient).
• Also, the dict.iterkeys(), dict.iteritems() and dict.itervalues() methods are no longer supported.

另外， .next（）方法已重命名为 .__ next __（），但字典视图不是迭代器。必须将 graph.iteritems（）。next（）的行转换为：

Also, the .next() method has been renamed to .__next__(), but dictionary views are not iterators. The line graph.iteritems().next() would have to be translated instead, to:

current = next(iter(graph.items()))

使用 iter（）将项目视图转换为可迭代的 next（）以从该可迭代中获取下一个值

which uses iter() to turn the items view into an iterable and next() to get the next value from that iterable.

您还必须在中重命名下一个变量，而 loop;使用它取代了你在这里需要的内置 next（）函数。使用下一个_ 。

You'll also have to rename the next variable in the while loop; using that replaces the built-in next() function which you need here. Use next_ instead.

下一个问题是您正在尝试使用 current 作为循环中的键，但当前是整数和列表的整数，使整个值不可哈希。我想你想要获得下一个键，在这种情况下， next（iter（dict））给你：

The next problem is that you are trying to use current as a key in cycles, but current is a tuple of an integer and a list of integers, making the whole value not hashable. I think you wanted to get just the next key instead, in which case next(iter(dict)) would give you that:

while graph:
    current = next(iter(graph))
    cycle = [current]
    cycles[current] = cycle
    while current in graph:
        next_ = graph[current][0]
        del graph[current][0]
        if len(graph[current]) == 0:
            del graph[current]
        current = next_
        cycle.append(next_)

然后产生一些输出：

>>> cycles
{0: [0, 3, 2, 1, 0], 2: [2, 6, 5, 4, 2], 6: [6, 8, 7, 9, 6]}

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