返回给定字典键的下一个键,python 3.6+

F. *_*. A 2 python dictionary iterator python-3.8

我正在尝试找到一种方法来获取 Python 3.6+ 的下一个密钥(已排序)

例如:

dict = {'one':'value 1','two':'value 2','three':'value 3'}
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我想要实现的是返回下一个键的功能。就像是:

next_key(dict, current_key='two')   # -> should return 'three' 
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这是我到目前为止所拥有的:

def next_key(dict,key):
    key_iter = iter(dict)  # create iterator with keys
    while k := next(key_iter):    #(not sure if this is a valid way to iterate over an iterator)
        if k == key:   
            #key found! return next key
            try:    #added this to handle when key is the last key of the list
                return(next(key_iter))
            except:
                return False
    return False

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好吧,这就是基本思想,我想我已经很接近了,但是这段代码给出了 StopIteration 错误。请帮忙。

谢谢你!

Kel*_*ndy 7

迭代器方式...

def next_key(dict, key):
    keys = iter(dict)
    key in keys
    return next(keys, False)
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演示:

>>> next_key(dict, 'two')
'three'
>>> next_key(dict, 'three')
False
>>> next_key(dict, 'four')
False
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Ry-*_*Ry- 6

循环while k := next(key_iter)无法正确停止\xe2\x80\x99。手动迭代iter可以通过捕获来完成StopIteration

\n\n
iterator = iter(some_iterable)\n\nwhile True:\n    try:\n        value = next(iterator)\n    except StopIteration:\n        # no more items\n
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或者通过传递默认值next并让它捕获StopIteration您,然后检查该默认值(但您需要选择一个不会出现在可迭代中的默认值\xe2\x80\x99!):

\n\n
iterator = iter(some_iterable)\n\nwhile True:\n    try:\n        value = next(iterator)\n    except StopIteration:\n        # no more items\n
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但迭代器本身是可迭代的,因此您可以跳过所有这些并使用普通的 ol\xe2\x80\x99 for 循环:

\n\n
iterator = iter(some_iterable)\n\nwhile (value := next(iterator, None)) is not None:\n    # \xe2\x80\xa6\n\n# no more items\n
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这翻译成你的例子是:

\n\n
iterator = iter(some_iterable)\n\nfor value in iterator:\n    # \xe2\x80\xa6\n\n# no more items\n
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