Return 给定字典键的下一个键，python 3.6+

Question

我正在尝试找到一种方法来获取 Python 3.6+（已订购）的下一个密钥

例如：

dict = {'one':'value 1','two':'value 2','three':'value 3'}

我想要实现的是 return 下一个键的功能。类似于：

next_key(dict, current_key='two')   # -> should return 'three'

这是我目前拥有的：

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

嗯，这是基本的想法，我想我很接近，但是这段代码给出了一个 StopIteration 错误。请帮忙。

谢谢！

Answer 1

您可以将字典的键作为列表获取，并使用index() 获取下一个键。您还可以使用 try/except 块检查 IndexError：

my_dict = {'one':'value 1','two':'value 2','three':'value 3'}

def next_key(d, key):
  dict_keys = list(d.keys())
  try:
    return dict_keys[dict_keys.index(key) + 1]
  except IndexError:
    print('Item index does not exist')
    return -1

nk = next_key(my_dict, key="two")
print(nk)

最好不要使用 dict、list 等作为变量名。

Answer 2

迭代器方式...

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

演示：

>>> next_key(dict, 'two')
'three'
>>> next_key(dict, 'three')
False
>>> next_key(dict, 'four')
False

Answer 3

循环 while k := next(key_iter) 没有正确停止。使用 iter 手动迭代是通过捕获 StopIteration:

来完成的

iterator = iter(some_iterable)

while True:
    try:
        value = next(iterator)
    except StopIteration:
        # no more items

或者通过将默认值传递给 next 并让它为您捕获 StopIteration，然后检查该默认值（但您需要选择一个不会出现在你的迭代！）：

iterator = iter(some_iterable)

while (value := next(iterator, None)) is not None:
    # …

# no more items

但是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' for 循环：

iterator = iter(some_iterable)

for value in iterator:
    # …

# no more items

将您的示例转换为：

def next_key(d, key):
    key_iter = iter(d)

    for k in key_iter:
        if k == key:
            return next(key_iter, None)

    return None

Answer 4

# Python3 code to demonstrate working of 
# Getting next key in dictionary Using list() + index()

# initializing dictionary 
test_dict = {'one':'value 1','two':'value 2','three':'value 3'}

def get_next_key(dic, current_key):
    """ get the next key of a dictionary.

    Parameters
    ----------
    dic: dict
    current_key: string

    Return
    ------
    next_key: string, represent the next key in dictionary.
    or
    False If the value passed in current_key can not be found in the dictionary keys,
    or it is last key in the dictionary
    """

    l=list(dic) # convert the dict keys to a list

    try:
        next_key=l[l.index(current_key) + 1] # using index method to get next key
    except (ValueError, IndexError):
        return False
    return next_key

get_next_key(test_dict, 'two')

'three'

get_next_key(test_dict, 'three')

错

get_next_key(test_dict, 'one')

'two'

get_next_key(test_dict, 'NOT EXISTS')

错

Return 给定字典键的下一个键，python 3.6+

Return next key of a given dictionary key, python 3.6+

python

dictionary

iterator

python-3.8