如何在Python中迭代Nlevel嵌套字典？

Question

我发现自己经常制作多层次词典。我总是需要编写非常详细的代码，以使用大量临时变量遍历所有级别的字典

是否有一种方法可以将此函数概括为迭代多个级别，而不是硬编码并手动指定有多少个级别？

def iterate_multilevel_dictionary(d, number_of_levels):
    # How to auto-detect number of levels? 
    # number_of_levels = 0
    if number_of_levels == 1:
        for k1, v1 in d.items():
            yield k1, v1
    if number_of_levels == 2:
        for k1, v1 in d.items():
            for k2, v2 in v1.items():
                yield k1, k2, v2
    if number_of_levels == 3:
        for k1, v1 in d.items():
            for k2, v2 in v1.items():
                for k3, v3 in v2.items():
                    yield k1, k2, k3, v3
                    
# Level 1
d_level1 = {"a":1,"b":2,"c":3}
for items in iterate_multilevel_dictionary(d_level1, number_of_levels=1):
    print(items)
# ('a', 1)
# ('b', 2)
# ('c', 3)

# Level 2
d_level2 = {"group_1":{"a":1}, "group_2":{"b":2,"c":3}}
for items in iterate_multilevel_dictionary(d_level2, number_of_levels=2):
    print(items)
#('group_1', 'a', 1)
#('group_2', 'b', 2)
#('group_2', 'c', 3)

# Level 3
d_level3 = {"collection_1":d_level2}
for items in iterate_multilevel_dictionary(d_level3, number_of_levels=3):
    print(items)
# ('collection_1', 'group_1', 'a', 1)
# ('collection_1', 'group_2', 'b', 2)
# ('collection_1', 'group_2', 'c', 3)

Answer 1

这里有一个快速而肮脏的解决方案：

d_level1 = {"a":1,"b":2,"c":3}
d_level2 = {"group_1":{"a":1}, "group_2":{"b":2,"c":3}}
d_level3 = {"collection_1":d_level2}

def flatten(d_in, base=()):
    for k in d_in:
        if type(d_in[k]) == dict:
            flatten(d_in[k], base+(k,))
        else:
            print(base + (k, d_in[k]))

flatten(d_level1)
# ('a', 1)
# ('b', 2)
# ('c', 3)

flatten(d_level2)
#('group_1', 'a', 1)
#('group_2', 'b', 2)
#('group_2', 'c', 3)

flatten(d_level3)
# ('collection_1', 'group_1', 'a', 1)
# ('collection_1', 'group_2', 'b', 2)
# ('collection_1', 'group_2', 'c', 3)

注意！！Python的递归限制约为1000！因此，在python中使用递归时，要非常仔细地考虑您要做什么，并准备好在调用这样的递归函数时捕获运行时错误

编辑： 通过评论，我意识到我犯了一个错误，没有将键添加到level1 dict输出，并且使用了可变结构作为默认参数。我在打印报表中添加了这些和参数，并重新发布。现在，输出与OP所需的输出相匹配，并使用更好、更现代的python

Answer 2

我是在看到@VoNWooDSoN的答案后写的。我把它变成了一个迭代器，而不是在函数内部打印，并做了一些修改，使它更具可读性。所以在这里看他的original answer

def flatten(d, base=()):
    for k, v in d.items():
        if isinstance(v, dict):
            yield from flatten(v, base + (k,))
        else:
            yield base + (k, v)

1-生产而不是印刷

2-isinstance()而不是type，因此dict的子类也可以工作。您还可以使用来自typing模块的MutableMapping而不是dict使其更通用

3-IMO，从.items()获取(k, v)对比k和d[k]可读性强得多

更通用

你想把它扩展到更一般的可以接受depths的数字以防万一吗

考虑这些例子：

d_level1 = {"a": 1, "b": 2, "c": 3}
d_level2 = {"group_1": {"a": 1}, "group_2": {"b": 2, "c": 3}}
d_level3 = {"collection_1": d_level2}

for items in flatten(d_level3):
    print(items)
print('------------------------------')
for items in flatten(d_level3, depth=0):
    print(items)
print('------------------------------')
for items in flatten(d_level3, depth=1):
    print(items)
print('------------------------------')
for items in flatten(d_level3, depth=2):
    print(items)

输出：

('collection_1', 'group_1', 'a', 1)
('collection_1', 'group_2', 'b', 2)
('collection_1', 'group_2', 'c', 3)
------------------------------
('collection_1', {'group_1': {'a': 1}, 'group_2': {'b': 2, 'c': 3}})
------------------------------
('collection_1', 'group_1', {'a': 1})
('collection_1', 'group_2', {'b': 2, 'c': 3})
------------------------------
('collection_1', 'group_1', 'a', 1)
('collection_1', 'group_2', 'b', 2)
('collection_1', 'group_2', 'c', 3)

^ }不考虑深度（仍然像你希望的那样在第一位置工作）。但是现在通过指定从0到2的深度，您可以看到我们可以迭代我们想要的深度。代码如下：

def flatten(d, base=(), depth=None):
    for k, v in d.items():
        if not isinstance(v, dict):
            yield base + (k, v)
        else:
            if depth is None:
                yield from flatten(v, base + (k,))
            else:
                if depth == 0:
                    yield base + (k, v)
                else:
                    yield from flatten(v, base + (k,), depth - 1)

Answer 3

试试这个代码

它还支持多个级别的组合

from typing import List, Tuple


def iterate_multilevel_dictionary(d: dict):
    dicts_to_iterate: List[Tuple[dict, list]] = [(d, [])]
    '''
    the first item is the dict object and the second object is the prefix keys 
    '''
    while dicts_to_iterate:
        current_dict, suffix = dicts_to_iterate.pop()
        for k, v in current_dict.items():
            if isinstance(v, dict):
                dicts_to_iterate.append((v, suffix + [k]))
            else:
                yield suffix + [k] + [v]


if __name__ == '__main__':
    d_level1 = {"a": 1, "b": 2, "c": 3}
    print(f"test for {d_level1}")
    for items in iterate_multilevel_dictionary(d_level1):
        print(items)
    d_level2 = {"group_1": {"a": 1}, "group_2": {"b": 2, "c": 3}}
    print(f"test for {d_level2}")
    for items in iterate_multilevel_dictionary(d_level2):
        print(items)

    d_level3 = {"collection_1": d_level2}
    print(f"test for {d_level3}")
    for items in iterate_multilevel_dictionary(d_level3):
        print(items)

    d_level123 = {}
    [d_level123.update(i) for i in [d_level1, d_level2, d_level3]]
    print(f"test for {d_level123}")
    for items in iterate_multilevel_dictionary(d_level123):
        print(items)

产出是：

test for {'a': 1, 'b': 2, 'c': 3}
['a', 1]
['b', 2]
['c', 3]
test for {'group_1': {'a': 1}, 'group_2': {'b': 2, 'c': 3}}
['group_2', 'b', 2]
['group_2', 'c', 3]
['group_1', 'a', 1]
test for {'collection_1': {'group_1': {'a': 1}, 'group_2': {'b': 2, 'c': 3}}}
['collection_1', 'group_2', 'b', 2]
['collection_1', 'group_2', 'c', 3]
['collection_1', 'group_1', 'a', 1]
test for {'a': 1, 'b': 2, 'c': 3, 'group_1': {'a': 1}, 'group_2': {'b': 2, 'c': 3}, 'collection_1': {'group_1': {'a': 1}, 'group_2': {'b': 2, 'c': 3}}}
['a', 1]
['b', 2]
['c', 3]
['collection_1', 'group_2', 'b', 2]
['collection_1', 'group_2', 'c', 3]
['collection_1', 'group_1', 'a', 1]
['group_2', 'b', 2]
['group_2', 'c', 3]
['group_1', 'a', 1]

使用递归是另一种方法，但我认为不使用递归的写作更具挑战性和效率：）