Question

我有下面的代码，目前只打印初始字典的值。但是，我想迭代嵌套字典的每个键，最初只打印名称。请参阅下面的代码：

Liverpool = {
    'Keepers':{'Loris Karius':1,'Simon Mignolet':2,'Alex Manninger':3},
    'Defenders':{'Nathaniel Clyne':3,'Dejan Lovren':4,'Joel Matip':5,'Alberto Moreno':6,'Ragnar Klavan':7,'Joe Gomez':8,'Mamadou Sakho':9}
}

for k,v in Liverpool.items():
    if k =='Defenders':
       print(v)

Answer 1

在其他答案中，您被指出如何解决给定dicts的任务，最大深度等级为2。这个程序将允许您循环遍历具有无限数量嵌套级别的dict的键值对（更通用的方法）：

def recursive_items(dictionary):
    for key, value in dictionary.items():
        if type(value) is dict:
            yield from recursive_items(value)
        else:
            yield (key, value)

a = {'a': {1: {1: 2, 3: 4}, 2: {5: 6}}}

for key, value in recursive_items(a):
    print(key, value)

打印

1 2
3 4
5 6

如果您只对最深层次的键值对感兴趣（当值不是dict时），那么这是相关的。如果您对值为dict的键值对感兴趣，请进行小编辑：

def recursive_items(dictionary):
    for key, value in dictionary.items():
        if type(value) is dict:
            yield (key, value)
            yield from recursive_items(value)
        else:
            yield (key, value)

a = {'a': {1: {1: 2, 3: 4}, 2: {5: 6}}}

for key, value in recursive_items(a):
    print(key, value)

打印

a {1: {1: 2, 3: 4}, 2: {5: 6}}
1 {1: 2, 3: 4}
1 2
3 4
2 {5: 6}
5 6

Answer 2

以下是打印所有团队成员的代码：

for k, v in Liverpool.items():
    for k1, v1 in v.items():
        print(k1)

因此，您只需逐个迭代每个内部字典并打印值。

Answer 3

Liverpool = {
    'Keepers':{'Loris Karius':1,'Simon Mignolet':2,'Alex Manninger':3},
    'Defenders':{'Nathaniel Clyne':3,'Dejan Lovren':4,'Joel Matip':5,'Alberto Moreno':6,'Ragnar Klavan':7,'Joe Gomez':8,'Mamadou Sakho':9}
}

for k,v in Liverpool.items():
    print(v.keys())

收率：

['Alberto Moreno', 'Joe Gomez', 'Dejan Lovren', 'Ragnar Klavan', 'Joel Matip', 'Nathaniel Clyne', 'Mamadou Sakho']
['Alex Manninger', 'Loris Karius', 'Simon Mignolet']

获取嵌套字典的所有键

3 个答案: