Kri*_*shn 12 python dictionary nested python-3.x
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)
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Dmi*_*rba 28
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.这是一个程序,它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)
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打印
1 2
3 4
5 6
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如果您只对最深层次的键值对感兴趣(当值不是dict时),则这是相关的.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)
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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
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小智 13
这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)
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因此,您只需逐个迭代每个内部字典并打印值.