如果字典的键在frozensets中,则检索值

时间:2015-04-05 15:49:28

标签: python dictionary set key frozenset

我使用frozensets来保持我的字典的键,以利用联合,差异和交叉操作。但是当我试图通过字典中的键通过dict.get()来检索值时,它会产生一个None值。

newDict = {'a': 1, 'b': 2, 'c': 3, 'd': True}
stKeys = set(newDict)
stA = frozenset('a')
stB = frozenset('b')
stC = frozenset('c')
stD = frozenset('d')

print(stKeys)
print(newDict.get(stA & stKeys))
print(newDict.get(stB & stKeys))
print(newDict.get(stC & stKeys))
print(newDict.get(stD & stKeys))

农产品:

>>>None
>>>None
>>>None
>>>None

甚至:

print(newDict.get(stA))
print(newDict.get(stB))
print(newDict.get(stC))
print(newDict.get(stD))

农产品:

>>>None
>>>None
>>>None
>>>None

如果您的密钥位于frozensets中,如何从字典中检索值?

  

感谢 Martijn Pieters !答案是DVO(字典视图   对象)和生成器表达式,如果要添加结果   列表()

3 个答案:

答案 0 :(得分:2)

如果您想查找设置交叉点,可以使用dictionary view objects

for key in newDict.viewkeys() & stA:
    # all keys that are in the intersection of stA and the dictionary

在Python 3中,返回字典视图是默认的;你可以在这里使用newDict.keys()

for key in newDict.keys() & stA:
    # all keys that are in the intersection of stA and the dictionary

Python 3上的演示:

>>> newDict = {'a': 1, 'b': 2, 'c': 3, 'd': True}
>>> stA = frozenset('a')
>>> stB = frozenset('b')
>>> stC = frozenset('c')
>>> stD = frozenset('d')
>>> newDict.keys() & stA
{'a'}
>>> for key in newDict.keys() & stA:
...     print(newDict[key])
... 
1

答案 1 :(得分:1)

要创建冻结集密钥,您需要实际创建冻结集并将其用作密钥:

newDict = {
    frozenset('a'): 1,
    frozenset('b'): 2,
    frozenset('c'): 3,
    frozenset('d'): True
}

测试:

>>> {frozenset('a'):1}[frozenset('a')]
1

答案 2 :(得分:0)

你实际上可以做你想做的事情,至少在3.6.1中我也怀疑2.7.x:

newDict = {frozenset('a'): 1, frozenset('b'): 2, frozenset('c'): 3, 'd': True}
stKeys = set(newDict)
stA = frozenset('a')
print(stA)
stB = frozenset('b')
stC = frozenset('c')
stD = 'd'

print(newDict[stA])
print(newDict[stB])
print(newDict[stC])
print(newDict[stD])

输出:

frozenset({'a'})
1
2
3
True

问题似乎是密钥被指定为字符串对象而不是冻结集,但搜索被设置为查找冻结集。