Question

这可能是一个愚蠢的问题，但由于某种原因，解决方案暂时逃脱了我。

我希望能够快速有效地访问列表格式的数据。例如，问题列表：

q = {}
q[1] = "my first string"
q[2] = "my second string"
q[3] = "my third string"

我可以通过q [2]轻松找到问题2的字符串。但我还想通过用字符串索引q来检索问题编号：

q["my second string"] -> gives 2 as answer

我想这样做而不迭代键（违背字典的目的）并且喜欢避免使用字符串作为键来定义第二个字典以避免浪费内存。这可能吗？

最终的原因是我想访问说q [2]或q [“我的第二个字符串”]并获取与问题2相关的数据，无论是使用数字还是字符串作为该数据的关键字。这是否可以在不必迭代所有键的同时避免数据重复？

Answer 1

将int和str混合为

是没有问题的

>>> q = {}
>>> q[1] = "my first string"
>>> q[2] = "my second string"
>>> q[3] = "my third string"
>>> q.update({v:k for k,v in q.items()})
>>> q["my second string"]
2

Answer 2

您可以使用OrderedDict，但对于其中一个指示，它不会像普通字典查找那样高效。

from collections import OrderedDict
q = OrderedDict()
q["my first string"] = 1
q["my second string"] = 2
q["my third string"] = 3
# Now you have normal key lookups on your string as a normal dict, and to get the order
q.values()[1]  # To get the second value out
# To get the key, value pair of the second entry
q.items()[1]
# Would return `('my second string', 2)`

Answer 3

有人问Efficient bidirectional hash table in Python?

答案保持不变 - 使用http://pypi.python.org/pypi/bidict

中的bidict

Answer 4

class MyDict(dict):
    def __init__(self, **kwargs):
        super(MyDict, self).__init__(**kwargs)
        for k, v in kwargs.iteritems():
            self[v] = k
    def __setitem__(self, key, val):
        super(MyDict, self).__setitem__(key, val)
        super(MyDict, self).__setitem__(val, key)

d = MyDict(a=1, b=2)
print d[1] # "a"
print d[2] # "b"
d['c'] = 3
print d[3] # "c"

在python中使用字符串或整数作为键的字典或映射？

4 个答案: