Python：defaultdict的defaultdict？

布满荆棘的人生 2022-11-27 07:24 184阅读 0赞

本文翻译自：[Python: defaultdict of defaultdict?][Python_ defaultdict of defaultdict]

*Is there a way to have a `defaultdict(defaultdict(int))` in order to make the following code work?* **有没有一种方法使`defaultdict(defaultdict(int))`可以使以下代码正常工作？**

for x in stuff:
        d[x.a][x.b] += x.c_int

*`d` needs to be built ad-hoc, depending on `xa` and `xb` elements.* **`d`需要临时构建，具体取决于`xa`和`xb`元素。**

*I could use:* **我可以使用：**

for x in stuff:
        d[x.a,x.b] += x.c_int

*but then I wouldn't be able to use:* **但后来我将无法使用：**

d.keys()
    d[x.a].keys()

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### \#1楼 ###

参考：[https://stackoom.com/question/l6vY/Python-defaultdict的defaultdict][https_stackoom.com_question_l6vY_Python-defaultdict_defaultdict]

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### \#2楼 ###

*Yes like this:* **是这样的：**

defaultdict(lambda: defaultdict(int))

*The argument of a `defaultdict` (in this case is `lambda: defaultdict(int)` ) will be called when you try to access a key that doesn't exist.* **当您尝试访问不存在的键时，将调用`defaultdict`的参数（在本例中为`lambda: defaultdict(int)` ）。** *The return value of it will be set as the new value of this key, which means in our case the value of `d[Key_doesnt_exist]` will be `defaultdict(int)` .* **它的返回值将设置为该键的新值，这意味着在我们的示例中`d[Key_doesnt_exist]`的值将为`defaultdict(int)` 。**

*If you try to access a key from this last defaultdict ie `d[Key_doesnt_exist][Key_doesnt_exist]` it will return 0, which is the return value of the argument of the last defaultdict ie `int()` .* **如果您尝试从最后一个defaultdict即`d[Key_doesnt_exist][Key_doesnt_exist]`访问密钥，它将返回0，这是最后一个defaultdict即`int()`的参数的返回值。**

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### \#3楼 ###

*The parameter to the defaultdict constructor is the function which will be called for building new elements.* **defaultdict构造函数的参数是用于构建新元素的函数。** *So let's use a lambda !* **因此，让我们使用lambda！**

>>> from collections import defaultdict
    >>> d = defaultdict(lambda : defaultdict(int))
    >>> print d[0]
    defaultdict(<type 'int'>, {})
    >>> print d[0]["x"]
    0

*Since Python 2.7, there's an [even better solution using Counter][] :* **从Python 2.7开始， [使用Counter][even better solution using Counter]有了一个[更好的解决方案][even better solution using Counter] ：**

>>> from collections import Counter
    >>> c = Counter()
    >>> c["goodbye"]+=1
    >>> c["and thank you"]=42
    >>> c["for the fish"]-=5
    >>> c
    Counter({'and thank you': 42, 'goodbye': 1, 'for the fish': -5})

*Some bonus features* **一些额外功能**

>>> c.most_common()[:2]
    [('and thank you', 42), ('goodbye', 1)]

*For more information see [PyMOTW - Collections - Container data types][] and [Python Documentation - collections][]* **有关更多信息，请参见[PyMOTW-集合-容器数据类型][PyMOTW - Collections - Container data types]和[Python文档-集合][Python Documentation - collections]**

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### \#4楼 ###

*I find it slightly more elegant to use `partial` :* **我发现使用`partial`稍微优雅`partial` ：**

import functools
    dd_int = functools.partial(defaultdict, int)
    defaultdict(dd_int)

*Of course, this is the same as a lambda.* **当然，这与lambda相同。**

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### \#5楼 ###

*Others have answered correctly your question of how to get the following to work:* **其他人已经正确回答了您有关如何使以下内容正常工作的问题：**

for x in stuff:
        d[x.a][x.b] += x.c_int

*An alternative would be to use tuples for keys:* **一种替代方法是使用元组作为键：**

d = defaultdict(int)
    for x in stuff:
        d[x.a,x.b] += x.c_int
        # ^^^^^^^ tuple key

*The nice thing about this approach is that it is simple and can be easily expanded.* **这种方法的好处是它很简单并且可以轻松扩展。** *If you need a mapping three levels deep, just use a three item tuple for the key.* **如果您需要三个层次的映射，只需使用一个三项元组作为键。**

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### \#6楼 ###

*For reference, it's possible to implement a generic nested `defaultdict` factory method through:* **作为参考，可以通过以下方式实现通用的嵌套`defaultdict`工厂方法：**

from collections import defaultdict
    from functools import partial
    from itertools import repeat
    
    
    def nested_defaultdict(default_factory, depth=1):
        result = partial(defaultdict, default_factory)
        for _ in repeat(None, depth - 1):
            result = partial(defaultdict, result)
        return result()

*The depth defines the number of nested dictionary before the type defined in `default_factory` is used.* **在使用`default_factory`定义的类型之前，深度定义了嵌套字典的数量。** *For example:* **例如：**

my_dict = nested_defaultdict(list, 3)
    my_dict['a']['b']['c'].append('e')

[Python_ defaultdict of defaultdict]: https://oldbug.net/q/l6vY/Python-defaultdict-of-defaultdict
[https_stackoom.com_question_l6vY_Python-defaultdict_defaultdict]: https://stackoom.com/question/l6vY/Python-defaultdict%E7%9A%84defaultdict
[even better solution using Counter]: http://docs.python.org/2.7/library/collections#collections.Counter
[PyMOTW - Collections - Container data types]: http://www.doughellmann.com/PyMOTW/collections/counter.html
[Python Documentation - collections]: http://docs.python.org/library/collections.html