从OrderedDict获取密钥计数，其中key是一个元组 [英] Get key count from OrderedDict where key is a tuple

问题描述

我有一个这样的字典：

my_dict=collections.OrderedDict([((123, 1), 'qwe'), ((232, 1), 'asd'), ((234, 2), 'zxc'), ((6745, 2), 'aaa'), ((456, 3), 'bbb')])

元组的组合始终是唯一的，我希望保持以下顺序插入，因此是OrderedDict。字典里我有超过1万个项目。如何有效维护一个计数器，该计数器给出元组中第二个元素的计数？基本上，每当我想在键中添加/删除项目时，我都需要知道计数。现在，我只是遍历 my_dict 并每次获得计数器 ，但是这样做似乎很昂贵。

The combination of the tuple is always unique and I would like to maintain the order of insertion, and hence OrderedDict. I've a well over ~10K items in the dict. How can I efficiently maintain a counter that gives the count of the second element in the tuple? Basically, I need to know the count whenever I would like to add/delete an item in the key. Right now I just iterate through my_dict and get the counter everytime but it seems to be very expensive to do that.

在上面的示例中，我希望输出为：

In the above example I want the output to be:

1:2 # As in 1 occurs 2 times 
2:2
3:1

现在我做了以下内容：

from collections import OrderedDict, Counter
my_dict = OrderedDict()
my_dict[(123,1)] = 'qwe'
my_dict[(232,1)] = 'asd'
my_dict[(234,2)] = 'zxc'
my_dict[(6745,2)] = 'aaa'
my_dict[(456,3)] = 'bbb'
cnt = []
for item in my_dict.keys():
    cnt.append(item[1])
print Counter(cnt)

我不确定这是最好的方法，但是有一种方法可以覆盖 = 运算符和 pop 函数，以便添加还是每次执行该操作时都减去一个计数？

I'm not sure if this is the best way but is there a way to override the the = operator and pop function, such that it adds or subtracts a count every time I do that operation?

推荐答案

获取 Counter 与 OrderedDict 配合使用可能需要一些子类。这可能会起作用（我只实现了 __ setitem __ 和 __ getitem __ ，但是如果您想要一个更强大的实现，让我知道）：

Getting a Counter to work nicely with an OrderedDict is probably going to require some subclassing. Here's something that might work (I've only implemented __setitem__ and __getitem__, but if you'd like a more robust implementation, let me know):

import collections

class CountedOrderedDict(collections.OrderedDict):
    def __init__(self, *args, **kwargs):
        self.counter = collections.Counter()
        super(CountedOrderedDict, self).__init__(*args, **kwargs)

    def __delitem__(self, key):
        super(CountedOrderedDict, self).__delitem__(key)
        self.counter[key[1]] -= 1

    def __setitem__(self, key, value):
        if key not in self:
            self.counter[key[1]] += 1

        super(CountedOrderedDict, self).__setitem__(key, value)

示例用法：

>>> my_dict = CountedOrderedDict({(123,1): 'sda', (232,1) : 'bfd', (234,2) : 'csd', (6745,2) : 'ds', (456,3) : 'rd'})
>>> my_dict.counter
Counter({'1': 2, '2': 2, '3': 1})
>>> del my_dict[(123,1)]
>>> my_dict.counter
Counter({'2': 2, '1': 1, '3': 1})
>>> my_dict[(150,1)] = "asdf"
>>> my_dict.counter
Counter({'1': 2, '2': 2, '3': 1})

这是一个更通用的 CountedOrderedDict 实现，它以键函数作为参数。

Here's a more general CountedOrderedDict implementation that takes a key function as a parameter.

import collections

class CountedOrderedDict(collections.OrderedDict):
    def __init__(self, key=lambda k: k, *args, **kwargs):
        self.counter = collections.Counter()
        self.key_transform = key
        super(CountedOrderedDict, self).__init__(*args, **kwargs)

    def __delitem__(self, key):
        super(CountedOrderedDict, self).__delitem__(key)
        self.counter[self.key_transform(key)] -= 1

    def __setitem__(self, key, value):
        if key not in self:
            self.counter[self.key_transform(key)] += 1

        super(CountedOrderedDict, self).__setitem__(key, value)

对于您的需求，您可以像这样实例化：

For your needs, you'd instantiate it like so:

my_dict = CountedOrderedDict(key=lambda k: k[1])

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