Python 把字典分成两半？

Split a dictionary in half?

提问by user1728853

What is the best way to split a dictionary in half?

将字典一分为二的最佳方法是什么？

d = {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5}

I'm looking to do this:

我正在寻找这样做：

d1 = {'key1': 1, 'key2': 2, 'key3': 3}
d2 = {'key4': 4, 'key5': 5}

It does not matter which keys/values go into each dictionary. I am simply looking for the simplest way to divide a dictionary into two.

哪个键/值进入每个字典并不重要。我只是在寻找将字典一分为二的最简单方法。

采纳答案by jone

This would work, although I didn't test edge-cases:

这会起作用，尽管我没有测试边缘情况：

>>> d = {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5}
>>> d1 = dict(d.items()[len(d)/2:])
>>> d2 = dict(d.items()[:len(d)/2])
>>> print d1
{'key1': 1, 'key5': 5, 'key4': 4}
>>> print d2
{'key3': 3, 'key2': 2}

In python3:

在python3中：

d = {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5}
d1 = dict(list(d.items())[len(d)//2:])
d2 = dict(list(d.items())[:len(d)//2])

Also note that order of items is not guaranteed

另请注意，不能保证项目的顺序

回答by Johan R?de

d1 = {key: value for i, (key, value) in enumerate(d.viewitems()) if i % 2 == 0}
d2 = {key: value for i, (key, value) in enumerate(d.viewitems()) if i % 2 == 1}

回答by Blckknght

Here's a way to do it using an iterator over the items in the dictionary and itertools.islice:

这是一种使用迭代器遍历字典中的项目的方法，并且itertools.islice：

import itertools

def splitDict(d):
    n = len(d) // 2          # length of smaller half
    i = iter(d.items())      # alternatively, i = d.iteritems() works in Python 2

    d1 = dict(itertools.islice(i, n))   # grab first n items
    d2 = dict(i)                        # grab the rest

    return d1, d2

回答by Gareth Latty

We can do this efficiently with itertools.zip_longest()(note this is itertools.izip_longest()in 2.x):

我们可以有效地做到这一点itertools.zip_longest()（注意这是itertools.izip_longest()在 2.x 中）：

from itertools import zip_longest
d = {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5}
items1, items2 = zip(*zip_longest(*[iter(d.items())]*2))
d1 = dict(item for item in items1 if item is not None)
d2 = dict(item for item in items2 if item is not None)

Which gives us:

这给了我们：

>>> d1
{'key3': 3, 'key1': 1, 'key4': 4}
>>> d2
{'key2': 2, 'key5': 5}

回答by David Bielen

If you use python +3.3, and want your splitted dictionaries to be the same across different python invocations, do not use .items, since the hash-values of the keys, which determines the order of .items()will change between python invocations. See Hash randomization

如果您使用python +3.3，并且希望您的拆分字典在不同的 Python 调用中保持相同，请不要使用.items，因为键的哈希值决定了.items()Python 调用之间的顺序变化。请参阅哈希随机化

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回答by Joe Cheng

Here is the function which can be used to split a dictionary to any divisions.

这是可用于将字典拆分为任何部分的函数。

import math

def linch_dict_divider(raw_dict, num):
    list_result = []
    len_raw_dict = len(raw_dict)
    if len_raw_dict > num:
        base_num = len_raw_dict / num
        addr_num = len_raw_dict % num
        for i in range(num):
            this_dict = dict()
            keys = list()
            if addr_num > 0:
                keys = raw_dict.keys()[:base_num + 1]
                addr_num -= 1
            else:
                keys = raw_dict.keys()[:base_num]
            for key in keys:
                this_dict[key] = raw_dict[key]
                del raw_dict[key]
            list_result.append(this_dict)

    else:
        for d in raw_dict:
            this_dict = dict()
            this_dict[d] = raw_dict[d]
            list_result.append(this_dict)

    return list_result

myDict = {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5}
print myDict
myList = linch_dict_divider(myDict, 2)
print myList

回答by user3548798

The answer by jonedid not work for me. I had to cast to a list before I could index the result of the .items() call. (I am running Python 3.6 in the example)

jone 的回答对我不起作用。在索引 .items() 调用的结果之前，我必须转换为列表。（我在示例中运行 Python 3.6）

d = {'one':1, 'two':2, 'three':3, 'four':4, 'five':5}
split_idx = 3
d1 = dict(list(d.items())[:split_idx])
d2 = dict(list(d.items())[split_idx:])

"""
output:
d1
{'one': 1, 'three': 3, 'two': 2}
d2
{'five': 5, 'four': 4}
"""

Note the dicts are not necessarily stored in the order of creation so the indexes may be mixed up.

请注意，dicts 不一定按创建顺序存储，因此索引可能会混淆。

回答by woodword

If you used numpy, then you could do this :

如果你使用 numpy，那么你可以这样做：

def divide_dict(dictionary, chunk_size):

'''
Divide one dictionary into several dictionaries

Return a list, each item is a dictionary
'''

import numpy, collections

count_ar = numpy.linspace(0, len(dictionary), chunk_size+1, dtype= int)
group_lst = []
temp_dict = collections.defaultdict(lambda : None)
i = 1
for key, value in dictionary.items():
    temp_dict[key] = value
    if i in count_ar:
        group_lst.append(temp_dict)
        temp_dict = collections.defaultdict(lambda : None)
    i += 1
return group_lst