问题描述

给定 n 个字典，编写一个函数，该函数将返回一个唯一字典，其中包含重复键的值列表.

Given n dictionaries, write a function that will return a unique dictionary with a list of values for duplicate keys.

例子:

d1 = {'a': 1, 'b': 2}
d2 = {'c': 3, 'b': 4}
d3 = {'a': 5, 'd': 6}

结果:

>>> newdict
{'c': 3, 'd': 6, 'a': [1, 5], 'b': [2, 4]}

到目前为止我的代码:

>>> def merge_dicts(*dicts):
...     x = []
...     for item in dicts:
...         x.append(item)
...     return x
...
>>> merge_dicts(d1, d2, d3)
[{'a': 1, 'b': 2}, {'c': 3, 'b': 4}, {'a': 5, 'd': 6}]

生成一个为这些重复键生成一个值列表的新字典的最佳方法是什么?

What would be the best way to produce a new dictionary that yields a list of values for those duplicate keys?

推荐答案

def merge_dicts(*dicts):
    d = {}
    for dict in dicts:
        for key in dict:
            try:
                d[key].append(dict[key])
            except KeyError:
                d[key] = [dict[key]]
    return d

这会返回:

{'a': [1, 5], 'b': [2, 4], 'c': [3], 'd': [6]}

这个问题略有不同.这里所有的字典值都是列表.如果长度为 1 的列表不希望这样做，则添加:

There is a slight difference to the question. Here all dictionary values are lists. If that is not to be desired for lists of length 1, then add:

    for key in d:
        if len(d[key]) == 1:
            d[key] = d[key][0]

在 return d 语句之前.但是，我真的无法想象您何时想要删除该列表.(考虑将列表作为值的情况；然后删除项目周围的列表会导致模棱两可的情况.)

before the return d statement. However, I cannot really imagine when you would want to remove the list. (Consider the situation where you have lists as values; then removing the list around the items leads to ambiguous situations.)

这篇关于合并字典，保留重复键的值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！