問題描述

給定 n 個字典，編寫一個函數，該函數將返回一個唯一字典，其中包含重復鍵的值列表.

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}

結果:

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

到目前為止我的代碼:

>>> 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}]

生成一個為這些重復鍵生成一個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?

推薦答案

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

這會返回:

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

這個問題略有不同.這里所有的字典值都是列表.如果長度為 1 的列表不希望這樣做，則添加:

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 語句之前.但是，我真的無法想象您何時想要刪除該列表.(考慮將列表作為值的情況；然后刪除項目周圍的列表會導致模棱兩可的情況.)

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.)

這篇關于合并字典，保留重復鍵的值的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！