将作为值存储在嵌套字典中的Python列表聚合到一个任意级别的列表中 [英] Aggregate Python lists stored as values in a nested dictionary into one list for arbitrary levels

问题描述

假设我有一个嵌套的字典，在某种程度上，终端值是列表.例如:

Suppose I have a nested dictionary where, at some level, the terminal values are lists. For example:

nested_dict = {1 : {'a' : [1,2,3], 'b' : [4,5]}, 
               2 : {'a' : [6],     'b' : [7,8,9]}}

我想将列表值聚合到[1,2,3,4,5,6,7,8,9]中.我有两个级别

I want to aggregate the list values into [1,2,3,4,5,6,7,8,9]. For two levels I have

values = []
for key1 in nested_dict.keys():
    for key2 in nested_dict[key1].keys():
        for value in nested_dict[key1][key2]:
            values.append(value)

如何使其更紧凑，并能够处理任意级别?也就是说，我知道所有列表都处于同一级别，但是我不知道它的深度，所以在我提供的代码中，我实际上需要不确定数量的for循环.

How can this be made more compact, and such that it handles arbitrary levels? That is, I know all the lists to be at the same level, but I don't know how deep, so in the code I provide I would effectively need an indeterminate number of for loops.

不是重复的

Not a duplicate of how to extract multi level dictionary keys/values in python because they only show up to two levels.

推荐答案

您可以使用递归:

nested_dict = {1 : {'a' :[1,2,3], 'b': [4,5]}, 2 : {'a' : [6], 'b' : [7,8,9]}}
def get_lists(d):
   r = [b if isinstance(b, list) else get_lists(b) for b in d.values()]
   return [i for b in r for i in b]

输出:

[1, 2, 3, 4, 5, 6, 7, 8, 9]

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