将嵌套列表的字典转换为元组列表 [英] Convert dict of nested lists to list of tuples

问题描述

我有dict个嵌套的list s:

d = {'a': [[('a1', 1, 1), ('a2', 1, 2)]], 'b': [[('b1', 2, 1), ('b2', 2, 2)]]}
print (d)
{'b': [[('b1', 2, 1), ('b2', 2, 2)]], 'a': [[('a1', 1, 1), ('a2', 1, 2)]]}

我需要创建tuple的list，例如:

[('b', 'b1', 2, 1), ('b', 'b2', 2, 2), ('a', 'a1', 1, 1), ('a', 'a2', 1, 2)]

我尝试过:

a = [[(k, *y) for y in v[0]] for k,v in d.items()]
a = [item for sublist in a for item in sublist]

我认为我的解决方案有些复杂.是否有一些更好，更Python化的解决方案，也许是一种解决方案?

I think my solution is a bit over-complicated. Is there some better, more pythonic, maybe one line solution?

推荐答案

您快到了:

[(k, *t) for k, v in d.items() for t in v[0]]

v[0]是必需的，因为您的值只是单元素列表，其中包含另一个列表.如果您想弄清楚上面的内容，可以将上面的内容扩展为以下嵌套的for循环:

The v[0] is needed because your values are just single-element lists with another list contained. The above can be expanded to the following nested for loops, if you wanted to figure out what it does:

for key, value in d.items():   # value is [[(...), (...), ...]]
    for tup in value[0]:  # each (...) from value[0]
        (key, *tup)  # produce a new tuple

演示:

>>> d = {'a': [[('a1', 1, 1), ('a2', 1, 2)]], 'b': [[('b1', 2, 1), ('b2', 2, 2)]]}
>>> [(k, *t) for k, v in d.items() for t in v[0]]
[('a', 'a1', 1, 1), ('a', 'a2', 1, 2), ('b', 'b1', 2, 1), ('b', 'b2', 2, 2)]

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