将混合的嵌套列表转换为嵌套元组 [英] Convert a mixed nested list to a nested tuple

问题描述

如果我有

easy_nested_list = [['foo', 'bar'], ['foofoo', 'barbar']]

并希望拥有

(('foo', 'bar'), ('foofoo', 'barbar'))

我能做

tuple(tuple(i) for i in easy_nested_list)

但是如果我有

mixed_nested_list = [['foo', 'bar'], ['foofoo', ['foo', 'bar']],'some', 2, 3]

，我想以此为基础建立元组，我不知道该如何开始.

and would like to build a tuple out of this, I don't know how to start.

很高兴得到:

(('foo', 'bar'), ('foofoo', ('foo', 'bar')), 'some', 2, 3)

第一个问题是Python将我的字符串转换为每个字符的元组.第二件事是我得到

The first problem is that Python turns my string into a tuple for each character. The second thing is I get

TypeError: 'int' object is not iterable

推荐答案

递归转换并测试列表:

def to_tuple(lst):
    return tuple(to_tuple(i) if isinstance(i, list) else i for i in lst)

这会为给定列表生成一个元组，但是会使用递归调用转换任何嵌套的list对象.

This produces a tuple for a given list, but converts any nested list objects using a recursive call.

演示:

>>> def to_tuple(lst):
...     return tuple(to_tuple(i) if isinstance(i, list) else i for i in lst)
... 
>>> mixed_nested_list = [['foo', 'bar'], ['foofoo', ['foo', 'bar']],'some', 2, 3]
>>> to_tuple(mixed_nested_list)
(('foo', 'bar'), ('foofoo', ('foo', 'bar')), 'some', 2, 3)

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