Python: Generating a quot;set of tuplesquot; from a quot;list of tuplesquot; that doesn#39;t take order into consideration(Python:生成“元组集;来自“元组列表;不考虑顺序)

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问题描述

如果我有如下的元组列表:

If I have the list of tuples as the following:

[('a', 'b'), ('c', 'd'), ('a', 'b'), ('b', 'a')]
我想删除重复的元组(在内容和内部项目的顺序方面重复),以便输出为: I would like to remove duplicate tuples (duplicate in terms of both content and order of items inside) so that the output would be: [('a', 'b'), ('c', 'd')] 或者 [('b', 'a'), ('c', 'd')] 我尝试将其转换为设置然后列表,但输出将同时保持 ('b', 'a')('a', 'b') 在结果集中! I tried converting it to set then to list but the output would maintain both ('b', 'a') and ('a', 'b') in the resulting set! 推荐答案 试试这个: a = [('a', 'b'), ('c', 'd'), ('a', 'b'), ('b', 'a')] b = list(set([ tuple(sorted(t)) for t in a ])) [('a', 'b'), ('c', 'd')] 让我们分解一下: 如果你对一个元组进行排序,它就会变成一个排序列表. If you sort a tuple, it becomes a sorted list. >>> t = ('b', 'a') >>> sorted(t) ['a', 'b'] 对于 a 中的每个元组 t,对其进行排序并将其转换回元组. For each tuple t in a, sort it and convert it back to a tuple. >>> b = [ tuple(sorted(t)) for t in a ] >>> b [('a', 'b'), ('c', 'd'), ('a', 'b'), ('a', 'b')] 将结果列表 b 转换为集合:值现在是唯一的.将其转换回列表. Convert the resulting list b to a set : values are now unique. Convert it back to a list. >>> list(set(b)) [('a', 'b'), ('c', 'd')] 等等!

请注意,您可以使用 b="noreferrer">生成器 而不是 列表理解.

Note that you can skip the creation of the intermediate list b by using a generator instead of a list comprehension.

>>> list(set(tuple(sorted(t)) for t in a))
[('a', 'b'), ('c', 'd')]

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