基于前两个元素从嵌套列表中删除重复项 [英] Removing Duplicates from Nested List Based on First 2 Elements

问题描述

仅在前两个元素相同的情况下，我才尝试从嵌套列表中删除重复项，而忽略了第三个...

I'm trying to remove duplicates from a nested list only if the first 2 elements are the same, ignoring the third...

列表:

L = [['el1','el2','value1'], ['el3','el4','value2'], ['el1','el2','value2'], ['el1','el5','value3']]

会返回:

L = [['el3','el4','value2'], ['el1','el2','value2'], ['el1','el5','value3']]

我找到了一种简单的方法来进行类似的操作在这里:

I found a simple way to do similar here:

dict((x[0], x) for x in L).values()

但这仅适用于第一个元素，不适用于第一个2元素，但这正是我想要的.

but this only works for the first element and not the first 2, but that is exactly what i want otherwise.

推荐答案

如果顺序无关紧要，则可以使用相同的方法，但是将第一和第二个元素的元组用作键:

If the order doesn't matter, you can use that same method but using a tuple of the first and second elements as the key:

dict(((x[0], x[1]), x) for x in L).values()

或者在Python 2.7和更高版本上:

Or on Python 2.7 and higher:

{(x[0], x[1]): x for x in L}.values()

您可以使用tuple(x[:2])代替(x[0], x[1])，只要您认为可读性更高即可.

Instead of (x[0], x[1]) you can use tuple(x[:2]), use whichever you find more readable.

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