对列表中相同元素的索引进行分组的有效方法 [英] Efficient way to group indices of the same elements in a list

问题描述

比方说，我有一个看起来像这样的列表:

Let's say I have a list that looks like:

[1, 2, 2, 5, 8, 3, 3, 9, 0, 1]

现在，我想对相同元素的索引进行分组，因此结果应类似于:

Now I want to group the indices of the same elements, so the result should look like:

[[0, 9], [1, 2], [3], [4], [5, 6], [7], [8]]

如何有效地做到这一点?我尽量避免使用循环，因此任何使用numpy/pandas函数的实现都很棒.

How do I do this in an efficient way? I try to avoid using loops so any implementations using numpy/pandas functions are great.

推荐答案

使用熊猫GroupBy.apply，这非常简单-使用数据对一系列索引进行分组.一个不错的好处是，您可以保持索引的顺序.

Using pandas GroupBy.apply, this is pretty straightforward—use your data to group on a Series of indices. A nice bonus here is you get to keep the order of your indices.

data = [1, 2, 2, 5, 8, 3, 3, 9, 0, 1]
pd.Series(range(len(data))).groupby(data, sort=False).apply(list).tolist()
# [[0, 9], [1, 2], [3], [4], [5, 6], [7], [8]]

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