删除 NumPy 数组中的连续重复项

问题描述

我想删除彼此跟随的重复项，但不删除整个数组中的重复项.另外，我想保持顺序不变.

I would like to remove duplicates which follow each other, but not duplicates along the whole array. Also, I want to keep the ordering unchanged.

所以如果输入是 [0 0 1 3 2 2 3 3] 输出应该是 [0 1 3 2 3]

So if the input is [0 0 1 3 2 2 3 3] the output should be [0 1 3 2 3]

我找到了一种使用 itertools.groupby() 的方法，但我正在寻找更快的 NumPy 解决方案.

I found a way using itertools.groupby() but I am looking for a faster NumPy solution.

推荐答案

a[np.insert(np.diff(a).astype(np.bool), 0, True)]
Out[99]: array([0, 1, 3, 2, 3])

一般的思路是使用diff来查找数组中两个连续元素之间的差异.然后我们只索引那些给出 non-zero 差异元素的元素.但是由于 diff 的长度短了 1.所以在索引之前，我们需要 insert 将 True 到 diff 数组的开头.

The general idea is to use diff to find the difference between two consecutive elements in the array. Then we only index those which give non-zero differences elements. But since the length of diff is shorter by 1. So before indexing, we need to insert the True to the beginning of the diff array.

说明:

In [100]: a
Out[100]: array([0, 0, 1, 3, 2, 2, 3, 3])

In [101]: diff = np.diff(a).astype(np.bool)

In [102]: diff
Out[102]: array([False,  True,  True,  True, False,  True, False], dtype=bool)

In [103]: idx = np.insert(diff, 0, True)

In [104]: idx
Out[104]: array([ True, False,  True,  True,  True, False,  True, False], dtype=bool)

In [105]: a[idx]
Out[105]: array([0, 1, 3, 2, 3])

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