删除NumPy数组中的连续重复项 [英] Remove consecutive duplicates in a NumPy array

问题描述

我想删除彼此重复的重复项，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.

因此，如果输入为[0 0 1 3 2 2 3 3]，则输出应为[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]

我找到了使用itertools.groupby()的方法，但是我正在寻找一种更快的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])

通常的想法是使用diff查找数组中两个连续元素之间的差异.然后，我们仅索引那些给出non-zero差异元素的元素.但是，由于diff的长度缩短了1.因此，在建立索引之前，我们需要insert 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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