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.
解决方案
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])
相关文章