問題描述
我想刪除彼此跟隨的重復項，但不刪除整個數組中的重復項.另外，我想保持順序不變.
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.
所以如果輸入是 [0 0 1 3 2 2 3 3] 輸出應該是 [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]
我找到了一種使用 itertools.groupby() 的方法，但我正在尋找更快的 NumPy 解決方案.
I found a way using itertools.groupby() but I am looking for a faster NumPy solution.
推薦答案
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來查找數組中兩個連續元素之間的差異.然后我們只索引那些給出 non-zero 差異元素的元素.但是由于 diff 的長度短了 1.所以在索引之前，我們需要 insert 將 True 到 diff 數組的開頭.
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.
說明:
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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