如果存在于另一个数组中，则从一个数组中删除元素，并保持重复-NumPy/Python [英] Remove elements from one array if present in another array, keep duplicates - NumPy / Python

问题描述

我有两个数组A(380万的镜头)和B(20k的镜头). 对于最小的示例，让我们来考虑这种情况:

I have two arrays A (len of 3.8million) and B (len of 20k). For the minimal example, lets take this case:

A = np.array([1,1,2,3,3,3,4,5,6,7,8,8])
B = np.array([1,2,8])

现在我希望结果数组为:

Now I want the resulting array to be:

C = np.array([3,3,3,4,5,6,7])

即如果在A中找到B中的任何值，请将其从A中删除，如果不保留它.

i.e. if any value in B is found in A, remove it from A, if not keep it.

我想知道是否有没有for循环的任何方法，因为它是一个冗长的数组，因此循环需要很长时间.

I would like to know if there is any way to do it without a for loop because it is a lengthy array and so it takes long time to loop.

推荐答案

使用searchsorted

通过排序B，我们可以使用 -

Using searchsorted

With sorted B, we can use searchsorted -

A[B[np.searchsorted(B,A)] !=  A]

从链接的文档中，searchsorted(a,v)将索引查找到排序后的数组a中，这样，如果v中的相应元素插入到索引之前，则将保留a的顺序.因此，假设idx = searchsorted(B,A)并使用B[idx]索引到B中，我们将获得与A中每个元素相对应的B映射版本.因此，将此映射版本与A进行比较会告诉我们A中的每个元素，如果B中是否存在匹配项.最后，索引到A以选择不匹配的内容.

From the linked docs, searchsorted(a,v) find the indices into a sorted array a such that, if the corresponding elements in v were inserted before the indices, the order of a would be preserved. So, let's say idx = searchsorted(B,A) and we index into B with those : B[idx], we will get a mapped version of B corresponding to every element in A. Thus, comparing this mapped version against A would tell us for every element in A if there's a match in B or not. Finally, index into A to select the non-matching ones.

一般情况(B未排序):

Generic case (B is not sorted) :

如果B尚未按照先决条件进行排序，则对其进行排序，然后使用建议的方法.

If B is not already sorted as is the pre-requisite, sort it and then use the proposed method.

或者，我们可以将sorter参数与searchsorted-

Alternatively, we can use sorter argument with searchsorted -

sidx = B.argsort()
out = A[B[sidx[np.searchsorted(B,A,sorter=sidx)]] != A]

更常见的情况(A的值大于B中的值):

More generic case (A has values higher than ones in B) :

sidx = B.argsort()
idx = np.searchsorted(B,A,sorter=sidx)
idx[idx==len(B)] = 0
out = A[B[sidx[idx]] != A]

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使用in1d/isin

我们还可以使用 np.in1d ，这很简单(链接的文档应该帮助澄清)，因为它在A中的每个元素中查找B中的任何匹配项，然后我们可以使用带有倒置掩码的布尔索引来查找不匹配项一个-

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Using in1d/isin

We can also use np.in1d, which is pretty straight-forward (the linked docs should help clarify) as it looks for any match in B for every element in A and then we can use boolean-indexing with an inverted mask to look for non-matching ones -

A[~np.in1d(A,B)]

与isin相同-

A[~np.isin(A,B)]

带有invert标志-

A[np.in1d(A,B,invert=True)]

A[np.isin(A,B,invert=True)]

这解决了B不一定要排序时的泛型问题.

This solves for a generic when B is not necessarily sorted.

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