问题
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])
Now I want the resulting array to be:
C = np.array([3,3,3,4,5,6,7])
i.e. if any value in B is found in A, remove it from A, if not keep it.
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.
回答1:
Using searchsorted
With sorted B, we can use searchsorted -
A[B[np.searchsorted(B,A)] != A]
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.
Generic case (B is not sorted) :
If B is not already sorted as is the pre-requisite, sort it and then use the proposed method.
Alternatively, we can use sorter argument with searchsorted -
sidx = B.argsort()
out = A[B[sidx[np.searchsorted(B,A,sorter=sidx)]] != A]
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)]
Same with isin -
A[~np.isin(A,B)]
With invert flag -
A[np.in1d(A,B,invert=True)]
A[np.isin(A,B,invert=True)]
This solves for a generic when B is not necessarily sorted.
回答2:
I am not very familiar with numpy, but how about using sets:
C = set(A.flat) - set(B.flat)
EDIT : from comments, sets cannot have duplicates values.
So another solution would be to use a lambda expression :
C = np.array(list(filter(lambda x: x not in B, A)))
来源：https://stackoverflow.com/questions/52417929/remove-elements-from-one-array-if-present-in-another-array-keep-duplicates-nu