如果存在于另一个数组中，则从一个数组中删除元素，保留重复项 - NumPy / Python

Question

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])

ie 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.

Answer 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]

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]

---

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.

Answer 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)))

Answer 3

Adding to Divakar's answer above -

if the original array A has a wider range than B, that will give you an 'index out of bounds' error. See:

A = np.array([1,1,2,3,3,3,4,5,6,7,8,8,10,12,14])
B = np.array([1,2,8])

A[B[np.searchsorted(B,A)] !=  A]
>> IndexError: index 3 is out of bounds for axis 0 with size 3

This will happen because np.searchsorted will assign index 3 (one-past-the-last in B) as the appropriate position for inserting in B the elements 10, 12 and 14 from A, in this example. Thus you get an IndexError in B[np.searchsorted(B,A)] .

To circumvent that, a possible approach is:

def subset_sorted_array(A,B):
    Aa = A[np.where(A <= np.max(B))]
    Bb = (B[np.searchsorted(B,Aa)] !=  Aa)
    Bb = np.pad(Bb,(0,A.shape[0]-Aa.shape[0]), method='constant', constant_values=True)
    return A[Bb]

Which works as follows:

# Take only the elements in A that would be inserted in B
Aa = A[np.where(A <= np.max(B))]

# Pad the resulting filter with 'Trues' - I split this in two operations for
# easier reading
Bb = (B[np.searchsorted(B,Aa)] !=  Aa)
Bb = np.pad(Bb,(0,A.shape[0]-Aa.shape[0]),  method='constant', constant_values=True)

# Then you can filter A by Bb
A[Bb]
# For the input arrays above:
>> array([ 3,  3,  3,  4,  5,  6,  7, 10, 12, 14])

Notice this will also work between arrays of strings and other types (for all types for which the comparison <= operator is defined).