Delete elements from numpy array by value [duplicate]

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I have two numpy arrays that represent 2D coordinates. Each row represents (x, y) pairs:

a = np.array([[1, 1], [2, 1], [3, 1], [3, 2], [3, 3], [5, 5]])
b = np.array([[1, 1], [5, 5], [3, 2]])

I would like to remove elements from a which are in b efficiently. So the result would be:

array([[2, 1], [3, 1], [3, 3]])

I can do it by looping and comparing, I was hoping I could do it easier.

This is a duplicate, I apologize, I did make a search, but was not able to find the answer. — enedene, Sep 30 '13 at 21:01
@enedene No worries, marking a question as a duplicate isn't an accusation or chastisement, it's really more of a pointer to a place where there are already answers. You asked the question more clearly, imo, but the answers already exist there. — askewchan, Sep 30 '13 at 21:32

hpaulj · Answer 1 · 2013-10-01 01:40:51Z

Python sets does a nice job of giving differences. It doesn't, though, maintain order

np.array(list(set(tuple(x) for x in a.tolist()).difference(set(tuple(x) for x in b.tolist()))))

Or to use boolean indexing, use broadcasting to create an outer equals, and sum with any and all

A = np.all((a[None,:,:]==b[:,None,:]),axis=-1)
A = np.any(A,axis=0)
a[~A,:]

Or make a and b complex:

ac = np.dot(a,[1,1j])
bc = np.dot(b,[1,1j])
A = np.any(ac==bc[:,None],axis=0)
a[~A,:]

or to use setxor1d

xx = np.setxor1d(ac,bc)
# array([ 2.+1.j,  3.+1.j,  3.+3.j])
np.array([xx.real,xx.imag],dtype=int).T

marked as duplicate by askewchan, Ophion, enedene, Jaime, Soner Gönül Oct 1 '13 at 14:27