Remove numpy rows contained in a list?

Question

I have a numpy array and a list. I want to remove the rows contained in the list.

a = np.zeros((3, 2))
a[0, :] = [1, 2]
l = [(1, 2), (3, 4)]

Cu

Answer 1

Approach #1 : Here's one with views (viewing each row as an element each with extended dtype) -

# https://stackoverflow.com/a/45313353/ @Divakar
def view1D(a, b): # a, b are arrays
    a = np.ascontiguousarray(a)
    b = np.ascontiguousarray(b)
    void_dt = np.dtype((np.void, a.dtype.itemsize * a.shape[1]))
    return a.view(void_dt).ravel(),  b.view(void_dt).ravel()

a1D,l1D = view1D(a,l)
out = a[np.in1d(a1D,l1D,invert=True)]

If you need to have unique rows only in the output as with set, use np.unique on the output obtained -

np.unique(out,axis=0)

Sample run outputs -

In [72]: a
Out[72]: 
array([[1, 2],
       [0, 0],
       [0, 0]])

In [73]: l
Out[73]: [(1, 2), (3, 4)]

In [74]: out
Out[74]: 
array([[0, 0],
       [0, 0]])
In [75]: np.unique(out,axis=0)
Out[75]: array([[0, 0]])

Approach #2 : With the same philosophy of reducing dimensionality, here's with matrix-multiplication specific to int dtype data -

l = np.asarray(l)
shp = np.maximum(a.max(0)+1,l.max(0)+1)
s = np.r_[shp[::-1].cumprod()[::-1][1:],1]
l1D = l.dot(s)
a1D = a.dot(s)
l1Ds = np.sort(l1D)
out = a[l1D[np.searchsorted(l1Ds,a1D)] != a1D]