Efficient way to merge two masked numpy arrays

Question

I have two numpy masked arrays that I want to combine. I am using the following code:

import numpy as np

a = np.zeros((10000, 10000), dtype=np.int16)
a[:5000, :5000] = 1
am = np.ma.masked_equal(a, 0)

b = np.zeros((10000, 10000), dtype=np.int16)
b[2500:7500, 2500:7500] = 2
bm = np.ma.masked_equal(b, 0)

arr = np.ma.array(np.dstack((am, bm)), mask=np.dstack((am.mask, bm.mask)))
arr = np.prod(arr, axis=2)
plt.imshow(arr)

The problem is that the operation is np.prod()very slow (4 seconds on my computer). Is there an alternative way to get the combined array in a more efficient way?

Answer 1

Instead of the last two lines using dstack()and prod()try the following:

arr = np.ma.array(am.filled(1) * bm.filled(1), mask=(am.mask * bm.mask))

Now you do not need prod()at all, and you do not have to completely allocate a 3D array.

Answer 2

, . 0 .

import numpy as np

a = np.zeros((1000, 1000), dtype=np.int16)
a[:500, :500] = 2
am = np.ma.masked_equal(a, 0)

b = np.zeros((1000, 1000), dtype=np.int16)
b[250:750, 250:750] = 3
bm = np.ma.masked_equal(b, 0)

c = np.zeros((1000, 1000), dtype=np.int16)
c[500:1000, 500:1000] = 5
cm = np.ma.masked_equal(c, 0)

bm.mask = np.logical_or(np.logical_and(am.mask, bm.mask), np.logical_not(am.mask))
am = np.ma.array(am.filled(0) + bm.filled(0), mask=(am.mask * bm.mask))

cm.mask = np.logical_or(np.logical_and(am.mask, cm.mask), np.logical_not(am.mask))
am = np.ma.array(am.filled(0) + cm.filled(0), mask=(am.mask * cm.mask))

plt.imshow(am)

, - -. , , . , - , .

. @morningsun 30% :

import numpy as np

a = np.zeros((1000, 1000), dtype=np.int16)
a[:500, :500] = 2
am = np.ma.masked_equal(a, 0)

b = np.zeros((1000, 1000), dtype=np.int16)
b[250:750, 250:750] = 3
bm = np.ma.masked_equal(b, 0)

c = np.zeros((1000, 1000), dtype=np.int16)
c[500:1000, 500:1000] = 5
cm = np.ma.masked_equal(c, 0)

am[am.mask] = bm[am.mask]
am[am.mask] = cm[am.mask]

plt.imshow(am)