Broadcasting multi-dimensional array indices of the same shape

Question

I have a mask array which represents a 2-dimensional binary image. Let's say it's simply:

mask = np.zeros((9, 9), dtype=np.uint8)
# 0 0 0 | 0 0 0 | 0 0 0
# 0 0 0 | 0 0 0 | 0 0 0
# 0 0 0 | 0 0 0 | 0 0 0
# ------+-------+------
# 0 0 0 | 0 0 0 | 0 0 0
# 0 0 0 | 0 0 0 | 0 0 0
# 0 0 0 | 0 0 0 | 0 0 0
# ------+-------+------
# 0 0 0 | 0 0 0 | 0 0 0
# 0 0 0 | 0 0 0 | 0 0 0
# 0 0 0 | 0 0 0 | 0 0 0

Suppose I want to flip the elements in the middle left ninth:

# 0 0 0 | 0 0 0 | 0 0 0
# 0 0 0 | 0 0 0 | 0 0 0
# 0 0 0 | 0 0 0 | 0 0 0
# ------+-------+------
# 1 1 1 | 0 0 0 | 0 0 0
# 1 1 1 | 0 0 0 | 0 0 0
# 1 1 1 | 0 0 0 | 0 0 0
# ------+-------+------
# 0 0 0 | 0 0 0 | 0 0 0
# 0 0 0 | 0 0 0 | 0 0 0
# 0 0 0 | 0 0 0 | 0 0 0

My incorrect approach was something like this:

x = np.arange(mask.shape[0])
y = np.arange(mask.shape[1])
mask[np.logical_and(y >= 3, y < 6), x < 3] = 1
# 0 0 0 | 0 0 0 | 0 0 0
# 0 0 0 | 0 0 0 | 0 0 0
# 0 0 0 | 0 0 0 | 0 0 0
# ------+-------+------
# 1 0 0 | 0 0 0 | 0 0 0
# 0 1 0 | 0 0 0 | 0 0 0
# 0 0 1 | 0 0 0 | 0 0 0
# ------+-------+------
# 0 0 0 | 0 0 0 | 0 0 0
# 0 0 0 | 0 0 0 | 0 0 0
# 0 0 0 | 0 0 0 | 0 0 0

(This is a simplification of the constraints I'm really dealing with, which would not be easily expressed as something like mask[:3,3:6] = 1 as in this case. Consider the constraints arbitrary, like x % 2 == 0 && y % 3 == 0 if you will.)

Numpy's behavior when the two index arrays are the same shape is to take them pairwise, which ends up only selecting the 3 elements above, rather than 9 I would like.

How would I update the right elements with constraints that apply to different axes? Given that the constraints are independent, can I do this by only evaluating my constraints N+M times, rather than N*M?

I can do something like mask[np.logical_and(y >= 3, y < 6)[:,np.newaxis] * (x < 3)] = 1 but this doesn't seem ideal. — rgov
– rgov, Commented Dec 4, 2019 at 3:39
as fancy as it looks, the above is not much different from mask[:3, 3:6]=1 — Quang Hoang
– Quang Hoang, Commented Dec 4, 2019 at 4:04

rgov · Accepted Answer · 2019-12-04 16:13:02Z

You can't broadcast the boolean arrays, but you can construct the equivalent numeric indices with ix_:

In [330]: np.ix_((y>=3)&(y<6), x<3)                                             
Out[330]: 
(array([[3],
        [4],
        [5]]), array([[0, 1, 2]]))

Applying it:

In [331]: arr = np.zeros((9,9),int)                                             
In [332]: arr[_330] = 1                                                         
In [333]: arr                                                                   
Out[333]: 
array([[0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0],
       [1, 1, 1, 0, 0, 0, 0, 0, 0],
       [1, 1, 1, 0, 0, 0, 0, 0, 0],
       [1, 1, 1, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0]])

Attempting to broadcast the booleans directly raises an error (too many indices):

arr[((y>=3)&(y<6))[:,None], x<3]

Quang Hoang · Accepted Answer · 2019-12-04 04:26:19Z

1

Per your comment, let's try this fancier example:

mask = np.zeros((90,90), dtype=np.uint8)

# criteria
def f(x,y): return ((x-20)**2 < 50) & ((y-20)**2 < 50)

# ranges
x,y = np.arange(90), np.arange(90)

# meshgrid
xx,yy = np.meshgrid(x,y)

zz = f(xx,yy)

# mask
mask[zz] = 1
plt.imshow(mask, cnap='gray')

Output:

edited Dec 4, 2019 at 4:26

answered Dec 4, 2019 at 3:59

Quang Hoang

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6 Comments

rgov Over a year ago

Thanks. In my "real" problem, the constraints are more complicated; the constraints could be something like (y - 4)**2 < 25 which would index a parabolic* section of the mask. So it's not so much the specific shape I'm trying to make here, but the problem of having constraints on different axes.

Quang Hoang Over a year ago

fyi, (y-4)**2<25 is not a circular shape. You probably want to update your example with what is closed to your real problem. I do think the solution is much similar to your given solution.

rgov Over a year ago

Thanks. I updated with another example and hopefully clarified. I guess the code becomes unreadable if I do it this way, though.

Quang Hoang Over a year ago

@rgov does the updated answer resemble your problem?

rgov Over a year ago

Let's say it was (x - 20)**2 < 50 and (y - 20)**2 < 50, rather than a single constraint (x-20)**2 + (y-20)**2 - 100 < 0. How would that change your answer?

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