填充numpy数组中的相邻元素

Question

Not sure whats the best way to title this question, but basically I would like to fill an existing numpy array with a value, based on the location provided and a specified distance. Assuming going diagonally is not valid.

For example, lets say we have an array with just 0s.

[[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]]

If I want (2,2) as the location with distance 1, it would fill the matrix with value 1, in the location that has distance one from the location provided, including itself. Thus the matrix would look like:

[[0 0 0 0 0]
 [0 0 1 0 0]
 [0 1 1 1 0]
 [0 0 1 0 0]
 [0 0 0 0 0]]

And if I provided a distance of 2, it would look like:

[[0 0 1 0 0]
 [0 1 1 1 0]
 [1 1 1 1 1]
 [0 1 1 1 0]
 [0 0 1 0 0]]

Basically everything within a distance of 2 from the location will be filled with the value 1. Assuming diagonal movement is not valid.

I also would like to support wrapping, where if the neighboring elements are out of bounds, it will wrap around.

For example, if the location provided is (4,4) with distance 1, the matrix should look like so:

[[0 0 0 0 1]
 [0 0 0 0 0]
 [0 0 0 0 0]
 [0 0 0 0 1]
 [1 0 0 1 1]]

I tried using np.ogrid along with a mask of where the 1's will be true but cant seem to get it working.

Answer 1

What you are trying to do is essentially binary dilation , but the wrapping poses a problem. Luckily, scipy 's grey dilation function has the wrap mode which we can leverage:

from scipy.ndimage.morphology import grey_dilation, generate_binary_structure, iterate_structure

st = generate_binary_structure(2,1)

# st essentially defines "neighbours", 
# and you can expand n times this using iterate_structure(st, n):

# >>> st
# array([[False,  True, False],
#        [ True,  True,  True],
#        [False,  True, False]])

# >>> iterate_structure(st,2)
# array([[False, False,  True, False, False],
#        [False,  True,  True,  True, False],
#        [ True,  True,  True,  True,  True],
#        [False,  True,  True,  True, False],
#        [False, False,  True, False, False]])


a = np.zeros((5,5))
a[4,4] = 1
dist = 1

dilated = grey_dilation(a, footprint = iterate_structure(st,dist), mode='wrap')

And as a function that creates your array for you:

from scipy.ndimage.morphology import grey_dilation, generate_binary_structure, iterate_structure

def create(size, dist, loc):
    a = np.zeros((size,size), dtype=int)
    a[loc] = 1
    st = generate_binary_structure(2,1)
    return grey_dilation(a, footprint = iterate_structure(st,dist), mode='wrap')

Examples : To reproduce your desired inputs and outputs:

>>> create(5, 1, (2,2))
array([[0, 0, 0, 0, 0],
       [0, 0, 1, 0, 0],
       [0, 1, 1, 1, 0],
       [0, 0, 1, 0, 0],
       [0, 0, 0, 0, 0]])

>>> create(5, 2, (2,2))
array([[0, 0, 1, 0, 0],
       [0, 1, 1, 1, 0],
       [1, 1, 1, 1, 1],
       [0, 1, 1, 1, 0],
       [0, 0, 1, 0, 0]])

>>> create(5, 1, (4,4))
array([[0, 0, 0, 0, 1],
       [0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0],
       [0, 0, 0, 0, 1],
       [1, 0, 0, 1, 1]])

Answer 2

def create(size, dist, loc):
    a = np.zeros((size, size))
    for i in range(-dist, dist + 1):
        for j in range(-dist + abs(i), dist - abs(i) + 1):
            i_ = (i + loc[0]) % size
            j_ = (j + loc[1]) % size
            a[i_, j_] = 1
    return a

create(5, 1, (4, 4))

returns

array([[0., 0., 0., 0., 1.],
       [0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 1.],
       [1., 0., 0., 1., 1.]])

Answer 3

This may not be the most efficient solution but you could try iterating through all the elements in the array, check if their distance to the location provided is what you want it to be and if it is, replace that element's value with the value specified. Basic code structure:

# declar my_arr
value = 1
distance = 2
centre_point = (4,4)
for row_index in range(len(my_arr)):
    for col_index in range(len(my_arr[row_index])):
        if distanceToPoint(row_index,col_index,centre_point) <= distance:
            my_arr[row_index][col_index] = value

The distanceToPoint function would be something like this:

def distanceToPoint(x,y,point):
   px,py = point
   dx,dy = px-x,py-y
   if x==px:
       return py-y
   if y==py:
       return px-x
   if abs(dx)==abs(dy):
       return dx
   else:
       return 1000000 #an arbitrarily large amount which should be bigger than distance