How to make a circular kernel?
Question:
For the Code below, I am wondering how to make a circular kernel instead of a rectangular one. I am currently looking at something circular, and I want to find the BGR average values for it. By adjusting my kernel, my data will be more accurate.
for center in c_1:
b = img2[center[0]-4: center[0]+5, center[1]-4: center[1]+5, 0]
g = img2[center[0]-4: center[0]+5, center[1]-4: center[1]+5, 1]
r = img2[center[0]-4: center[0]+5, center[1]-4: center[1]+5, 2]
Answers:
Get the circle region when given the center, you could try the following function:
def circleAverage(center, r = 4):
"""
"""
for i in range(center[0]-r, center[0]+r):
for j in range(center[1]-r, center[1] + r):
if (center[0] - i) ** 2 + (center[1] - j) ** 2 <= r**2:
// do your computation here.
Hope this helps you.
We manually created a structuring elements in the previous examples with help of Numpy. It is rectangular shape. But in some cases, you may need elliptical/circular shaped kernels. So for this purpose, OpenCV has a function, cv2.getStructuringElement(). You just pass the shape and size of the kernel, you get the desired kernel.
# Elliptical Kernel
>>> cv2.getStructuringElement(cv2.MORPH_ELLIPSE,(5,5))
array([[0, 0, 1, 0, 0],
[1, 1, 1, 1, 1],
[1, 1, 1, 1, 1],
[1, 1, 1, 1, 1],
[0, 0, 1, 0, 0]], dtype=uint8)
Came here to find how to make a circular (symmetric) kernel. Ended up with my own implementation.
import numpy as np
def get_circular_kernel(diameter):
mid = (diameter - 1) / 2
distances = np.indices((diameter, diameter)) - np.array([mid, mid])[:, None, None]
kernel = ((np.linalg.norm(distances, axis=0) - mid) <= 0).astype(int)
return kernel
Note that for low diameters, behavior is perhaps unexpected. Variable mid
when used for the second time can for example be replaced by diameter / 2
.
I’ve implemented it in a following way:
r = 16
kernel = np.fromfunction(lambda x, y: ((x-r)**2 + (y-r)**2 <= r**2)*1, (2*r+1, 2*r+1), dtype=int).astype(np.uint8)
Extra type conversion is needed to avoid overflow
For the Code below, I am wondering how to make a circular kernel instead of a rectangular one. I am currently looking at something circular, and I want to find the BGR average values for it. By adjusting my kernel, my data will be more accurate.
for center in c_1:
b = img2[center[0]-4: center[0]+5, center[1]-4: center[1]+5, 0]
g = img2[center[0]-4: center[0]+5, center[1]-4: center[1]+5, 1]
r = img2[center[0]-4: center[0]+5, center[1]-4: center[1]+5, 2]
Get the circle region when given the center, you could try the following function:
def circleAverage(center, r = 4):
"""
"""
for i in range(center[0]-r, center[0]+r):
for j in range(center[1]-r, center[1] + r):
if (center[0] - i) ** 2 + (center[1] - j) ** 2 <= r**2:
// do your computation here.
Hope this helps you.
We manually created a structuring elements in the previous examples with help of Numpy. It is rectangular shape. But in some cases, you may need elliptical/circular shaped kernels. So for this purpose, OpenCV has a function, cv2.getStructuringElement(). You just pass the shape and size of the kernel, you get the desired kernel.
# Elliptical Kernel
>>> cv2.getStructuringElement(cv2.MORPH_ELLIPSE,(5,5))
array([[0, 0, 1, 0, 0],
[1, 1, 1, 1, 1],
[1, 1, 1, 1, 1],
[1, 1, 1, 1, 1],
[0, 0, 1, 0, 0]], dtype=uint8)
Came here to find how to make a circular (symmetric) kernel. Ended up with my own implementation.
import numpy as np
def get_circular_kernel(diameter):
mid = (diameter - 1) / 2
distances = np.indices((diameter, diameter)) - np.array([mid, mid])[:, None, None]
kernel = ((np.linalg.norm(distances, axis=0) - mid) <= 0).astype(int)
return kernel
Note that for low diameters, behavior is perhaps unexpected. Variable mid
when used for the second time can for example be replaced by diameter / 2
.
I’ve implemented it in a following way:
r = 16
kernel = np.fromfunction(lambda x, y: ((x-r)**2 + (y-r)**2 <= r**2)*1, (2*r+1, 2*r+1), dtype=int).astype(np.uint8)
Extra type conversion is needed to avoid overflow