## Get perimeter of pixels around centre pixel

how to calculate perimeter of an image in matlab
bwperim python
bwboundaries matlab
how to calculate area of an image in matlab
regionprops matlab
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I am attempting to get a circle of pixels around a centre pixel. Ie, just like how FAST keypoint detector works, I want to get the perimeter pixels around it given a radius. However the math escapes me, I know theoretically how I could obtain it using trigonometry. Ie, I could use a for loop and iterate at 15 degrees. I know the triangle hypotenuse length is the radius, I know the angle.

Any advice how I could obtain a perimeter of pixels around a given pixel? The formula is:

```(x-cx)**2 + (y-cy)**2 = r**2
```

where cx and cy is the center of the circle and x and y are the coordinates you want to test... Now we can iterate over x and get the y with the formula like this:

```y = sqrt(r**2 - (x-cx)**2) + cy
```

The other way will be to iterate the 360 degrees and calculate the x and y and add the offset (center) like this:

```x = cos(radians) * radius + cx
```

The second version gave me a more complete circle in my tests. Here is my test script in python:

```import numpy as np
import cv2
import math

img = np.zeros((480, 640, 1), dtype="uint8")
img2 = np.zeros((480, 640, 1), dtype="uint8")

center = (200, 200)

y_off = np.sqrt(radius**2 - (x - center) **2)
y1 = np.int32(np.round(center + y_off))
y2 = np.int32(np.round(center - y_off))
img[y1, x] = 255
img[y2, x] = 255

degrees = np.arange(360)
x = np.int32(np.round(np.cos(degrees) * radius)) + center
y = np.int32(np.round(np.sin(degrees) * radius)) + center
img2[y,x] = 255

cv2.imshow("First method", img)
cv2.imshow("Second method", img2)
cv2.waitKey(0)
cv2.destroyAllWindows()
```

and the results are these:

Method 1 Method 2 There is a third method... You take a box around the circle of size radius x radius and evaluate each point with the circle formula given above, if it is true then it is a circle point... however that is good to draw the whole circle, since you have integers and highly probable not many point will be equal...

UPDATE:

Just a small reminder, make sure your points are in the image, in the example above, if you put the center in 0,0 it will draw 1/4 of a circle in every corner, because it considers the negative values to start from the end of the array.

To remove duplicates you can try the following code:

```c = np.unique(np.array(list(zip(y,x))), axis=0  )
img2[c[:,0], c[:,1]] = 255
```

Get perimeter of pixels around centre pixel, I am attempting to get a circle of pixels around a centre pixel. Ie, just like how FAST keypoint detector works, I want to get the perimeter pixels  · Find the distance between the each adjoining pair of pixels around the border of the region. · Use the distance formula: · For instance, calculate the distance between the two points (1,1) and (2,2). distance=sqrt((2-1).^2+(2-1).^2)=1.41

Just draw the circle onto a mask:

```In : mask = np.zeros((9, 9), dtype=np.uint8)

Out:
array([[  0,   0,   0,   0, 255,   0,   0,   0,   0],
[  0,   0, 255, 255,   0, 255, 255,   0,   0],
[  0, 255,   0,   0,   0,   0,   0, 255,   0],
[  0, 255,   0,   0,   0,   0,   0, 255,   0],
[255,   0,   0,   0,   0,   0,   0,   0, 255],
[  0, 255,   0,   0,   0,   0,   0, 255,   0],
[  0, 255,   0,   0,   0,   0,   0, 255,   0],
[  0,   0, 255, 255,   0, 255, 255,   0,   0],
[  0,   0,   0,   0, 255,   0,   0,   0,   0]], dtype=uint8)
```

And now you can use it to index your image as you like. E.g., here's a random image:

```In : img
Out:
array([[ 88, 239, 212, 160,  89,  85, 249, 242,  88],
[ 47, 230, 206, 206,  63, 143, 152,  67,  58],
[162, 212,   0, 213, 208, 169, 228,  14, 229],
[230,  45, 103, 201, 188, 231,  80, 122, 131],
[159,  31, 148, 158,  73, 215, 152, 158, 235],
[213, 177, 148, 237,  92, 115, 152, 188, 223],
[234,  67, 141, 173,  14,  18, 242, 208, 147],
[ 53, 194, 229, 141,  37, 215, 230, 167,  82],
[ 72,  78, 152,  76, 230, 128, 137,  25, 168]], dtype=uint8)
```

Here's the values on the perimeter:

```In : img[np.nonzero(mask)]
Out:
array([ 89, 206, 206, 143, 152, 212,  14,  45, 122, 159, 235, 177, 188,
67, 208, 229, 141, 215, 230, 230], dtype=uint8)
```

Setting the value of the image at the perimeter of the circle to 0:

```In : img[np.nonzero(mask)] = 0

In : img
Out:
array([[ 88, 239, 212, 160,   0,  85, 249, 242,  88],
[ 47, 230,   0,   0,  63,   0,   0,  67,  58],
[162,   0,   0, 213, 208, 169, 228,   0, 229],
[230,   0, 103, 201, 188, 231,  80,   0, 131],
[  0,  31, 148, 158,  73, 215, 152, 158,   0],
[213,   0, 148, 237,  92, 115, 152,   0, 223],
[234,   0, 141, 173,  14,  18, 242,   0, 147],
[ 53, 194,   0,   0,  37,   0,   0, 167,  82],
[ 72,  78, 152,  76,   0, 128, 137,  25, 168]], dtype=uint8)
```

You can easily get the coordinates as well:

```In : np.where(mask)
Out:
(array([0, 1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 7, 7, 8]),
array([4, 2, 3, 5, 6, 1, 7, 1, 7, 0, 8, 1, 7, 1, 7, 2, 3, 5, 6, 4]))
```

Find perimeter of objects in binary image, BW2 = bwperim( BW , conn ) specifies the pixel connectivity, conn . This example shows how to find the perimeter pixels in a binary image using the The 1 -valued elements define neighborhood locations relative to the center element of  Then count up the boxes as a whole unit each. Reinforce the difference between perimeter as a measure of length and square unites as a measure of area. Give students the option to do their name, or to do their own image, based on pixels. One pixel per square graph unit.

Assume `img` is your image, `radius` is the radius of the circle and `x, y` are the coordinates of the center around which you want to focus.

The the `focus_img` can be obtained using

```offset = math.ceil(radius * math.sqrt(2))
focus_img = img[y-offset:y+offset, x-offset:x+offset]
```

Discrete Geometry for Computer Imagery: 19th IAPR International , pixels, placed at nearly equal-spaced angular intervals, around the perimeter of intensity of the rotated test image centre pixel and perimeter pixels to better  BW2 = bwperim (BW) returns a binary image that contains only the perimeter pixels of objects in the input image BW. A pixel is part of the perimeter if it is nonzero and it is connected to at least one zero-valued pixel. BW2 = bwperim (BW,conn) specifies the pixel connectivity, conn.

Advances in Signal Processing and Intelligent Recognition Systems: , Initially the LBP operators convolves on image and finds the centre pixel To acquire neighbourhood pixels the sampling points P on the perimeter of the The centre pixel then compared with the points on the perimeter of the circle to get the circular neighbourhood of P sampling points and radius R around point ðx,yÞ. Kids love pixels. Capitalize on this trend with pixel art to teach area and perimeter. I suggested students use a winter theme, since we worked on these right before the holidays, and pulled up some sample images. Pixels are represented in kid-art, but you also find images used in cross-stitching and game development.

How to Calculate area of partial pixels in a square image cut off by a , and for each square (pixel) that gets crossed by the blue perimeter line, he wants to know the area that is inside the blue circle . He says "The  I have a binary image which I have extracted the perimeter with bwperim() function. Now that I have the perimeter image I am trying to extract the coordinates of pixels in this perimeter.Here are the images:

Intelligent Autonomous Systems: IAS-4 : Proceedings of the , A 3x3 mask is moved through out the image to find the edge candidate pixel and the superfluous pixel. The edge pixels which are in the normal direction of the centre edge pixel and having Perimeter = total no. of pixels in the boundary 3. I need to trace the boundary curve of an image region enumerated by measure.regionprops, similar to bwboundaries in Matlab.. By boundary curve, I mean a list of border pixels of the region in, say, clockwise direction around the region's perimeter, such that I can, for example, represent the region with a polygon.

• Both functions could be vectorized pretty easily. The second one in particular---just replace the `math` functions with `numpy` functions and use an array for `d`. You can directly index the image with the two pairs of corresponding arrays for the x and y coordinates (after rounding and explicitly casting to an int).
• @api55 maybe the following is correct? Instead of `for d in range(360):` it would be `for d in range(0, 360, math.floor(360 / (radius * 4))):` replace `4` for `8` depending.