CODEWITHSUNDEEP
pythonopencvcomputer-vision
paulbarber
paulbarber

Reputation: 21

python cv2.circle sub-pixel drawing

Does anyone know how to use cv2.circle with the sub-pixel "shift" parameter?

import cv2
import numpy as np

i1 = np.zeros((256, 256, 3), np.float32)
cv2.circle(i1, (199,199), 10, (1,0,0), -1)
cv2.imshow('1', i1)
i2 = np.zeros((256, 256, 3), np.float32)
cv2.circle(i2, (199,199), 10, (1,0,0), -1, shift=1)
cv2.imshow('2', i2)

cv2.waitKey(10000)

I would expect the circle in the second circle to be at (99.5,99.5), 199*(2^-1) = 99.5 For me it appears in the same place as the first. Cheers,

Upvotes: 2

Views: 5197

Answers (2)

Amir
Amir

Reputation: 2021

The shift parameter is like working with everything multiplied by (1 << shift) Thus if you want to draw a circle at (199.5, 199.5) you need to do something like this:

i2 = np.zeros((256, 256, 3), np.float32)
shift = 3
factor = (1 << shift)
cv2.circle(i2, (int(199.5 * factor + 0.5),int(199.5 * factor + 0.5)), 10 * factor, (1,0,0), -1, shift=shift)
cv2.imshow('2', i2)

Upvotes: 2

DavidT
DavidT

Reputation: 79

I did not encounter this issue as of today (openCV 3.1.0). However you have to be carefull: The input radius is shifted as well! Thus in your example the 2nd circle is half the size.

Upvotes: 1

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