CODEWITHSUNDEEP
pythonopencv
mrgloom
mrgloom

Reputation: 21622

cv2.boundingRect rect format documentation

Why this code produce (1, 1, 1, 1) result, as I understand cv2.boundingRect return rect in format x,y,w,h, but why w,h is not zero in this case?

import cv2
import numpy as np

cv2.__version__
'4.4.0'

np.__version__
'1.19.2'

a = np.ones((8,2))
x, y, w, h = cv2.boundingRect(a.astype(np.float32))

There is no much information in the bindings docs:

def boundingRect(array): # real signature unknown; restored from __doc__
    """
    boundingRect(array) -> retval
    .   @brief Calculates the up-right bounding rectangle of a point set or non-zero pixels of gray-scale image.
    .   
    .   The function calculates and returns the minimal up-right bounding rectangle for the specified point set or
    .   non-zero pixels of gray-scale image.
    .   
    .   @param array Input gray-scale image or 2D point set, stored in std::vector or Mat.
    """
    pass

Update:

Seems in C++ it returned like return Rect(xmin, ymin, xmax - xmin + 1, ymax - ymin + 1);

Upvotes: 1

Views: 1960

Answers (1)

fmw42
fmw42

Reputation: 53081

Not in the correct format. One way is to compute the points in Python/OpenCV.

import cv2
import numpy as np

a = np.ones((8,2))
points = np.column_stack(np.where(a.transpose() > 0))

x, y, w, h = cv2.boundingRect(points)
print(x,y,w,h)

Returns:

0 0 2 8

Upvotes: 3

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