如何在图像周围添加圆形边框？

Question

我有一个矩形图像，我想将其角变圆，然后为其添加黑色边框（因此边框也是圆形的）。

有没有一种简单的方法可以实现它？

这将是所需的输出：

类似的未回答问题

Answer 1

在我的第一个答案的评论中与 Mark 进行一些讨论后，我决定使用 OpenCV 和 NumPy 制作另一个解决方案，它能够轻松地将一些真实图像（例如照片）提供给该方法，并获取包含圆角边框的图像，以及边界外的透明度！

import cv2
import numpy as np


def rect_with_rounded_corners(image, r, t, c):
    """
    :param image: image as NumPy array
    :param r: radius of rounded corners
    :param t: thickness of border
    :param c: color of border
    :return: new image as NumPy array with rounded corners
    """

    c += (255, )

    h, w = image.shape[:2]

    # Create new image (three-channel hardcoded here...)
    new_image = np.ones((h+2*t, w+2*t, 4), np.uint8) * 255
    new_image[:, :, 3] = 0

    # Draw four rounded corners
    new_image = cv2.ellipse(new_image, (int(r+t/2), int(r+t/2)), (r, r), 180, 0, 90, c, t)
    new_image = cv2.ellipse(new_image, (int(w-r+3*t/2-1), int(r+t/2)), (r, r), 270, 0, 90, c, t)
    new_image = cv2.ellipse(new_image, (int(r+t/2), int(h-r+3*t/2-1)), (r, r), 90, 0, 90, c, t)
    new_image = cv2.ellipse(new_image, (int(w-r+3*t/2-1), int(h-r+3*t/2-1)), (r, r), 0, 0, 90, c, t)

    # Draw four edges
    new_image = cv2.line(new_image, (int(r+t/2), int(t/2)), (int(w-r+3*t/2-1), int(t/2)), c, t)
    new_image = cv2.line(new_image, (int(t/2), int(r+t/2)), (int(t/2), int(h-r+3*t/2)), c, t)
    new_image = cv2.line(new_image, (int(r+t/2), int(h+3*t/2)), (int(w-r+3*t/2-1), int(h+3*t/2)), c, t)
    new_image = cv2.line(new_image, (int(w+3*t/2), int(r+t/2)), (int(w+3*t/2), int(h-r+3*t/2)), c, t)

    # Generate masks for proper blending
    mask = new_image[:, :, 3].copy()
    mask = cv2.floodFill(mask, None, (int(w/2+t), int(h/2+t)), 128)[1]
    mask[mask != 128] = 0
    mask[mask == 128] = 1
    mask = np.stack((mask, mask, mask), axis=2)

    # Blend images
    temp = np.zeros_like(new_image[:, :, :3])
    temp[(t-1):(h+t-1), (t-1):(w+t-1)] = image.copy()
    new_image[:, :, :3] = new_image[:, :, :3] * (1 - mask) + temp * mask

    # Set proper alpha channel in new image
    temp = new_image[:, :, 3].copy()
    new_image[:, :, 3] = cv2.floodFill(temp, None, (int(w/2+t), int(h/2+t)), 255)[1]

    return new_image


img = cv2.imread('path/to/your/image.png')
cv2.imshow('img', img)

new_img = rect_with_rounded_corners(img, 50, 20, (0, 0, 0))
cv2.imshow('new_img', new_img)

cv2.waitKey(0)
cv2.destroyAllWindows()

这与我的其他答案中使用的概念相同，其中包含一些有关正确透明度内容的更多代码。

一些示例性输入：

相应的输出：

另一个输入和参数集：

new_img = rect_with_rounded_corners(img, 20, 10, (0, 0, 128))

输出：

希望这也有帮助！

----------------------------------------
System information
----------------------------------------
Platform:    Windows-10-10.0.16299-SP0
Python:      3.8.1
NumPy:       1.18.1
OpenCV:      4.2.0
----------------------------------------