填充一个轮廓的圆圈

Question

我有一组图像，其中一个圆圈绘制为白色轮廓。但是，我想用白色填充整个圆圈。什么是快速的方法？以下是图片示例：

我曾尝试使用嵌套循环来实现这一点，但这需要很长时间，而且我有大约 150 万张图像。以下是我的代码：

roundRobinIndex = 0
new_image = np.zeros((img_w, img_h))
for row in range(540):
    for column in range(800):
        if image[row,column] == 255:
            roundRobinIndex = (roundRobinIndex + 1) % 2
        if roundRobinIndex == 1:
            new_image[row, column] = 255

Answer 1

使用cv2.fillPoly()填充圆形轮廓

import cv2

image = cv2.imread('1.png', 0)
thresh = cv2.threshold(image, 0, 255, cv2.THRESH_BINARY + cv2.THRESH_OTSU)[1]
cnts = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if len(cnts) == 2 else cnts[1]
cv2.fillPoly(image, cnts, [255,255,255])

cv2.imshow('image', image)
cv2.waitKey()

注意：由于输入图像已经是二值图像，可以去除 Otsu 的阈值以获得稍快的性能，您可以直接在灰度图像上找到轮廓

Answer 2

我尝试找到白色轮廓的边界框，并找到它的中心，然后从那里向外填充白色。

#!/usr/bin/env python3

import cv2

def findfill(image):
    thresh = cv2.threshold(image, 0, 255, cv2.THRESH_BINARY + cv2.THRESH_OTSU)[1]
    cnts = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
    cnts = cnts[0] if len(cnts) == 2 else cnts[1]
    cv2.fillPoly(image, cnts, [255,255,255])

def me(image):
    x,y,w,h = cv2.boundingRect(image)
    cv2.floodFill(image,None,(int(x+w/2),int(y+h/2)),255)
    return image

image = cv2.imread('BLYmz.png', 0)

%timeit findfill(image)
%timeit me(image)

这似乎给出了相同的结果并且运行速度提高了 2.5 倍：

findfill
810 µs ± 2.94 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

me
343 µs ± 1.06 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

当然，如果你有 150 万个任务要做，我也会推荐一些并行处理 :-)

Answer 3

对于真正任意的形状，我建议使用洪水填充。但是，由于您有保证的凸形，您可以进行一些优化。具体来说，图像的每一行/每一列都将遵循以下三种模式之一：

1. 全黑
2. 黑、白、黑
3. 黑、白、黑、白、黑

从技术上讲，还有更多选项，因为选项 2 和 3 中的一个或两个黑色边距可能会丢失。目标是填充选项 3 中的中间黑色区域。这可以通过一些简单的 numpy 掩码和花哨的索引来完成。

基本算法是：

1. 计算每个白色段的起始索引
2. 制作包含两个起始索引的行的行掩码
3. 制作一个包含原始数据的完整掩码，索引之间的元素也设置为True。

def fill_convex(image):
    mask = image.astype(np.bool)
    # mask out elements that are 1, but the previous is 0
    start = (mask[:, 1:] & ~mask[:, :-1])
    # find rows that have exactly two runs of True
    row_mask = (np.count_nonzero(start, axis=1) == 2)
    # get the pairs of column indices that correspond to the masked elements
    cols = np.nonzero(start[row_mask, :])[1].reshape(-1, 2)
    # create a row of column indices the same size as a row
    count = np.arange(image.shape[1])
    # fill in the elements between start and stop indices for each row
    # the None indices are used to trigger broadcasting
    to_fill = ((count[None, :] >= cols[:, 0, None]) & (count[None, :] <= cols[:, 1, None]))
    # update the mask
    mask[row_mask, :] |= to_fill
    # fill in the image
    image[mask] = 255
    return image

时机

这种方法的速度大约是@nathancy's 的两倍，比@MarkSetchell's 慢10 倍以上。在这一点上，我基本上把它留在这里。

$ python -m timeit -s 'import q58174115' 'q58174115.nathancy(q58174115.image)'
500 loops, best of 5: 437 usec per loop
$ python -m timeit -s 'import q58174115' 'q58174115.MarkSetchell(q58174115.image.copy())'
5000 loops, best of 5: 62.9 usec per loop
$ python -m timeit -s 'import q58174115' 'q58174115.MadPhysicist(q58174115.image.copy())'
500 loops, best of 5: 779 usec per loop

这里，q58174115.py 是

import cv2
import numpy as np

def nathancy(image):
    thresh = cv2.threshold(image, 0, 255, cv2.THRESH_BINARY + cv2.THRESH_OTSU)[1]
    cnts = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
    cnts = cnts[0] if len(cnts) == 2 else cnts[1]
    cv2.fillPoly(image, cnts, [255,255,255])
    return image

def MarkSetchell(image):
    x,y,w,h = cv2.boundingRect(image)
    cv2.floodFill(image,None,(int(x+w/2),int(y+h/2)),255)
    return image

def MadPhysicist(image):
    mask = image.astype(np.bool)
    # mask out elements that are 1, but the previous is 0
    start = (mask[:, 1:] & ~mask[:, :-1])
    # find rows that have exactly two runs of True
    row_mask = (np.count_nonzero(start, axis=1) == 2)
    # get the pairs of column indices that correspond to the masked elements
    cols = np.nonzero(start[row_mask, :])[1].reshape(-1, 2)
    # create a row of column indices the same size as a row
    count = np.arange(image.shape[1])
    # fill in the elements between start and stop indices for each row
    # the None indices are used to trigger broadcasting
    to_fill = ((count[None, :] >= cols[:, 0, None]) & (count[None, :] <= cols[:, 1, None]))
    # update the mask
    mask[row_mask, :] |= to_fill
    # fill in the image
    image[mask] = 255
    return image

image = cv2.imread('58174115.png', 0)