Python 在二进制图像中找到一个能够容纳形状的空白空间

对于这个问题，这里有一个简单而幼稚的解决方案.它使用UnquoteQuote中提到的2D卷积.

import random
import numpy as np
import scipy.signal as sig
import matplotlib.pyplot as plt
import cv2

# load images
image = (cv2.imread('image.jpg').mean(axis=2) > 127).astype(np.float32)
shape = (cv2.imread('shape.jpg').mean(axis=2) > 127).astype(np.float32)

# perform 2D convolution
conv = sig.convolve2d(image, shape, mode='valid')
solutions = np.where(conv == 0)

# draw some solutions
plt.figure(figsize=(16, 4))
for i in range(4):
    r = random.randint(0, solutions[0].shape[0] - 1)
    x, y = solutions[0][r], solutions[1][r]

    solution_plot = np.zeros((*image.shape, 3))
    solution_plot[:, :, 0] = image
    solution_plot[x:x + shape.shape[0], y:y + shape.shape[1], 1] = shape

    plt.subplot(1, 4, i + 1)
    plt.imshow(solution_plot)

plt.show()

Example results:

该算法可以找到所有可能的解决方案.如果你只需要一个，你可以优化它，使它得到一个随机的(x, y)点，并执行形状和裁剪图像区域[x:x+shape_width, y:y+shape_height]的点积，以判断是否有空间，直到你找到正确的点.

例如，可以这样做:

while True:
    x = random.randint(0, image.shape[0] - shape.shape[0])
    y = random.randint(0, image.shape[1] - shape.shape[1])

    if np.sum(shape*image[x:x + shape.shape[0], y:y + shape.shape[1]]) == 0:
        break

# x, y is the solution

与卷积相比，这种方法要快得多(但这取决于解的数量):

• 卷积:6.65 s ± 21.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
• 随机搜索:1.31 ms ± 31.2 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)