python-opencv 关于ny.array的创建以及轮廓的查找

本文介绍了一个使用Python结合OpenCV库实现轮廓检测与绘制的应用案例。该案例通过读取图像文件,利用findContours函数找到图像中的所有轮廓,并通过创建滑动条的方式让用户选择要绘制的具体轮廓。文中展示了如何设置回调函数来更新显示的轮廓,并提供了完整的代码示例。

import numpy as np
import cv2
if __name__ == '__main__':
    import sys
    try: fn = sys.argv[1]
    except: fn = 'E:\\1.BMP'
    print __doc__
    global m    #全局变量
    def callback(*argv):
         m=cv2.getTrackbarPos('select', 'forshow')
         forshow=np.zeros((h,w,3),np.uint8)#cv2接口创建矩阵
         cv2.drawContours( forshow, contours0,m, (0,0,255), 1, cv2.CV_AA, hierarchy,3  )#m为所描绘的轮廓 m = hierarchy[idx][0]
         cv2.imshow('forshow',forshow)
    img = cv2.imread(fn, 2)
    h,w=img.shape[0:2]
    contours0, hierarchy = cv2.findContours( img.copy(), cv2.RETR_TREE, cv2.CHAIN_APPROX_NONE)#参数分别为src,mode,method
                                                                                                #CV_RETR_TREE 即所有轮廓 CV_RETR_EXTERNAL 外部轮廓
                                                                                                #CV_CHAIN_APPROX_SIMPLE 只留下脚点 CV_CHAIN_APPROX_NONE 所有点
    
    forshow=np.zeros((h,w,3), np.uint8)
    cv2.namedWindow('forshow')
    cv2.createTrackbar('select', 'forshow', 0, 20, callback)#设置回调
    m=cv2.getTrackbarPos('select', 'forshow')
    cv2.drawContours( forshow, contours0,m, (0,0,255), 1, cv2.CV_AA, hierarchy,3  )   #m为所描绘的轮廓 m = hierarchy[idx][0]
    cv2.imshow('forshow',forshow)
    print hierarchy
    while True:
        ch = 0xFF & cv2.waitKey(0)
        if ch == 27:
             break

        cv2.destroyAllWindows()

通过回调函数callback来改变golba m的值,从而描绘我们所需要的轮廓

要将迷宫图片数字化,您需要执行以下步骤: 1. 使用OpenCV加载迷宫图像并将其转换为灰度图像。 2. 对图像进行二值化,以便仅包含黑色和白色像素。 3. 使用形态学转换(例如膨胀和腐蚀)来填充迷宫中的空隙并消除不必要的噪声。 4. 找到迷宫的入口和出口。这可以通过查找轮廓并选择最长的两个轮廓来完成。 5. 使用霍夫线变换找到迷宫中的所有水平和垂直线。 6. 使用线段交点检测找到所有交点。 7. 将交点与入口和出口相匹配。 8. 创建一个表示迷宫的矩阵,其中表示墙壁的像素被设置为1,表示通道的像素被设置为0。 9. 根据找到的交点和线段,将墙壁添加到矩阵中。 10. 使用路径搜索算法(例如广度优先搜索或Dijkstra算法)找到从入口到出口的最短路径。 以下是一个示例代码,演示了如何实现这些步骤: ``` python import cv2 import numpy as np # Load the maze image and convert it to grayscale maze = cv2.imread('maze.png') gray = cv2.cvtColor(maze, cv2.COLOR_BGR2GRAY) # Threshold the image to get a binary image thresh = cv2.threshold(gray, 0, 255, cv2.THRESH_BINARY_INV + cv2.THRESH_OTSU)[1] # Apply morphological transformations to fill gaps and remove noise kernel = cv2.getStructuringElement(cv2.MORPH_RECT, (3, 3)) thresh = cv2.morphologyEx(thresh, cv2.MORPH_CLOSE, kernel) # Find the contours of the maze and select the two longest contours contours, hierarchy = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE) contours = sorted(contours, key=cv2.contourArea, reverse=True)[:2] # Find the entrance and exit points of the maze entrance, exit = None, None for contour in contours: x, y, w, h = cv2.boundingRect(contour) if w > 2 * h: if entrance is None or x < entrance[0]: entrance = (x, y) if exit is None or x > exit[0]: exit = (x, y) elif h > 2 * w: if entrance is None or y < entrance[1]: entrance = (x, y) if exit is None or y > exit[1]: exit = (x, y) # Detect horizontal and vertical lines in the maze edges = cv2.Canny(thresh, 50, 150) lines = cv2.HoughLines(edges, 1, np.pi / 180, 150) horizontal_lines, vertical_lines = [], [] for line in lines: rho, theta = line[0] a, b = np.cos(theta), np.sin(theta) x0, y0 = a * rho, b * rho if abs(a) < 0.1: # Vertical line vertical_lines.append((int(x0), int(y0))) elif abs(b) < 0.1: # Horizontal line horizontal_lines.append((int(x0), int(y0))) # Find the intersection points of the lines intersections = [] for hl in horizontal_lines: for vl in vertical_lines: x, y = int(vl[0]), int(hl[1]) intersections.append((x, y)) # Match the entrance and exit points to the nearest intersection point entrance = min(intersections, key=lambda p: np.linalg.norm(np.array(p) - np.array(entrance))) exit = min(intersections, key=lambda p: np.linalg.norm(np.array(p) - np.array(exit))) # Create a matrix representation of the maze maze_matrix = np.zeros(gray.shape[:2], dtype=np.uint8) for hl in horizontal_lines: x0, y0 = hl for vl in vertical_lines: x1, y1 = vl if x1 <= x0 + 5 and x1 >= x0 - 5 and y1 <= y0 + 5 and y1 >= y0 - 5: # This is an intersection point maze_matrix[y1, x1] = 0 elif x1 < x0: # This is a vertical wall maze_matrix[y1, x1] = 1 elif y1 < y0: # This is a horizontal wall maze_matrix[y1, x1] = 1 # Find the shortest path from the entrance to the exit using BFS queue = [(entrance[1], entrance[0])] visited = np.zeros(maze_matrix.shape[:2], dtype=np.bool) visited[entrance[1], entrance[0]] = True prev = np.zeros(maze_matrix.shape[:2], dtype=np.int32) while queue: y, x = queue.pop(0) if (y, x) == exit: # We have found the shortest path break for dy, dx in [(1, 0), (-1, 0), (0, 1), (0, -1)]: ny, nx = y + dy, x + dx if ny >= 0 and ny < maze_matrix.shape[0] and nx >= 0 and nx < maze_matrix.shape[1] \ and maze_matrix[ny, nx] == 0 and not visited[ny, nx]: queue.append((ny, nx)) visited[ny, nx] = True prev[ny, nx] = y * maze_matrix.shape[1] + x # Reconstruct the shortest path path = [] y, x = exit while (y, x) != entrance: path.append((y, x)) p = prev[y, x] y, x = p // maze_matrix.shape[1], p % maze_matrix.shape[1] path.append((y, x)) path.reverse() # Draw the shortest path on the maze image output = maze.copy() for i in range(len(path) - 1): cv2.line(output, path[i][::-1], path[i + 1][::-1], (0, 0, 255), 2) # Display the output image cv2.imshow('Output', output) cv2.waitKey(0) cv2.destroyAllWindows() ``` 此示例代码假定您的迷宫是一个黑色的正方形,并且在其中只有一个入口和一个出口。如果您的迷宫有其他形状或有多个入口/出口,则需要根据需要进行修改。
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