# OpenCV_Laser_Cross **Repository Path**: alimteach/OpenCV_Laser_Cross ## Basic Information - **Project Name**: OpenCV_Laser_Cross - **Description**: OpenCV激光十字识别， 识别端点还有交叉点。 - **Primary Language**: Unknown - **License**: MIT - **Default Branch**: master - **Homepage**: None - **GVP Project**: No ## Statistics - **Stars**: 1 - **Forks**: 1 - **Created**: 2020-10-27 - **Last Updated**: 2021-07-12 ## Categories & Tags **Categories**: Uncategorized **Tags**: None ## README # OpenCV激光中心与端点识别-1Z实验室 出品：1Z实验室 （1ZLAB： Make Things Easy） ## 概要 下面的这张图片是十字激光的图像， 当前我们的任务就是要识别十字的四个端点还有十字中心。 这个缺口就是螺丝的孔位，所以你获得的图像可能是片段的，不一定连续。 ![shizi](./image/shizi.bmp) 我们用不同的颜色标示不同的端点， 空心圆代表交点。 ![laser cross](./image/cross-laser-demo.png) ## 算法流程讲解 ### 1-图像二值化 ![binary](./image/binary.png) 二值化排除其他干扰。 ```python binary = cv2.inRange(gray, 200, 255) ``` ### 2-拟合线段与直线 用`HoughLineP` 进行线段拟合。 ```python line_segs = cv2.HoughLinesP(binary, rho=2,theta=0.1, threshold=100) len(line_segs) ``` `HoughLineP`返回的是数组，每个元素是`Tuple`类型的数据。还是要打印一下这个数据结构。 ```python import math for lseg in line_segs: # x1,y1,x2,y2 = lseg[0] # 计算权重 weight = math.sqrt(math.pow(x1-x2, 2) + math.pow(y1-y2, 2)) print('x1: {}, y1: {}, x2: {}, y2: {}, weight: {}'.format(x1, y1, x2, y2, weight)) ``` **打印结果** 线段的标示方法是记录线段的两个端点， 从`(x1, y1)` 点到`（x2, y2)` 点。 数组的个数，取决于你的调参，从十几个到几百个不等。 ``` x1: 817, y1: 408, x2: 1068, y2: 164, weight: 350.0528531522061 x1: 525, y1: 93, x2: 868, y2: 468, weight: 508.20665088131227 x1: 630, y1: 202, x2: 818, y2: 408, weight: 278.89065957826557 x1: 488, y1: 56, x2: 814, y2: 412, weight: 482.7131653477042 x1: 816, y1: 407, x2: 960, y2: 267, weight: 200.83824337013107 x1: 714, y1: 520, x2: 714, y2: 520, weight: 0.0 x1: 845, y1: 391, x2: 1113, y2: 131, weight: 373.39523296367884 x1: 698, y1: 275, x2: 817, y2: 404, weight: 175.50498568416796 .... ``` 把线段的**长度** 作为这个线段的**权重 Weight**。 ```python weight = math.sqrt(math.pow(x1-x2, 2) + math.pow(y1-y2, 2)) ``` 每个线段都可以求解出这个线段所在直线的`k`还有`b`. ```python y = k*x + b ``` 求解`k`，`b`的算法比较简单，需要借助初中所学的知识。 ```python def calculate_line(x1, y1, x2, y2): ''' 计算直线 如果直线水平或者垂直，统一向一个方向倾斜特定角度。 TODO 这里面没有考虑水平或者垂直的情况 ''' if x1 > x2: x1,y1,x2,y2 = x2,y2,x1,y1 if x1 == x2 or y1 == y2: # 有时候会出现单个像素点 x1 = x2 而且 y1 = y2 print('x1:{} y1:{} x2:{} y2:{}'.format(x1, y1, x2, y2)) k = (y1 - y2) / (x1 - x2) b = (y2 * x1 - y1*x2) / (x1 - x2) return k,b ``` ### 3-合并线段 我们需要把这几百个线段拟合成**两条直线**， 这个**2** 属于我们的先验知识。 遍历所有的线段，求解其`k`， `b` 还有对应的`weight`。 然后我们通过字典数据结构来存放合并过后的**直线**。 数据结构细节如下： | 参数 | 备注 | | ---------- | ----------------------------- | | cur_k | 当前合并的直线的K值 | | cur_b | 当期合并的直线的b值 | | k_sum | K值的带权累加和 | | b_sum | b值的带权累加和 | | weight_sum | 权重和 | | x1 | 合并后大线段的左侧端点的x坐标 | | y1 | 合并后大线段的左侧端点的y坐标 | | x2 | 合并后大线段的右侧端点的x坐标 | | y2 | 合并后大线段的右侧端点的y坐标 | 所有合并后的直线/线段，存放在`lines`里面， 每个线段在合并的时候，遍历lines里面合并后的线段，通过`k`的差值（`max_k_distance`）判断是否属于同一个直线，如果里面都没有的话就另外添加一个。 ```python lines = [] # 最小权值 min_weight = 20 # 相同k之间最大的差距 max_k_distance = 0.3 for lseg in line_segs: # 获取线段端点值 x1,y1,x2,y2 = lseg[0] if x1 > x2: x1, y1, x2, y2 = x2, y2, x1, y1 # 计算权重 weight = math.sqrt(math.pow(x1 - x2, 2) + math.pow(y1 - y2, 2)) if weight != 0 and weight > min_weight: # 计算K与b k, b = calculate_line(x1, y1, x2, y2) # print('k: {:.2f}, b: {:.2f}, weight: {:.2f}'.format(k, b, weight)) if len(lines) == 0: # 初次填充line line = {} line['cur_k'] = k line['cur_b'] = b line['k_sum'] = k * weight line['b_sum'] = b * weight line['weight_sum'] = weight line['x1'] = x1 line['y1'] = y1 line['x2'] = x2 line['y2'] = y2 lines.append(line) continue # 根据k的差异做加权 # 首先获取lines数组里面k举例最近的那个 neighbor_line = min(lines, key=lambda line:abs(line['cur_k'] - k)) if abs(neighbor_line['cur_k'] - k) < max_k_distance: # 小于最大k差值，认为是同一条线 neighbor_line['weight_sum'] += weight neighbor_line['k_sum'] += k * weight neighbor_line['b_sum'] += b * weight neighbor_line['cur_k'] = neighbor_line['k_sum'] / neighbor_line['weight_sum'] neighbor_line['cur_b'] = neighbor_line['b_sum'] / neighbor_line['weight_sum'] if neighbor_line['x1'] > x1: neighbor_line['x1'] = x1 neighbor_line['y1'] = y1 if neighbor_line['x2'] < x2: neighbor_line['x2'] = x2 neighbor_line['y2'] = y2 else: # 添加另外一条线 # 初次填充line line = {} line['cur_k'] = k line['cur_b'] = b line['k_sum'] = k * weight line['b_sum'] = b * weight line['weight_sum'] = weight line['x1'] = x1 line['y1'] = y1 line['x2'] = x2 line['y2'] = y2 lines.append(line) ``` 做完上述的操作之后，我们获取的`lines`的长度可能不是2, 可能会大于2. 所以我们可以根据`weight_sum`对`lines`进行重新排序， 然后截取前两个，作为激光十字的两条直线。 ```python # 根据权重对lines数组进行排序， 取前两个（lines的长度有可能大于2） sorted_lines = sorted(lines, key=lambda line: line['weight_sum'])[::-1] line1 = sorted_lines[0] line2 = sorted_lines[1] ``` ```python [{'b_sum': -3304027.8377846032, 'cur_b': -482.1075439824276, 'cur_k': 1.0900334200603314, 'k_sum': 7470.32650483927, 'weight_sum': 6853.300428555484, 'x1': 478, 'x2': 1001, 'y1': 54, 'y2': 597}, {'b_sum': 8948293.312710544, 'cur_b': 1209.7121822845368, 'cur_k': -0.9799324921216083, 'k_sum': -7248.603010345705, 'weight_sum': 7397.043233715087, 'x1': 599, 'x2': 1113, 'y1': 607, 'y2': 129}] ``` ### 4-计算交点 ```python def calculate_intersection(line1, line2): a1 = line1['y2'] - line1['y1'] b1 = line1['x1'] - line1['x2'] c1 = line1['x2'] * line1['y1'] - line1['x1'] * line1['y2'] a2 = line2['y2'] - line2['y1'] b2 = line2['x1'] - line2['x2'] c2 = line2['x2'] * line2['y1'] - line2['x1'] * line2['y2'] if (a1 * b2 - a2 * b1) != 0 and (a2 * b1 - a1 * b2) != 0: cross_x = int((b1*c2-b2*c1)/(a1*b2-a2*b1)) cross_y = int((c1*a2-c2*a1)/(a1*b2-a2*b1)) return (cross_x, cross_y) return None ``` 计算交点： ```python (cx, cy) = calculate_intersection(line1, line2) print('cx: {} cy: {}'.format(cx, cy)) ``` ``` cx: 816 cy: 405 ``` ### 5-信息可视化 在画面上绘制四个端点与中心交点： ```python canvas = cv2.cvtColor(binary, cv2.COLOR_GRAY2BGR) # 绘制第一条线 pt_radius = 20 cv2.circle(canvas, (line1['x1'], line1['y1']),pt_radius, (255, 0, 0), thickness=-1) cv2.circle(canvas, (line1['x2'], line1['y2']),pt_radius, (0, 255, 0), thickness=-1) cv2.circle(canvas, (line2['x1'], line2['y1']),pt_radius, (0, 255, 255), thickness=-1) cv2.circle(canvas, (line2['x2'], line2['y2']),pt_radius, (0, 0, 255), thickness=-1) cv2.circle(canvas, (cx, cy), 40, (255, 0, 255), thickness=20) plt.imshow(cv2.cvtColor(canvas, cv2.COLOR_BGR2RGB)) plt.show() ``` ![demo](./image/cross-laser-demo.png) ## 推广 出品：1Z实验室 （1ZLAB： Make Things Easy） 扫码关注微信公众号**1Z实验室**， 回复关键词**激光十字源码** 获取源码。 1Z实验室 Make Things Easy . 致力于在机器人+计算机视觉+人工智能的重叠区域, 制作小白友好的教程. ![tuiguang](https://upload-images.jianshu.io/upload_images/1199728-589a80ff77f380d8.png?imageMogr2/auto-orient/strip%7CimageView2/2/w/700)