需求：计算下图中7个白色矩形的平均宽度大小

讯享网

方法一：提取边白色矩形与背景的交界线，通过直线拟合，利用算子计算两直线距离。

代码：

dev_close_window () dev_open_window (0, 0, 512, 512, 'black', WindowHandle) dev_update_off () count_seconds (T1) read_image (Image1, '图片的位置') *判断图像的通道数，来决定是否将其转换为灰度图 count_channels (Image1, Channels) if (Channels==3) rgb1_to_gray (Image1, Image1) endif *将需要测量距离的地方提取出来 threshold (Image1, Region, 128, 255) connection (Region, ConnectedRegions) *形状选择 select_shape (ConnectedRegions, SelectedRegions, ['rect2_len1','rect2_len2'], 'and', [100,50], [150,80]) select_shape (SelectedRegions, SelectedRegions1, 'convexity', 'and', 0.8, 1) *正确提取后应该有七个目标块 count_obj (SelectedRegions1, Number) if (Number!=7) stop () endif *将七个目标块合并 union1 (SelectedRegions1, RegionUnion) *裁剪出ROI smallest_rectangle1 (RegionUnion, Row1, Column1, Row2, Column2) gen_rectangle1 (Rectangle, Row1+30, Column1-10, Row2-30, Column2+10) reduce_domain (Image1, Rectangle, ImageReduced) *提取边 edges_sub_pix (ImageReduced, Edges, 'canny', 1, 20, 40) *合并轮廓 union_adjacent_contours_xld (Edges, UnionContours, 10, 1, 'attr_keep') select_shape_xld (UnionContours, SelectedXLD, 'contlength', 'and', 150, 99999) *正确提取后应该有14个线段 count_obj (SelectedXLD, Number1) if (Number1!=14) stop () endif *按列对轮廓进行排序 sort_contours_xld (SelectedXLD, SortedContours, 'upper_left', 'true', 'column') *边缘遍历 a:=[] for i := 1 to Number1 by 2 read_image (Image, 'printer_chip/printer_chip_01') select_obj (SortedContours, Obj1, i) *最小二乘拟合直线 fit_line_contour_xld (Obj1, 'tukey', -1, 0, 5, 2, RowBegin, ColBegin, RowEnd, ColEnd, Nr, Nc, Dist) *生成轮廓 gen_contour_polygon_xld (Contour1, [RowBegin,RowEnd], [ColBegin,ColEnd]) select_obj (SortedContours, Obj2, i+1) fit_line_contour_xld (Obj2, 'tukey', -1, 0, 5, 2, RowBegin, ColBegin, RowEnd, ColEnd, Nr, Nc, Dist) gen_contour_polygon_xld (Contour2, [RowBegin,RowEnd], [ColBegin,ColEnd]) *求距离 distance_cc (Contour1, Contour2, 'point_to_point', DistanceMin, DistanceMax) *存储所求距离的不同值 a:=[a,DistanceMin] endfor *求距离的平均值 tuple_mean (a, Mean) count_seconds (T2) T:=T2-T1 dev_display (Image1) disp_message (WindowHandle, Mean, 'window', 0, 0, 'black', 'true') disp_message (WindowHandle, T+'秒', 'window', 50, 0, 'black', 'true')

方法二：通过灰度投影形成函数，找出“明暗”交界线的两个点的横坐标做差，便可求得矩形宽度。

代码：

讯享网dev_close_window () dev_open_window (0, 0, 512, 512, 'black', WindowHandle) dev_update_off () count_seconds (T1) read_image (Image1, '图片的位置') dev_close_window () dev_open_window (0, 0, 512, 512, 'black', WindowHandle) dev_update_off () *开始时间 count_seconds (T1) *读取图像 read_image (Image1, '图片的位置') dev_display (Image1) *生成空区域 gen_empty_obj (rect) *生成7个矩形区域 for i := 1 to 7 by 1 dev_set_line_width(3) dev_set_draw ('margin') draw_rectangle1 (WindowHandle, Row1, Column1, Row2, Column2) gen_rectangle1 (Rectangle, Row1, Column1, Row2, Column2) concat_obj (rect, Rectangle, rect) endfor stop () *结果数组 a:=[] for Index := 1 to 7 by 1 *选择区域 select_obj (rect, ObjectSelected, Index) *裁剪 reduce_domain (Image1, ObjectSelected, ImageReduced) *计算竖直方向上的灰度投影 gray_projections (ObjectSelected, ImageReduced, 'simple', HorProjection, VertProjection) *通过一个一维数组创建一个离散一维函数 create_funct_1d_array (VertProjection, Function) *平滑一维函数 smooth_funct_1d_gauss (Function, 2, SmoothedFunction) *一阶导数 derivate_funct_1d (SmoothedFunction, 'first', Derivative) *平滑函数，获得函数的极值点 smooth_funct_1d_gauss (Derivative, 2, SmoothedFunction1) *获取所有波峰波谷对应的x值 local_min_max_funct_1d (SmoothedFunction1, 'strict_min_max', 'true', Min, Max) x:=[Min,Max] *获取所有x值对应的y值 get_y_value_funct_1d (SmoothedFunction1, x, 'constant', Y) *找出最大值、最小值 result:=[] for i := 0 to |x|-1 by 1 if (abs(Y[i])>10) result:=[result,x[i]] endif endfor *计算宽度 d:=abs(result[0]-result[1]) a:=[a,d] stop () endfor *求平均值 tuple_mean (a, Mean) disp_message (WindowHandle, Mean, 'window', 0, 0, 'black', 'true')

注:这段代码“draw_rectangle1 (WindowHandle, Row1, Column1, Row2, Column2)”执行时需要从左至右框选七个矩形。矩形大小以刚好框柱白色矩形但又含有边缘交界为宜，如下图所示。