對于這樣的圖片：

摳出其中的黑色區(qū)域，效果如下：

import cv2 import numpy as np import matplotlib.pyplot as plt import timedef findUnicomArea(img):#先二值化ret,threshold = cv2.threshold(img,128,255,cv2.THRESH_BINARY)img_flag = np.zeros(threshold.shape,np.int8)count = 0findpoint = []#首先遍歷圖像找到所有的聯(lián)通區(qū)for x in range(threshold.shape[0]):for y in range(threshold.shape[1]):if(threshold[x][y] == 0 and img_flag[x][y] == 0):#這里表示已經(jīng)找到了一個沒有標志過的黑點，是一個新的聯(lián)通區(qū)count += 1img_flag[x][y] = countfindpoint.append((x,y))while len(findpoint) > 0:xx,yy = findpoint.pop()if xx > 0 :#上面if threshold[xx-1][yy] == 0 and img_flag[xx-1][yy] == 0:findpoint.append((xx-1,yy))img_flag[xx-1][yy] = countif xx < img.shape[0]:#下面if threshold[xx + 1][yy] == 0 and img_flag[xx + 1][yy] == 0:findpoint.append((xx + 1, yy))img_flag[xx+1][yy] = countif yy > 0:#左面if threshold[xx][yy-1] == 0 and img_flag[xx][yy-1] == 0:findpoint.append((xx, yy-1))img_flag[xx][yy-1] = countif yy < img.shape[1]:#右面if threshold[xx][yy+1] == 0 and img_flag[xx][yy+1] == 0:findpoint.append((xx, yy+1))img_flag[xx][yy+1] = count#聯(lián)通區(qū)任然存在一張表中，需要將其分離coutours = []for num in range(1,count+1):coutours.append([])for x in range(img_flag.shape[0]):for y in range(img_flag.shape[1]):if img_flag[x][y] == num:coutours[num-1].append([x,y,img_flag[x][y]])desCoutous = np.empty(len(coutours),np.object)for num in range(len(coutours)):#將分離后的圖像提取出來。計算出聯(lián)通區(qū)所在的范圍tmp = np.mat(coutours[num])minX = np.min(tmp[:,0])maxX = np.max(tmp[:,0])minY = np.min(tmp[:,1])maxY = np.max(tmp[:,1])desCoutous[num] = img[minX:maxX,minY:maxY]return desCoutous#測試程序如下 img = cv2.imread("testpic/connected_test.png",0); cost = time.time() unicoms = findUnicomArea(img) print("cost",time.time()-cost) for i in range(len(unicoms)):cv2.imshow("test"+str(i),unicoms[i])cv2.waitKey(0)

目前只是實現(xiàn)了功能，效率還不高，還有很大的優(yōu)化空間。

這個程序顯然有重復計算，第二次循環(huán)完全可以略去，可以合并到第一個循環(huán)中。另外，只需要遍歷1~shape[0]-1的空間就能找到所有的點，所以邊界的比較也可以省去，如此，程序如下：

def findUnicomArea(img):#先二值化ret,threshold = cv2.threshold(img,128,255,cv2.THRESH_BINARY)img_flag = np.zeros(threshold.shape,np.int8)count = 0findpoint = []coutours = []#首先遍歷圖像找到所有的聯(lián)通區(qū)for x in range(1,threshold.shape[0]-1):for y in range(1,threshold.shape[1]-1):if(threshold[x][y] == 0 and img_flag[x][y] == 0):#這里表示已經(jīng)找到了一個沒有標志過的黑點，是一個新的聯(lián)通區(qū)count += 1#新增一個聯(lián)通區(qū)存儲點coutours.append([])img_flag[x][y] = countfindpoint.append((x,y))while len(findpoint) > 0:xx,yy = findpoint.pop()#上面if threshold[xx-1][yy] == 0 and img_flag[xx-1][yy] == 0:findpoint.append((xx-1,yy))img_flag[xx-1][yy] = countcoutours[count - 1].append([xx, yy, img_flag[x][y]])#下面if threshold[xx + 1][yy] == 0 and img_flag[xx + 1][yy] == 0:findpoint.append((xx + 1, yy))img_flag[xx+1][yy] = countcoutours[count - 1].append([xx, yy, img_flag[x][y]])#左面if threshold[xx][yy-1] == 0 and img_flag[xx][yy-1] == 0:findpoint.append((xx, yy-1))img_flag[xx][yy-1] = countcoutours[count - 1].append([xx, yy, img_flag[x][y]])#右面if threshold[xx][yy+1] == 0 and img_flag[xx][yy+1] == 0:findpoint.append((xx, yy+1))img_flag[xx][yy+1] = countcoutours[count - 1].append([xx, yy, img_flag[x][y]])desCoutous = np.empty(len(coutours),np.object)for num in range(len(coutours)):#將分離后的圖像提取出來。計算出聯(lián)通區(qū)所在的范圍tmp = np.mat(coutours[num])minX = np.min(tmp[:,0])maxX = np.max(tmp[:,0])minY = np.min(tmp[:,1])maxY = np.max(tmp[:,1])desCoutous[num] = img[minX:maxX,minY:maxY]return desCoutous

這樣程序性能從1.3秒提升至0.8秒，程序有了顯著的提升。
如果我們只是摳出圖中的聯(lián)通區(qū)的話，只需要檢測邊界聯(lián)通區(qū)的外邊界就可以了，這樣效率能進一步提升。

def findUnicomBoundry(img):#先二值化ret,threshold = cv2.threshold(img,128,255,cv2.THRESH_BINARY)img_flag = np.zeros(threshold.shape,np.int8)count = 0findpoint = []coutours = []desCoutous = []existCoutous = []#首先遍歷圖像找到所有的聯(lián)通區(qū)for x in range(1,threshold.shape[0]-1):for y in range(1,threshold.shape[1]-1):if(threshold[x][y] == 0 and (threshold[x-1][y]==255 or threshold[x+1][y]==255 or threshold[x][y-1]==255 or threshold[x][y+1]==255) and img_flag[x][y] == 0):#是邊界的條件是中心點為0，四周至少要有一個白點#這里表示已經(jīng)找到了一個邊界點，并且是一個新的聯(lián)通區(qū)的邊界點# 將分離后的圖像提取出來。計算出聯(lián)通區(qū)所在的范圍if count > len(desCoutous):desCoutous.append([])tmp = np.mat(coutours[count - 1])minX = np.min(tmp[:, 0])maxX = np.max(tmp[:, 0])minY = np.min(tmp[:, 1])maxY = np.max(tmp[:, 1])existCoutous.append([minX, maxX, minY, maxY])desCoutous[count - 1] = img[minX:maxX, minY:maxY]desCoutous[count-1] = np.mat(desCoutous[count-1])isChildArea = Falsefor num in range(len(existCoutous)):if x>existCoutous[num][0] and x < existCoutous[num][1] and y > existCoutous[num][2] and y < existCoutous[num][3] :isChildArea = Trueif not isChildArea:count += 1#新增一個聯(lián)通區(qū)的邊界存儲點coutours.append([])img_flag[x][y] = countfindpoint.append((x,y))while len(findpoint) > 0:#xx,yy肯定是一個邊界點了，現(xiàn)在尋找下一個邊界點xx,yy = findpoint.pop()#上面if threshold[xx-1][yy] == 0 and (threshold[xx-2][yy]==255 or threshold[xx-1][yy-1]==255 or threshold[xx-1][yy+1]==255) and img_flag[xx-1][yy] == 0:findpoint.append((xx-1,yy))img_flag[xx-1][yy] = countcoutours[count - 1].append([xx-1, yy, img_flag[xx-1][yy]])#下面if threshold[xx + 1][yy] == 0 and (threshold[xx+1][yy]==255 or threshold[xx+1][yy-1]==255 or threshold[xx+1][yy+1]==255) and img_flag[xx + 1][yy] == 0:findpoint.append((xx + 1, yy))img_flag[xx+1][yy] = countcoutours[count - 1].append([xx+1, yy, img_flag[xx][yy]])#左面if threshold[xx][yy-1] == 0 and (threshold[xx][yy-2]==255 or threshold[xx-1][yy-1]==255 or threshold[xx+1][yy-1]==255) and img_flag[xx][yy-1] == 0:findpoint.append((xx, yy-1))img_flag[xx][yy-1] = countcoutours[count - 1].append([xx, yy-1, img_flag[xx][yy-1]])#右面if threshold[xx][yy+1] == 0 and (threshold[xx][yy+2]==255 or threshold[xx-1][yy+1]==255 or threshold[xx+1][yy+1]==255) and img_flag[xx][yy+1] == 0:findpoint.append((xx, yy+1))img_flag[xx][yy+1] = countcoutours[count - 1].append([xx, yy+1, img_flag[xx][yy+1]])#左上if threshold[xx-1][yy-1] == 0 and (threshold[xx-2][yy-1]==255 or threshold[xx][yy]==255 or threshold[xx-1][yy-2]==255 or threshold[xx-1][yy]==255) and img_flag[xx-1][yy-1] == 0:findpoint.append((xx-1, yy-1))img_flag[xx-1][yy-1] = countcoutours[count - 1].append([xx-1, yy-1, img_flag[xx-1][yy-1]])#右上if threshold[xx-1][yy+1] == 0 and (threshold[xx-2][yy+1]==255 or threshold[xx][yy+1]==255 or threshold[xx-1][yy]==255 or threshold[xx-1][yy+1]==255) and img_flag[xx-1][yy+1] == 0:findpoint.append((xx-1, yy+1))img_flag[xx-1][yy+1] = countcoutours[count - 1].append([xx-1, yy+1, img_flag[xx-1][yy+1]])#左下if threshold[xx+1][yy-1] == 0 and (threshold[xx][yy-1]==255 or threshold[xx+2][yy-1]==255 or threshold[xx+1][yy-2]==255 or threshold[xx+1][yy]==255) and img_flag[xx+1][yy-1] == 0:findpoint.append((xx+1, yy-1))img_flag[xx+1][yy-1] = countcoutours[count - 1].append([xx+1, yy-1, img_flag[xx+1][yy-1]])#右下if threshold[xx+1][yy+1] == 0 and (threshold[xx][yy+1]==255 or threshold[xx+2][yy+1]==255 or threshold[xx+1][yy]==255 or threshold[xx+1][yy+2]==255) and img_flag[xx+1][yy+1] == 0:findpoint.append((xx+1, yy+1))img_flag[xx+1][yy+1] = countcoutours[count - 1].append([xx+1, yy+1, img_flag[xx+1][yy+1]])return desCoutous

程序的運行時間縮小到0.6秒

c++基于opencv的實現(xiàn)：

#include<opencv2/opencv.hpp> #include<iostream> using namespace std; using namespace cv;Mat findContinousArea(Mat& img) {int rows = img.rows;int cols = img.cols;Mat coutinous = Mat(img.rows,img.cols,img.type(),Scalar(0));queue<Point> tmp;int cCount = 0;for(int i=0;i<rows;i++){for(int j=0;j<cols;j++){if(img.at<char>(i,j)==0 && coutinous.at<char>(i,j)==0){cCount+=30;tmp.push(Point(i,j));coutinous.at<char>(i,j) = cCount;while(!tmp.empty()){Point p = tmp.front();tmp.pop();if(p.x>0 && img.at<char>(p.x-1,p.y)==0 && coutinous.at<char>(p.x-1,p.y)==0){tmp.push(Point(p.x-1,p.y));coutinous.at<char>(p.x-1,p.y) = cCount;}if(p.x<rows && img.at<char>(p.x+1,p.y)==0 && coutinous.at<char>(p.x+1,p.y)==0){tmp.push(Point(p.x+1,p.y));coutinous.at<char>(p.x+1,p.y) = cCount;}if(p.y>0 && img.at<char>(p.x,p.y-1)==0 && coutinous.at<char>(p.x,p.y-1)==0){tmp.push(Point(p.x,p.y-1));coutinous.at<char>(p.x,p.y-1) = cCount;}if(p.y<cols && img.at<char>(p.x,p.y+1)==0 && coutinous.at<char>(p.x,p.y+1)==0){tmp.push(Point(p.x,p.y+1));coutinous.at<char>(p.x,p.y+1) = cCount;}}}}}return coutinous; }int main() {Mat frame = imread("/Users/jinweiliu/Pictures/coutinous.png",0);int rows = frame.rows;int cols = frame.cols;//二值化for(int i=0;i<rows;i++){for(int j=0;j<cols;j++){if((frame.at<char>(i,j)&0xff)<128){frame.at<char>(i,j) = 0;}else{frame.at<char>(i,j) = (char)255;}}}TickMeter tm;tm.start();Mat coutinous = findContinousArea(frame);tm.stop();cout<<tm.getTimeMilli()<<endl;imshow("test",coutinous);waitKey();return 0; }

原圖：

結果如下：

注意，結果是用不同的數(shù)字表示不同的聯(lián)通區(qū)的，比如，第一個聯(lián)通區(qū)全部標志為30，第二個全部標志為60,以此類推。之所以是30，60，90等而不是1，2，3等是因為顯示出來看著比較明顯。這里只是把不同的聯(lián)通區(qū)標示出來，并沒有做切割之類的，僅供參考。最后，c++的這段代碼耗時只有3ms,性能還是挺不錯的。

總結

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