交叉邮寄here
假设我有一个这样的二进制图像
Mat img(1000, 2000, CV_8UC1);
randu(img, Scalar(0), Scalar(255));
inRange(img, Scalar(160), Scalar(200), img);
我将使用 connectedComponentsWithStats
来标记二进制 img
Mat labels, stats, centroids;
int counts_img = connectedComponentsWithStats(img, labels, stats, centroids);
我将指定一些要从 labels
中删除的标签。
vector<int> drop_label(counts_img - 3);
iota(drop_label.begin(), drop_label.end(), 1);
当然我可以用下面的方法来做:
Mat img(1000, 2000, CV_8UC1);
randu(img, Scalar(0), Scalar(255));
inRange(img, Scalar(160), Scalar(200), img);
Mat labels, stats, centroids;
int counts_img = connectedComponentsWithStats(img, labels, stats, centroids);
vector<int> drop_label(counts_img - 3);
iota(drop_label.begin(), drop_label.end(), 1);
//start to count the time.
double start_time = (double)getTickCount();
Mat select = img.clone() = 0;
int img_height = img.rows, img_width = img.cols;
for (int i = 0; i < img_height; i++) {
int*plabels = labels.ptr<int>(i);
uchar*pselect = select.ptr<uchar>(i);
for (int j = 0; j < img_width; j++) {
if (find(drop_label.begin(), drop_label.end(), plabels[j]) != drop_label.end())
pselect[j] = 255;
}
}
Mat result = img - select;
//total time
double total_time = ((double)getTickCount() - start_time) / getTickFrequency();
cout << "total time: " << total_time << "s" << endl;
total time: 96.8676s
如您所见,我确实可以做到,而且据我所知,.ptr
是最快的方法。但我不得不说,我无法忍受函数find
耗费了我那么多时间。任何人都可以告诉我最快的方法吗?
最佳答案
正如 CrisLuengo 的评论 here ,我现在将 99s
增加到 0.004s
Mat img(1000, 2000, CV_8UC1);
randu(img, Scalar(0), Scalar(255));
inRange(img, Scalar(160), Scalar(200), img);
Mat labels, stats, centroids;
int counts_img = connectedComponentsWithStats(img, labels, stats, centroids);
vector<int> drop_label(counts_img - 3);
iota(drop_label.begin(), drop_label.end(), 1);
vector<int> replace_table(counts_img);
iota(replace_table.begin(), replace_table.end(), 0);
for (int i : drop_label)
replace_table[i] = 0;
//start to count the time.
double start_time = (double)getTickCount();
Mat select = img.clone() = 0;
int img_height = img.rows, img_width = img.cols;
for (int i = 0; i < img_height; i++) {
int*plabels = labels.ptr<int>(i);
uchar*pselect = select.ptr<uchar>(i);
for (int j = 0; j < img_width; j++) {
if (replace_table[plabels[j]] != 0)
pselect[j] = 255;
}
}
Mat result = img - select;
//total time
double total_time = ((double)getTickCount() - start_time) / getTickFrequency();
cout << "total time: " << total_time << "s" << endl;
关于c++ - 如何以最快的方法删除那些指定的标签组件,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48664698/