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聲明Mat圖像並分配值。單通道OpenCV Mat.at <>給出錯誤值
Mat magnitude = Mat(gradient_columns.cols, gradient_columns.rows, CV_64FC1);
for(int i = 0; i < gradient_columns.cols; i++)
{
for(int j = 0; j < gradient_columns.rows; j++)
{
magnitude.at<double>(Point(i,j)) = (double)hypot(gradient_columns.at<double>(Point(i,j)), gradient_rows.at<double>(Point(i,j)));
}
}
打印上述墊:
cout << "M = " << magnitude << endl;
結果:
M = [0, 0, 0.1257399049164523, 12.36316814720732, 12.50461780753356, 0.2674320707434279, 10.39230484541326, 12.03299037437945, 5.430256687374658,
12.03299037437945, 4.684492386402418, 4.72934192083521, 12.16431633381293, 5.397674732957373, 12.30042244512288, 10.25834261322606, 0.3944487245360109,
12.16431633381293, 11.84297951775486, 12.44187210544911, 12.10132213098092,
0.4088817310696626, 10.15078660267586, 12.09573607646389, 2.076275433221507, 0, 0.1257399049164523, 0, 0.1257399049164523, 0;
.....
.....]
上述結果是完全正確的和預期。
但是,如果我嘗試打印單個值我得到錯誤的結果:
cout.precision(20);
cout << "CHANNELS: " << magnitude.channels() << endl;
cout << magnitude.at<double>(Point(0, 2)) << endl;
cout << magnitude.at<double>(Point(0, 3)) << endl;
cout << magnitude.at<double>(Point(0, 4)) << endl;
cout << magnitude.at<double>(Point(0, 5)) << endl;
Result | Actual Value:
CHANNELS: 1
0.062870 | 0.1257399049164523,
0.000000 | 12.36316814720732,
0.031404 | 12.50461780753356,
0.000000 | 0.2674320707434279
我理解它的一些數據類型轉換問題,但如果任何人都可以提出任何解決辦法嗎?
謝謝。
不是你的問題的答案,而是代替'Mat.at(Point(i,j))',我通常會發現它更清晰地寫出'Mat.at (j,i)' –
Sunreef
'一個額外的評論:爲了獲得更好的性能,你應該交換兩個循環(迭代外部循環中的行) – Micka