cv :: Mat矩阵,如何减少cv :: Mat中小数点右边的数字？

Question

我有一个应用程序在iPhone屏幕上打印3x3 cv :: Mat.我需要减少小数,因为屏幕不是那么大,请参阅:

[1.004596557012473,-0.003116992336797859,5.936915104939593; -0.007241746117066327,0.99973985665720294,-0.2118670500989478; 1.477734234970711e-05,-1.03363734495053e-05,1.00000089074805124]

所以我想减少小数.4或.6或六位小数.有任何想法吗？

干杯

Answer 1

直接OpenCV版本

根据需要选择基本格式化程序,根据
http://docs.opencv.org/3.1.0/d3/da1/classcv_1_1Formatter.html
然后调整函数set64fPrecision(或set32fPrecision)
然后cout:

 cv::Ptr<cv::Formatter> formatMat=Formatter::get(cv::Formatter::FMT_DEFAULT);
 formatMat->set64fPrecision(3);
 formatMat->set32fPrecision(3);
 std::cout << "matrix:" << std::endl << formatMat->format( sim ) << std::endl;

Answer 2

如果您正在使用printf

cv::Mat data(3, 3, CV_64FC1);
for (int y = 0; y < data.rows; ++y) {
  for (int x = 0;x < data.cols; ++x) {
    printf("%.6f ", data.at<double>(y, x));
  }
}

如果您正在使用std::cout

cv::Mat data(3, 3, CV_64FC1);
std::cout.setf(std::ios::fixed, std:: ios::floatfield);
std::cout.precision(6);
for (int y = 0; y < data.rows; ++y) {
  for (int x = 0;x < data.cols; ++x) {
    std::cout<<data.at<double>(y, x)<<" ";
  }
}