本文介绍了如何从OpenCV的SiftDescriptorExtractor转换描述符值？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 29岁程序员，3月因学历无情被辞！ 我有一个关于SiftDescriptorExtractor作业最后一部分的问题， 我正在做以下操作： SiftDescriptorExtractor extractor; Mat descriptors_object; extractor.compute（img_object，keypoints_object，descriptors_object）; 现在我想检查一个descriptors_object对象的元素： std :: cout< descriptors_object.row（1）<< std :: endl; 输出如下所示： [0,0,0,0,0,0,0,0,0,0,0,0,0,0,3,3,0,0,0,0,0， 0，0，0，0，0，0，0，0，0，0，0，0，0， 51，154，20，0，0，0，0，0,154,154,1,2,1,0,0,05,1,48,18,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0， 2，154,61,0,0,0,0,5,60,154,30,0,0,0,0,34,70,6,15,3,2,1,0,14,16,18,19,20,21,22,23,24,25,26， 2,0,0,0,0,05,04,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 6，1，0，1，0，0，0] 但在 Lowe纸说：因此，来自特征向量的数字应不大于0.2的值。 问题是，如何在Mat对象中转换这些值？ p> 解决方案 否。该论文说SIFT描述符是： 规范化（使用L2范数） code> 0.2 作为阈值（即循环正常化值并在适当时截断） 再次规范 因此，在理论上，任何SIFT描述符组件都在 [0，1] 之间，问题是，这些值是如何在Mat对象中转换的？ $ b（） 将它们从浮点值转换为 unsigned char -s。 这里是来自OpenCV的相关部分 modules / nonfree / src / sift.cpp calcSIFTDescriptor 方法： float nrm2 = 0; len = d * d * n; for（k = 0; k nrm2 + = dst [k] * dst [k] float thr = std :: sqrt（nrm2）* SIFT_DESCR_MAG_THR; for（i = 0，nrm2 = 0; i { float val = std :: min（dst [i]，thr） dst [i] = val; nrm2 + = val * val; } nrm2 = SIFT_INT_DESCR_FCTR / std :: max（std :: sqrt（nrm2），FLT_EPSILON）; for（k = 0; k { dst [k] = saturate_cast< uchar>（dst [k] * nrm2） } 具有： static const float SIFT_INT_DESCR_FCTR = 512.f;这是因为经典的SIFT实现将归一化的浮点值量化为 unsigned char（）。 整数乘以512乘法因子，这等效于考虑任何SIFT分量在 [0,1 / 2] 之间变化，松散精度试图编码完整的 [0，1] 范围。 I have a question about the last part of the SiftDescriptorExtractor job,I'm doing the following: SiftDescriptorExtractor extractor; Mat descriptors_object; extractor.compute( img_object, keypoints_object, descriptors_object );Now I want to check the elements of a descriptors_object Mat object:std::cout<< descriptors_object.row(1) << std::endl;output looks like:[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 0, 0, 0, 0, 0, 0, 32, 15, 0, 0, 0, 0, 0, 0, 73, 33, 11, 0, 0, 0, 0, 0, 0, 5, 114, 1, 0, 0, 0, 0, 51, 154, 20, 0, 0, 0, 0, 0, 154, 154, 1, 2, 1, 0, 0, 0, 154, 148, 18, 1, 0, 0, 0, 0, 0, 2, 154, 61, 0, 0, 0, 0, 5, 60, 154, 30, 0, 0, 0, 0, 34, 70, 6, 15, 3, 2, 1, 0, 14, 16, 2, 0, 0, 0, 0, 0, 0, 0, 154, 84, 0, 0, 0, 0, 0, 0, 154, 64, 0, 0, 0, 0, 0, 0, 6, 6, 1, 0, 1, 0, 0, 0]But in Lowe paper it is stated that:So the numbers from the feature vector should be no larger than 0.2 value.The question is, how these values have been converted in a Mat object? 解决方案 No. The paper says that SIFT descriptors are:normalized (with L2 norm)truncated using 0.2 as a threshold (i.e. loop over the normalized values and truncate when appropriate)normalized againSo in theory any SIFT descriptor component is between [0, 1], even though in practice the effective range observed is smaller (see below).They are converted from floating-point values to unsigned char-s.Here's the related section from OpenCV modules/nonfree/src/sift.cpp calcSIFTDescriptor method:float nrm2 = 0;len = d*d*n;for( k = 0; k < len; k++ ) nrm2 += dst[k]*dst[k];float thr = std::sqrt(nrm2)*SIFT_DESCR_MAG_THR;for( i = 0, nrm2 = 0; i < k; i++ ){ float val = std::min(dst[i], thr); dst[i] = val; nrm2 += val*val;}nrm2 = SIFT_INT_DESCR_FCTR/std::max(std::sqrt(nrm2), FLT_EPSILON);for( k = 0; k < len; k++ ){ dst[k] = saturate_cast<uchar>(dst[k]*nrm2);}With:static const float SIFT_INT_DESCR_FCTR = 512.f;This is because classical SIFT implementations quantize the normalized floating point values into unsigned char integer through a 512 multiplying factor, which is equivalent to consider that any SIFT component varies between [0, 1/2], and thus avoid to loose precision trying to encode the full [0, 1] range. 这篇关于如何从OpenCV的SiftDescriptorExtractor转换描述符值？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！