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是否有任何 API 可以实现我自己的 RANSAC 模型?即,OpenCV 是否有一个我可以继承的通用 RANSAC 引擎,或者我可以在哪里编码我自己的观察模型?

如果没有,依赖重用某些 RANSAC OpenCV 代码的最简单方法是什么?

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1 回答 1

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据我所知,但这里的代码中有一个 RANSAC 的实现estimateRigidTransform-

https://github.com/opencv/opencv/blob/master/modules/video/src/lkpyramid.cpp

// RANSAC stuff:
// 1. find the consensus
for( k = 0; k < RANSAC_MAX_ITERS; k++ )
{
    int idx[RANSAC_SIZE0];
    Point2f a[RANSAC_SIZE0];
    Point2f b[RANSAC_SIZE0];
    // choose random 3 non-complanar points from A & B
    for( i = 0; i < RANSAC_SIZE0; i++ )
    {
        for( k1 = 0; k1 < RANSAC_MAX_ITERS; k1++ )
        {
            idx[i] = rng.uniform(0, count);
            for( j = 0; j < i; j++ )
            {
                if( idx[j] == idx[i] )
                    break;
                // check that the points are not very close one each other
                if( fabs(pA[idx[i]].x - pA[idx[j]].x) +
                    fabs(pA[idx[i]].y - pA[idx[j]].y) < FLT_EPSILON )
                    break;
                if( fabs(pB[idx[i]].x - pB[idx[j]].x) +
                    fabs(pB[idx[i]].y - pB[idx[j]].y) < FLT_EPSILON )
                    break;
            }
            if( j < i )
                continue;
            if( i+1 == RANSAC_SIZE0 )
            {
                // additional check for non-complanar vectors
                a[0] = pA[idx[0]];
                a[1] = pA[idx[1]];
                a[2] = pA[idx[2]];
                b[0] = pB[idx[0]];
                b[1] = pB[idx[1]];
                b[2] = pB[idx[2]];
                double dax1 = a[1].x - a[0].x, day1 = a[1].y - a[0].y;
                double dax2 = a[2].x - a[0].x, day2 = a[2].y - a[0].y;
                double dbx1 = b[1].x - b[0].x, dby1 = b[1].y - b[0].y;
                double dbx2 = b[2].x - b[0].x, dby2 = b[2].y - b[0].y;
                const double eps = 0.01;
                if( fabs(dax1*day2 - day1*dax2) < eps*std::sqrt(dax1*dax1+day1*day1)*std::sqrt(dax2*dax2+day2*day2) ||
                    fabs(dbx1*dby2 - dby1*dbx2) < eps*std::sqrt(dbx1*dbx1+dby1*dby1)*std::sqrt(dbx2*dbx2+dby2*dby2) )
                    continue;
            }
            break;
        }
        if( k1 >= RANSAC_MAX_ITERS )
            break;
    }
    if( i < RANSAC_SIZE0 )
        continue;
    // estimate the transformation using 3 points
    getRTMatrix( a, b, 3, M, fullAffine );
    const double* m = M.ptr<double>();
    for( i = 0, good_count = 0; i < count; i++ )
    {
        if( std::abs( m[0]*pA[i].x + m[1]*pA[i].y + m[2] - pB[i].x ) +
            std::abs( m[3]*pA[i].x + m[4]*pA[i].y + m[5] - pB[i].y ) < std::max(brect.width,brect.height)*0.05 )
            good_idx[good_count++] = i;
    }
    if( good_count >= count*RANSAC_GOOD_RATIO )
        break;
}
if( k >= RANSAC_MAX_ITERS )
    return Mat();
于 2016-08-16T18:57:43.950 回答