也许这可以帮助某人。
变量是
img：原始图像
角度：度
规模
dst：目标图像

int width = img.size().width, 
    height = img.size().height;
Mat rot = getRotationMatrix2D(Point2f(0,0), angle, scale)/scale; //scale later
double sinv = rot.at<double>(0,1),
       cosv = rot.at<double>(0,0);
rot.at<double>(1,2) = width*sinv;  //adjust row offset
Size dstSize(width*cosv + height*sinv, width*sinv + height*cosv);
Mat dst;
warpAffine(img, dst, rot, dstSize);
resize(dst, dst, Size(), scale, scale);  //scale now

我意识到您已经找到了其他更快的解决方案（90度旋转应该非常快，并且不需要warpAffine的所有机制），但是我想为遇到此问题的任何人解决黑边问题。

warpAffine还能做什么？指定目标图像的宽度大于其高度，并且仿射变换仅指定旋转（围绕图像中心），而不缩放。那正是它所做的。在任何地方都没有信息告诉warpAffine应该在这些黑色边框中绘制什么，因此将其留为黑色。

直接物理实验：在桌子上放一张纸。在其周围绘制轮廓（这是您指定要使结果与原始形状/大小相同时执行的操作）。现在，将该工作表绕其中心旋转90度。查看表上轮廓所界定的区域。如果它是黑表，它将看起来完全像您的结果。

我找到了不涉及warpAffine() 的解决方案 。

但是在此之前，我需要声明（供将来参考）我的怀疑是正确的，您需要在调用warpAffine()时传递目标的大小：

warpAffine(image, rotated_img, rot_matrix, rotated_img.size());

据我所知，此函数绘制的黑色边框（由偏移产生）似乎是其标准行为。我注意到使用C接口以及使用2.3.1a和2.3.0版本在Mac和Linux上运行的OpenCV的C ++接口。

我最终使用的解决方案比所有这些曲解 要简单得多。您可以使用cv::transpose()和cv::flip()将图像旋转90度。这里是：

Mat src = imread(argv[1], 1);

cv::Mat dst;
cv::transpose(src, dst);
cv::flip(dst, dst, 1);

imwrite("rotated90.jpg", dst);

---- I>

许多人由于偏移等原因在旋转图像或图像块时遇到了问题。因此，我发布了一个解决方案，允许您旋转图像的一个区域（或整个区域）并将其粘贴到另一个图像中或具有该功能计算一切都适合的图像。

// ROTATE p by R
/**
 * Rotate p according to rotation matrix (from getRotationMatrix2D()) R
 * @param R     Rotation matrix from getRotationMatrix2D()
 * @param p     Point2f to rotate
 * @return      Returns rotated coordinates in a Point2f
 */
Point2f rotPoint(const Mat &R, const Point2f &p)
{
    Point2f rp;
    rp.x = (float)(R.at<double>(0,0)*p.x + R.at<double>(0,1)*p.y + R.at<double>(0,2));
    rp.y = (float)(R.at<double>(1,0)*p.x + R.at<double>(1,1)*p.y + R.at<double>(1,2));
    return rp;
}

//COMPUTE THE SIZE NEEDED TO LOSSLESSLY STORE A ROTATED IMAGE
/**
 * Return the size needed to contain bounding box bb when rotated by R
 * @param R     Rotation matrix from getRotationMatrix2D()
 * @param bb    bounding box rectangle to be rotated by R
 * @return      Size of image(width,height) that will compleley contain bb when rotated by R
 */
Size rotatedImageBB(const Mat &R, const Rect &bb)
{
    //Rotate the rectangle coordinates
    vector<Point2f> rp;
    rp.push_back(rotPoint(R,Point2f(bb.x,bb.y)));
    rp.push_back(rotPoint(R,Point2f(bb.x + bb.width,bb.y)));
    rp.push_back(rotPoint(R,Point2f(bb.x + bb.width,bb.y+bb.height)));
    rp.push_back(rotPoint(R,Point2f(bb.x,bb.y+bb.height)));
    //Find float bounding box r
    float x = rp[0].x;
    float y = rp[0].y;
    float left = x, right = x, up = y, down = y;
    for(int i = 1; i<4; ++i)
    {
        x = rp[i].x;
        y = rp[i].y;
        if(left > x) left = x;
        if(right < x) right = x;
        if(up > y) up = y;
        if(down < y) down = y;
    }
    int w = (int)(right - left + 0.5);
    int h = (int)(down - up + 0.5);
    return Size(w,h);
}

/**
 * Rotate region "fromroi" in image "fromI" a total of "angle" degrees and put it in "toI" if toI exists.
 * If toI doesn't exist, create it such that it will hold the entire rotated region. Return toI, rotated imge
 *   This will put the rotated fromroi piece of fromI into the toI image
 *
 * @param fromI     Input image to be rotated
 * @param toI       Output image if provided, (else if &toI = 0, it will create a Mat fill it with the rotated image roi, and return it).
 * @param fromroi   roi region in fromI to be rotated.
 * @param angle     Angle in degrees to rotate
 * @return          Rotated image (you can ignore if you passed in toI
 */
Mat rotateImage(const Mat &fromI, Mat *toI, const Rect &fromroi, double angle)
{
    //CHECK STUFF
    // you should protect against bad parameters here ... omitted ...

    //MAKE OR GET THE "toI" MATRIX
    Point2f cx((float)fromroi.x + (float)fromroi.width/2.0,fromroi.y +
               (float)fromroi.height/2.0);
    Mat R = getRotationMatrix2D(cx,angle,1);
    Mat rotI;
    if(toI)
        rotI = *toI;
    else
    {
        Size rs = rotatedImageBB(R, fromroi);
        rotI.create(rs,fromI.type());
    }

    //ADJUST FOR SHIFTS
    double wdiff = (double)((cx.x - rotI.cols/2.0));
    double hdiff = (double)((cx.y - rotI.rows/2.0));
    R.at<double>(0,2) -= wdiff; //Adjust the rotation point to the middle of the dst image
    R.at<double>(1,2) -= hdiff;

    //ROTATE
    warpAffine(fromI, rotI, R, rotI.size(), INTER_CUBIC, BORDER_CONSTANT, Scalar::all(0)); 

    //& OUT
    return(rotI);
}

我发现的一个问题是，将warpAffine目标图像大小设置为image.size()而不是image.size() rotated_img.size() 。但是，在扭曲之后，它仍然在x和y中转换得太远了...我尝试了完全相同的扭曲

[ 6.123031769111886e-17 1                     163.9999999999999;
 -1                     6.123031769111886e-17 1132;
  0                     0                     1]

从MatC的OpenCV的getRotationMatrix2D中获取，并且运行良好。我开始闻到warpAffine可能存在的错误...