What is the fastest method for rotating an image w

问题:

There are some looooong and hungry algorithms for doing so, but as of yet I haven't come up with or found anything particularly fast.

回答1:

The fastest way is to this is to use unsafe calls to manipulate the image memory directly using LockBits. It sounds scary but it's pretty straight forward. If you search for LockBits you'll find plently of examples such as here.

The interesting bit is:

BitmapData originalData = originalBitmap.LockBits(
     new Rectangle(0, 0, originalWidth, originalHeight), 
     ImageLockMode.ReadOnly, 
     PixelFormat.Format32bppRgb);

Once you have the BitmapData you can pass the pixels and map them into a new image (again using LockBits). This is significantly quicker than using the Graphics API.

回答2:

Here's what I ended up doing (after an extensive amount of continued research, and the helpful link provided by TheCodeKing):

public Image RotateImage(Image img, float rotationAngle)
    {
        // When drawing the returned image to a form, modify your points by 
        // (-(img.Width / 2) - 1, -(img.Height / 2) - 1) to draw for actual co-ordinates.

        //create an empty Bitmap image 
        Bitmap bmp = new Bitmap((img.Width * 2), (img.Height *2));

        //turn the Bitmap into a Graphics object
        Graphics gfx = Graphics.FromImage(bmp);

        //set the point system origin to the center of our image
        gfx.TranslateTransform((float)bmp.Width / 2, (float)bmp.Height / 2);

        //now rotate the image
        gfx.RotateTransform(rotationAngle);

        //move the point system origin back to 0,0
        gfx.TranslateTransform(-(float)bmp.Width / 2, -(float)bmp.Height / 2);

        //set the InterpolationMode to HighQualityBicubic so to ensure a high
        //quality image once it is transformed to the specified size
        gfx.InterpolationMode = InterpolationMode.HighQualityBicubic;

        //draw our new image onto the graphics object with its center on the center of rotation
        gfx.DrawImage(img, new PointF((img.Width / 2), (img.Height / 2)));

        //dispose of our Graphics object
        gfx.Dispose();

        //return the image
        return bmp;
    }

Cheers!

回答3:

void Graphics.RotateTransform(float angle);

This should rotate the image in C#. What is it doing instead?

I haven't experimented too much with GDI+. Remember to reverse the rotation after the image is drawn.

回答4:

This answer returns both the offset it should be drawn on and the image which has been rotated.It works by recreating the new image to the size it should be without clipping the angles. Originally written by Hisenburg from the #C# IRC chatroom and Bloodyaugust.

   public static double NormalizeAngle(double angle)
    {
        double division = angle / (Math.PI / 2);
        double fraction = Math.Ceiling(division) - division;

        return (fraction * Math.PI / 2);
    }


    public static Tuple<Image,Size> RotateImage(Image img, double rotationAngle)
    {

        double normalizedRotationAngle = NormalizeAngle(rotationAngle);

        double widthD = img.Width, heightD = img.Height;
        double newWidthD, newHeightD;



        newWidthD = Math.Cos(normalizedRotationAngle) * widthD + Math.Sin(normalizedRotationAngle) * heightD;
        newHeightD = Math.Cos(normalizedRotationAngle) * heightD + Math.Sin(normalizedRotationAngle) * widthD;

        int newWidth, newHeight;
        newWidth = (int)Math.Ceiling(newWidthD);
        newHeight = (int)Math.Ceiling(newHeightD);

        Size offset = new Size((newWidth - img.Width) / 2,(newHeight - img.Height) / 2);

        Bitmap bmp = new Bitmap(newWidth, newHeight);
        Graphics gfx = Graphics.FromImage(bmp);
        //gfx.Clear(Color.Blue);
        gfx.TranslateTransform((float)bmp.Width / 2, (float)bmp.Height / 2);
        gfx.RotateTransform((float)(rotationAngle / Math.PI * 180));
        gfx.TranslateTransform(-(float)bmp.Width / 2, -(float)bmp.Height / 2);
        gfx.InterpolationMode = InterpolationMode.HighQualityBicubic;
        gfx.DrawImage(img, new PointF((bmp.Width / 2 - img.Width / 2), (bmp.Height / 2 - img.Height / 2)));
        gfx.Dispose();  
        return new Tuple<Image,Size>(bmp,offset);
    }

回答5:

System.Drawing.Image imageToRotate = System.Drawing.Image.FromFile(imagePath);
switch (rotationAngle.Value)
{
    case "90":
        imageToRotate.RotateFlip(RotateFlipType.Rotate90FlipNone);
        break;
    case "180":
        imageToRotate.RotateFlip(RotateFlipType.Rotate180FlipNone);
        break;
    case "270":
        imageToRotate.RotateFlip(RotateFlipType.Rotate270FlipNone);
        break;
    default:
        throw new Exception("Rotation angle not supported.");
}
imageToRotate.Save(imagePath, ImageFormat.Jpeg);