I'm using the following code to Transform a small rectangle coordinates to a larger one ie: A rectangle position on a small image to the same position on the larger resolution of the same image
Rectangle ConvertToLargeRect(Rectangle smallRect, Size largeImageSize, Size smallImageSize)
{
double xScale = (double)largeImageSize.Width / smallImageSize.Width;
double yScale = (double)largeImageSize.Height / smallImageSize.Height;
int x = (int)(smallRect.X * xScale + 0.5);
int y = (int)(smallRect.Y * yScale + 0.5);
int right = (int)(smallRect.Right * xScale + 0.5);
int bottom = (int)(smallRect.Bottom * yScale + 0.5);
return new Rectangle(x, y, right - x, bottom - y);
}
But there seems to be a problem with some images.The transformed rectangle coordinates seems to be off the image.
UPDATE:
img.Draw(rect, new Bgr(232, 3, 3), 2);
Rectangle transret= ConvertToLargeRect(rect, orgbitmap.Size, bit.Size);
target = new Bitmap(transret.Width, transret.Height);
using (Graphics g = Graphics.FromImage(target))
{
g.SmoothingMode = SmoothingMode.HighQuality;
g.DrawImage(orgbitmap, new Rectangle(0, 0, target.Width, target.Height),
transret, GraphicsUnit.Pixel);
}
Rectangle Drawn on small resolution Image
{X=190,Y=2,Width=226,Height=286}
Rectangle Transformed into Orginal Large Resolution Image {X=698,Y=7,Width=830,Height=931}
Original Image
First of all, if you resize the shape it shouldn't move position. That's not what one would expect out of enlarging a shape. This means the X,Y point of the top-left corner shouldn't be transformed.
Second, you shouldn't be adding 0.5 manually to operations, that's not a clean way to proceed. Use the ceiling function as suggested by #RezaAghaei
Third, you should not substract X/Y from the height/width, your calculations should be done as width * scale.
Please correct those mistakes, and if it doesn't work I'll update the answer with extra steps.
Related
I'm working on an image viewer in C# WPF that can zoom and crop images. I'd like to be able to make the two work together, but I don't know how to calculate that.
To zoom an image, it is done by modifying TranslateTransform and ScaleTransform X & Y values. Inspired by this answer https://stackoverflow.com/a/6782715/2923736
The crop function is done by calculating TopLeftX, TopRightY, BottomRightX, BottomLeftY that has been drawn on screen.
The size and aspect ratio calculation is done like so:
maxWidth = Math.Min(ScreenWidth, imageWidth);
maxHeight = Math.Min(ScreenHeight, imageHeight);
AspectRatio = Math.Min(maxWidth / imageWidth, maxHeight / imageHeight);
ImageWidth = imageWidth * AspectRatio;
ImageHeight = imageHeight * AspectRatio;
I use this code to calculate the cropped area, but don't know how to put TranslateTransform and ScaleTransform X & Y values into the formula.
internal static Int32Rect? GetCrop()
{
// Contains the dimensions and coordinates of cropped area
var cropArea = CropService.GetCroppedArea();
if (cropArea == null) { return null; }
int x, y, width, height;
x = Convert.ToInt32(cropArea.CroppedRectAbsolute.X / AspectRatio);
y = Convert.ToInt32(cropArea.CroppedRectAbsolute.Y / AspectRatio);
switch (Rotateint) // Degress the image has been rotated by
{
case 0:
case 180:
width = Convert.ToInt32(cropArea.CroppedRectAbsolute.Width / AspectRatio);
height = Convert.ToInt32(cropArea.CroppedRectAbsolute.Height / AspectRatio);
break;
default:
width = Convert.ToInt32(cropArea.CroppedRectAbsolute.Height / AspectRatio);
height = Convert.ToInt32(cropArea.CroppedRectAbsolute.Width / AspectRatio);
break;
}
return new Int32Rect(x, y, width, height);
}
Example of image being cropped, where it has been resized to fit the window and zoomed in by 20%.
I need to calculate that, so it can be saved on the user's hard disk.
I am working on a program that takes a Bitmap and converts it into circular form. The code is as follows:
public static Image CropToCircle(Image srcImage, Color backGround)
{
Image dstImage = new Bitmap(srcImage.Width, srcImage.Height, srcImage.PixelFormat);
Graphics g = Graphics.FromImage(dstImage);
using (Brush br = new SolidBrush(backGround)) {
g.FillRectangle(br, 0, 0, dstImage.Width, dstImage.Height);
}
GraphicsPath path = new GraphicsPath();
path.AddEllipse(0, 0, dstImage.Width, dstImage.Height);
g.SetClip(path);
g.DrawImage(srcImage, 0, 0);
return dstImage;
}
It returns the image in circular shape; however I need to read an image wedge in the form of degrees; that is, the circle has 360 degrees and I am trying to write a function that will accept a degree (e.g. 10) and will return the pixels of the image that fall in 10th degree. Such that entire image will be readable in 1 to 360 degrees.
Since my hint was actually rather misleading, let me make up by giving you a working code:
// collect a list of colors from a bitmap with a cetner c and radius r
List<Color> getColorsByAngle(Bitmap bmp, Point c, int r, float angle)
{
List<Color> colors = new List<Color>();
for (int i = 0; i < r; i++)
{
int x = (int)(Math.Sin(angle / 180f * Math.PI) * i);
int y = (int)(Math.Cos(angle / 180f * Math.PI) * i);
colors.Add(bmp.GetPixel(c.X + x, c.Y + y));
}
return colors;
}
Here it is at work:
(The gif is rather quantized for size..)
Note that
Pixels close to the center will be read multiple time, the center itself even each time
To collect all outer pixels you need to read as many angles as the circumference of the circle has pixels, ie 2 * PI * radius. So for a circle with a radius of 300 pixels you need to step the angle in 360° / (600 * 3.14) or about 0.2°..
Also note the the coordinate systems in GDI and in geometry are not the same, neither in the direction of the axes nor the angles. Adapting this is left for you..
The original version didn't mention a 'wedge area'. To read an area or the whole image simply loop over an angle range in suitable steps!
I want to crop from an image using user-drawn rectangles on a canvas. The rectangles can be moved, re-sized, and rotated.
When the user selects "Get Cropped Image", the area inside the rectangle should be saved in a second image location on the page, which I can do perfectly well, so long as the rectangle is not rotated. (Straight-forward use of CroppedBitmap.) However, when the rectangle is at an angle I do not know how to perform the crop.
This is what I want to do (forgive my poor MS Paint skills):
My questions are:
1) How do I correctly track or calculate the points of the rectangle?
and,
2) Once I have the points, how do I crop the rotated rectangle?
EDIT:
Thanks to user Rotem, I believe that I have the answer to the second question. Using code modified from the following answers: Answer 1, Answer 2, I am seeing good results. Unfortunately, I am still unable to track the correct location points for the rectangle, so I cannot fully test this as of yet.
public static Bitmap CropRotatedRect(Bitmap source, System.Drawing.Rectangle rect, float angle, bool HighQuality)
{
Bitmap result = new Bitmap((int)rect.Width, (int)rect.Height);
using (Graphics g = Graphics.FromImage(result))
{
g.InterpolationMode = HighQuality ? InterpolationMode.HighQualityBicubic : InterpolationMode.Default;
using (Matrix mat = new Matrix())
{
mat.Translate(-rect.Location.X, -rect.Location.Y);
mat.RotateAt(-(angle), rect.Location);
g.Transform = mat;
g.DrawImage(source, new System.Drawing.Point(0, 0));
}
}
return result;
}
EDIT:
The answer to the first point is much easier than I had originally thought. You can always get the top-left corner of the rectangle by calling—
double top = Canvas.GetTop(rect);
double left = Canvas.GetLeft(rect);
You can then calculate the rest of the points using the width and the height—
Point topLeft = new Point(left, top);
Point topRight = new Point(left + rect.Width, top);
Point bottomLeft = new Point(left, top + rect.Height);
Point bottomRight = new Point(left + rect.Width, top + rect.Height);
Point centerPoint = new Point(left + (rect.Width / 2), top + (rect.Height / 2));
If your rectangle is rotated, then you have to translate these points to determine where they truly lie on the canvas—
public Point TranslatePoint(Point center, Point p, double angle)
{
// get the point relative to (0, 0) by subtracting the center of the rotated shape.
Point relToOrig = new Point(p.X - center.X, p.Y - center.Y);
double angleInRadians = angle * Math.PI / 180;
double sinOfA = Math.Sin(angleInRadians);
double cosOfA = Math.Cos(angleInRadians);
Point translatedPoint = new Point(relToOrig.X * cosOfA - relToOrig.Y * sinOfA,
relToOrig.X * sinOfA + relToOrig.Y * cosOfA);
return new Point(translatedPoint.X + center.X, translatedPoint.Y + center.Y);
}
Once you are able to translate the top-left corner, you can use Rotem's cropping method. You can also calculate the position of the rest of the rectangle, so you are able to determine if the rectangle is within the bounds of the image, if it is touching an edge, or any other thing that you might want to do in regards to the position.
I discovered the answer to my own question(s), and made the appropriate edits along the way. Please see above for the answer.
I am trying to fill an image inside a rectangle. I was able to set the image position correctly to the leftmost corner of the rectangle. However the scaling does not work as expected. Any help on this is appreciated. Below is my code. This is a 1290*1990 dimensions image.
Cairo.Rectangle imageRectangle = new Cairo.Rectangle(50, 100, width, height);
ctx.NewPath();
Cairo.ImageSurface imgSurface = new Cairo.ImageSurface("C:/Temp/Image.png");
ctx.SetSource(imgSurface, topLeftPoint); //topLeft is (50,100)
float xScale = (float)imageRectangle.Width / (float)imgSurface.Width;
float yScale = (float)imageRectangle.Height / (float)imgSurface.Height;
//Reposition the image to the rectangle origin
ctx.Translate(imageRectangle.X, imageRectangle.Y);
ctx.Scale(xScale, yScale);
ctx.Paint();
Thanks!!
I found the solution. I was setting the source at the wrong place. Below is the right code
Cairo.Rectangle imageRectangle = new Cairo.Rectangle(50, 100, width, height);
ctx.NewPath();
Cairo.ImageSurface imgSurface = new Cairo.ImageSurface("C:/Temp/Image.png");
float xScale = (float)imageRectangle.Width / (float)imgSurface.Width;
float yScale = (float)imageRectangle.Height / (float)imgSurface.Height;
//Reposition the image to the rectangle origin
ctx.Translate(imageRectangle.X, imageRectangle.Y);
ctx.Scale(xScale, yScale);
ctx.SetSource(imgSurface);
ctx.Paint();
Thanks!
I want to scale an image so that the image is always the size of the screen no matter how it is rotated. Does anyone have any ideas on going about this? By the way I am programing this in C# with xna.
Perhaps something similiar to this, although I'm unsure how you expect to draw the texture. It would be easiest by using triangles and texture wrapping them.
This is how I got the new width and new height after rotating:
Matrix origin = Matrix.CreateTranslation(0, 0, 0);
Matrix scale = Matrix.CreateScale(1f);
Matrix rotation = Matrix.CreateRotationZ(MathHelper.ToRadians(rotate));
Matrix translation = Matrix.CreateTranslation(0, 0, 0);
Vector2 pos1 = Vector2.Transform(new Vector2(Texture.Width / 2, Texture.Height / 2), origin * scale * rotation * origin);
Vector2 pos2 = Vector2.Transform(new Vector2(Texture.Width, Texture.Height), origin * scale * rotation * translation);
int width = (int)Math.Abs(pos2.X - pos1.X) * 2;
int height = (int)Math.Abs(pos2.Y - pos1.Y) * 2;
float scaleX = (graphics.PreferredBackBufferWidth / width);
float scaleY = (graphics.PreferredBackBufferHeight / height);
You will probably figure out the best way to draw this, because an image flipped 45 degrees will look weird drawn on the screen so you probably have to scale it up so it fits the screen but still be rotated. That you left out, an image rotated 180 degrees or 90 degrees should work better.
You can apply a RotateTransform to transform the image, then enclose that in a LayoutTransform to fill the dimensions of the container (the screen in this case).