I was wondering what the unit is of my boundRect[].tl() output.
topleft = boundRect[largest_contour_index].tl();
My assumption is that it is in pixels.
If so, do I need to look at the pixels of my camera and the format it outputs to calculate the position of my object?
Or do the pixels that the function outputs change due to the fact that OpenCV converts the image to an 8-bit image? I can imagine that the amount of pixels where the image consists of becomes smaller when the image is converted to 8 bit.
Please correct me if I'm wrong.
Thank you!
First of all, the BoundingRect returns x,y coordinates, width and height. you can refer to its documentation: docs.opencv.org/2.4/modules/core/doc/basic_structures.html#rect
second, the 8-bit image conversion was based on pixel value of color and doesn't have a direct relation with pixel count. So converting a 100x100 image to 8-bit image will still be 100x100 px
Related
As we all know,we can use the function cv::getOptimalNewCameraMatrix() with alpha = 1 to get a new CameraMatrix. Then we use the function cv::undistort() with the new CameraMatrix can get the image after dedistortion. However, I find the image after distortion is as large as the original image and some part of the image after distortion covered by black.
So my question is ：Does this mean that the original image pixel is lost? and is there any way to avoid pixel loss or get the image whose size larger than origin image with opencv?
cv::Mat NewKMatrixLeft = cv::getOptimalNewCameraMatrix(KMatrixLeft,DistMatrixLeft ,cv::Size(image.cols,image.rows),1);
cv::undistort(image, show_image, KMatrixLeft, DistMatrixLeft,NewKMatrixLeft);
The size of image and show_image are both 640*480,however from my point of view,the size of image after distortion should be larger than 640*480 because some part of it is meaningless.
Thanks!
In order to correct distortion, you basically have to reverse the process that caused the initial distortion. This implies that pixels are stretched and squashed along various directions to correct the distortion. In some cases, this would move the pixels away from the image edge. In OpenCV, this is handled by inserting black pixels. There is nothing wrong with this approach. You can then choose how to crop it to remove the black pixels at the edges.
I was wondering how to determine the equivalent of RGB values for a grayscale image. The original image is grayscale and everything I have found online is converting an RGB image pixel values to the grayscale pixel values. I already can read in the image. Ideally, this would be for xCode.
I was wondering if there was a class which would do this for me. If so, and you could point me to it, that would be great. I will read on it.
Any help is greatly appreciated.
NOTE: I am a beginner in C++ and do not have time to learn everything formally; I have to learn all of my programming on the fly.
You need more information to transform from a simple Greyscale to RGB, when you do reverse operation, the color information is "lost", as the three channels are set to same value(depending on the algorithm each channel will have a different/same weight in the final color computation).
Digital cameras, usually store more information per pixel, 12 bits per channel in 35mm and 14 bits per channel in medium format (those bits number are the average, some products offer less or even more quality).
Thanks to those additional bits per channel, the camera can compute the "real" color, or what it thinks is the real color based on some parameters.
TL;DR: You can't without more data from your source, in this case the image.
You can convert a gray value to RGB by setting each component of the RGB value to the gray value:
ColorRGB myColorRGB = ColorRGBMake(myGrayValue, myGrayValue, myGrayValue);
I have a matrix (Mat) constituted by double, in the range [0,1].
When I save it by means of command imwrite, the resulting image is totally black.
I suppose the problem is a casting problem, but I don't know how to solve it.
Thanks
The only way for OpenCV to store array of doubles without converting them to other formats (and losing information) is by using FileStorage. imwrite is restricted to arrays of 'char' or 'short'.
You get the totally black image is because all images pixels are within range [0,1] (actually either 0 or 1 when saving to image), which is approaching total black (either for gray-scale image or color image).
To save the matrix to the image with normal color, you need first to transform the double matrix to range [0, 255] by multiplying each value by 255. Remember to transform back if you later load the matrix from this image by dividing each value by 255.
I was reading this paper for a project work using imagmagick and C++.
We train on 1.6 million 32*32 color images that have been preprocessed
by subtracting from each pixel its mean value over all images and then
dividing by the standard deviation of all pixels over all images.
I've trouble distinguishing between "from each pixel its mean value over all images" and "standard deviation of all pixels over all images".
Since, I'm dealing with color images, can I just take rgb values of each pixel as one value or should I calculate the mean and SD for every color differently?
For example if I have r=255, g=255, b=255, can I take pixel value as (in binary), (r<<16)+(g<<8)+b ?
Color channel values should be used independently. If you would use 32 bit representation of the pixels, you would get big value differences between very near colors which differ in red or green channel.
I wrote an simple kinect application where I'm accessing the depth values to detect some objects. I use the following code to get the depth value
depth = NuiDepthPixelToDepth(pBufferRun);
this will give me the depth value for each pixel. Now I want to subselect a region of the image, and get the RGB camera values of this corresponding region.
What I'm not sure about:
do I need to open a color image stream?
or is it enough to just convert the depth into color?
how do I use NuiImageGetColorPixelCoordinateFrameFromDepthPixelFrameAtResolution?
I'm fine with the simplest solution where I have a depth frame and a color frame, so that I can select a ROI with opencv and then crop the color frame accordingly.
do I need to open a color image stream?
Yes. You can get the coordinates in the colour frame without opening the stream, but you won't be able to do anything useful with them because you'll have no colour data to index into!
or is it enough to just convert the depth into color?
There's no meaningful conversion of distance into colour. You need two image streams, and a co-ordinate conversion function.
how do I use NuiImageGetColorPixelCoordinateFrameFromDepthPixelFrameAtResolution?
That's a terribly documented function. Go take a look at NuiImageGetColorPixelCoordinatesFromDepthPixelAtResolution instead, because the function arguments and documentation actually make sense! Depth value and depth (x,y) coordinate in, RGB (x,y) coordinate out. Simple.
To get the RGB data at some given coordinates, you must first grab an RGB frame using NuiImageStreamGetNextFrame to get an INuiFrameTexture instance. Call LockRect on this to get a NUI_LOCKED_RECT. The pBits property of this object is a pointer to the first pixel of the raw XRGB image. This image is stored row wise, in top-to-bottom left-to-right order, with each pixel being represented by 4 sequential bytes representing a padding byte then R, G and B follwing it.
The pixel at position (100, 200) is therefore at
lockedRect->pBits[ ((200 * width * 4) + (100 * 4) ];
and the byte representing the red channel should be at
lockedRect->pBits[ ((200 * width * 4) + (100 * 4) + 1 ];
This is a standard 32bit RGB image format, and the buffer can be freely passed to your image manipulation library of choice... GDI, WIC, OpenCV, IPL, whatever.
(caveat... I'm not totally certain I have the pixel byte ordering correct. I think it is XRGB, but it could be XBGR or BGRX, for example. Testing for which one is actually being returned should be trivial)