I need to store a float image in OpenCV. Converting it to a CV8U image as suggested by #tomriddle_1234 still stores a black png.
reference.type() = 5
reference.channels() = 1
reference.depth() = 5
How can I convert the image to a 8bit or 16bit so that imwrite can store the image, while maintaining it's float property i.e: the stored image is not "washed out colours" due to conversion/loss of precision!
imshow("5t aligned Mean", reference); //Displays the correct image
//reference.convertTo(reference, CV_8U); //Convert image to 8Bit INCORRECT
reference.convertTo(reference, CV_8U, 255.0, 1/255.0); //Correct image
imwrite(subject.c_str(), reference); //Stores a completely black png
Any suggestions are much appreciated!
You can convert to 16bit by multiplying each float pixel by 2^16-1. Floating point images are stored with values between [0,1] which you want to map to the range [0,2^16-1]
opencv will save 16bit uncompressed in PNG and TIFF with the normal imwrite().
(It will also save them as JPEG although I've had less luck finding things that read 16bit jpeg)
normalize the image before converting between 0 and 255 using CV_NORM_MINMAX
Related
I am trying to convert some bitmap files into custom images (exr, pfm, whatever), and after that, back to bitmap:
CImg<float> image(_T("D:\\Temp\\test.bmp"));
image.normalize(0.0, 1.0);
image.save_exr(_T("D:\\Temp\\test.exr"));
and goes fine (same for .pfm file), I mean the exr file is ok, same for pfm file.
But when this exr, or pfm file I trying to convert back to bitmap:
CImg<float> image;
image.load_exr(_T("D:\\Temp\\test.exr")); // image.load_pfm(_T("D:\\Tempx\\test.pfm"));
image.save_bmp(_T("D:\\Temp\\test2.bmp"));
the result, test2.bmp is black. Complete. Why ? What I am doing wrong ?
Some image formats support saving as float, but most formats save as unsigned 8 bit integer (or uint8), meaning normal image values are from 0 to 255. If you try to save an array that is made up of floats from 0 to 1 into a format that does not support floats, your values will most likely be converted to integers. When you display your image with most image-viewing software, it'll appear entirely black since 0 is black and 1 is almost black.
Most likely when you save your image to bitmap it is trying to convert the values to uint8 but not scaling properly. You can fix this by multiplying normalized values between 0 and 1 by 255. img = int(img*255) or using numpy img = (img*255).astype(np.uint8).
It is also possible that somehow your save function is able to preserve floating point values in the bitmap format. However your image viewing software might not know how to view/display a float image. Perhaps use some imshow function (matplotlib.pyplot can easily display floating point grayscale arrays) between each line of code to check if the arrays are consistent with what you expect them to be.
I wrote a code for watershed segmentation in C API. Now I am converting all those into C++. so, cvsaveimage becomes imwrite. But when I use imwrite ,all i get is a black image.
this is the code:-
Mat img8bit;
Mat img0;
img0 = imread("source.png", 1);
Mat wshed(img0.size(), CV_32S);
wshed.setTo(cv::Scalar::all(0));
////after performing watershed segmentation and
// displaying the watershed image from wshed//
wshed.convertTo(img8bit, CV_32FC3, 255.0);
imwrite("Watershed.png", img8bit);
The original image that I want to save is in wshed. I saw suggestions from the net that we need to convert it to 16 bit or higher so that the imwrite saves it right. Like you see,I tried that. But the wshed image is being displayed correctly when using imshow.The img0 is grey image/black and white while the wshed image is coloured. any help on this?
Edit- I changed the 4th line to
Mat wshed(img0.size(), CV_32FC3);
When calling Mat::convertTo() with a scalar (255 in your case), the values of every matrix item will be multiplied by this scalar value. This will cause all most every result pixel values exceed 255 (i.e. white pixels) except those of 0s where they remain 0 (i.e. black pixels). This is why you will get the black-white pixel in the end.
To make it work, simply change it to:
wshed.convertTo(img8bit, CV_32FC3);
You said:
The original image that I want to save is in wshed. I saw suggestions
from the net that we need to convert it to 16 bit or higher so that
the imwrite saves it right.
If saving the image does not work you should keep in mind that the image data has to be either 8-Bits or 16-Bit unsigned when using the imwrite Function, not 16-Bits or higher.
This is stated in the documentation:
The function imwrite saves the image to the specified file. The image
format is chosen based on the filename extension (see imread() for the
list of extensions). Only 8-bit (or 16-bit unsigned (CV_16U) in case
of PNG, JPEG 2000, and TIFF) single-channel or 3-channel (with ‘BGR’
channel order) images can be saved using this function. If the format,
depth or channel order is different, use Mat::convertTo() , and
cvtColor() to convert it before saving. Or, use the universal
FileStorage I/O functions to save the image to XML or YAML format.
I know this question can be wired for experts, but I want to access pixels of grayscale image in openCV. I am using the following code:
cv::Mat img1 = cv::imread("bauckhage.jpg", CV_LOAD_IMAGE_GRAYSCALE);
Now if I want to print any pixel(say at position 255,255) using img1.at<float>(255,255) I get 0, which is not actually black pixel. I thought the image has been read as 2d Matrix.
Actually I want to do some calculation on each pixel, for that I need to access each pixel explicitly.
Any help or suggestion will be appreciated.
You will have to do this
int x = (int)(img.at<uchar>(j,i))
This is because of the way the image is stored and the way img.at works. Grayscale image is stored in CV_8UC1 format and hence it is basically an array of uchar type. So the return value is a uchar which you typecast into an int to operate on it as ints.
Also, this is similar to your questions:
accessing pixel value of gray scale image in OpenCV
float is the wrong type here.
if you read an image like this:
cv::Mat img1 = cv::imread("bauckhage.jpg", CV_LOAD_IMAGE_GRAYSCALE);
then it's type will be CV_8U, (uchar, a single grayscale byte per pixel). so, to access it:
uchar &pixel = img1.at<uchar>(row,col);
I need to convert a CV_8U image with 3 channels to an image which must be a single channel CV_32S. But when I'm trying to do so, the image I get is all black. I don't understand why my code is not working.
I'm dealing with a grayscale image, this is why I split the 3 channels image into a vector of single channel image, and then process only the first channel.
//markers->Image() returns a valid image, so this is not the problem
cv::Mat dst(markers->Image().size(), CV_32SC1);
dst = cv::Scalar::all(0);
std::vector<cv::Mat> vectmp;
cv::split(markers->Image(), vectmp);
vectmp.at(0).convertTo(dst, CV_32S);
//vectmp.at(0) is ok, but dst is black...?
Thank you in advance.
Have you tried to get values of result image? Like this:
for (int i=0; i<result.rows; i++)
{
for (int j=0; j<result.cols; j++)
{
cout << result.at<int>(i,j) << endl;
}
}
I have converted (also used convertTo) random gray-scale single-channel image to CV_32S (it is a signed 32bit integer value for each pixel) my output was like this:
80
111
132
And when I tried to show it I also get black image. From documentation:
If the image is 16-bit unsigned or 32-bit integer, the pixels are
divided by 256. That is, the value range [0,255*256] is mapped to
[0,255].
So if you divide these small numbers to 255 than you will get 0 (int type). That's why imshow displays black image.
If you want to display your 32-bit image and see a meaningful result, you need to multiply all of its elements by 256 prior to calling imshow. Otherwise, imshow will scale your values down to zero and you will get a black image (as Astor has pointed out).
Since the original values are 8 bit unsigned, they must be less than 255. Therefore multiplying them by 256 is safe and will not overflow a 32-bit integer.
EDIT I just realized your output type is a signed 32-bit integer, but the original type is unsigned 8-bit integer. In that case, you need to scale your values appropriately (have a look at scaleAdd).
Finally, you may want to make sure your image is in YCbCr format before you start throwing away image channels.
I had the same problem, solved it indirectly by trying to convert a 8UC1 to 32S instead of 8UC3.
RgbToGray accept to create a gray image using 8UC3 or 8UC1 element type.
8UC1 image is my marker image.
I've done this in Opencvsharp :
Mat buf3 = new Mat(iplImageMarker);
buf3.ConvertTo(buf3, MatType.CV_32SC1);
iplImageMarker= (IplImage)buf3;
iplImageMarker=iplImageMarker* 256;
I believe this is what you are looking for. Convert your image to this, 8 bit, single channel. CV_8UC1. You are starting with a 8 bit image and changing it to 32 bit single channel? Why? Keep it 8 bit.
i have converted an image from RGB to B/W then i want to convert it back to RGB but i have a problem on that:
my code:
int width=zoomedImage->width;
int height=zoomedImage->height;
TempImage=cvCreateImage(cvSize(width,height),IPL_DEPTH_8U,1);
cvCvtColor(zoomedImage, TempImage,CV_RGB2GRAY);
cvThreshold( TempImage, TempImage,128,256,CV_THRESH_BINARY);
cvCvtColor( TempImage,zoomedImage,CV_GRAY2RGB);
this->pictureBox1->Image=(gcnew
System::Drawing::Bitmap(zoomedImage->width,zoomedImage->height,zoomedImage->widthStep,
System::Drawing::Imaging::PixelFormat::Format24bppRgb,(System::IntPtr)zoomedImage->imageData));
here i'm displaying zoomedImage as a B/W image,in another action i want to display zoomedImage as an RGB image the major problem here is that i can't change the image that will be draw as another parts of my code is depending on this sequence, i wrote that in the other action:
cvCvtColor( TempImage,zoomedImage,CV_GRAY2RGB);
this->pictureBox1->Image=(gcnew
System::Drawing::Bitmap(zoomedImage->width,zoomedImage->height,zoomedImage->widthStep,
System::Drawing::Imaging::PixelFormat::Format24bppRgb,(System::IntPtr)zoomedImage->imageData));
but zoomedImage still dispalyed as B/W, i heared that when a true color image is converted to gray it can't be returned again as a true color image, so what does CV_GRAY2RGB do?
When you convert an RGB image to a gray level image, color information is lost, and this information cannot be recovered fom the gray level image again.
When you try to convert B/W image to RGB you only make a 3 channel image, but all channels contain the same intensity data. Hence you get a gray level image with 3 channels. Nothing more.
i have solved my problem as following:
Convert Original image to B/W
int width=zoomedImage->width;
int height=zoomedImage->height;
ColorSaver=cvCreateImage(cvSize(width,height),zoomedImage->depth,zoomedImage->nChannels);
ColorSaver=cvCloneImage(zoomedImage);
TempImage=cvCreateImage(cvSize(width,height),IPL_DEPTH_8U,1);
cvCvtColor(zoomedImage, TempImage,CV_RGB2GRAY);
cvThreshold( TempImage, TempImage,128,256,CV_THRESH_BINARY);
cvCvtColor( TempImage,zoomedImage,CV_GRAY2RGB);
this->pictureBox1->Image=(gcnew
System::Drawing::Bitmap(zoomedImage->width,zoomedImage->height,zoomedImage->widthStep,
System::Drawing::Imaging::PixelFormat::Format24bppRgb,(System::IntPtr)zoomedImage->imageData));
return back the original image to RGB:
zoomedImage=cvCloneImage(ColorSaver);
this->pictureBox1->Image=(gcnew
System::Drawing::Bitmap( zoomedImage->width, zoomedImage->height, zoomedImage->widthStep,
System::Drawing::Imaging::PixelFormat::Format24bppRgb,(System::IntPtr) zoomedImage->imageData));