I have one image cv::Mat fooImage 1000*1000 pixels in CV_32F format.
Now I want to show the image, I use
fooImage.convertTo(displayImage,CV_8UC1)
However, it takes about 5ms just for this line. Is this normal?? How can I quickly convert a CV_32F Mat image to CV_8UC1?
Thanks!
That sounds slow but convertTo() probably isn't particularly optomised to use SSE2 or anything.
You are reading 4Mb from RAM, allocating 1Mb, doing 4Million floating point ops and writing 1Mb back to RAM so a millisecond isn't unreasonable.
You could write a simple loop to convert the image data into uchar yourself by simply multiplying each value by 255.0
Are you including the time to display the image? YOu are creating a "displayImage" that is still 8bit greyscale, this will have to be converted into a RGB or RGBA image when it is displayed
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 came across this code:
image.convertTo(temp_image,CV_16SC3);
I saw the description of the convertTo() function from here, but what confuses me is image. How can we read the above code? What would be the relation between image and temp_image?
Thanks.
The other answers here are correct, but lack some details. Let me try.
image.convertTo(temp_image,CV_16SC3);
You have a source image image, and a destination image temp_image. You didn't specify the type of image, but probably is CV_8UC3 or CV_32FC3, i.e. a 3 channel image (since convertTo doesn't change the number of channels), where each channel has depth 8 bit (unsigned char, CV_8UC3) or 32 bit (float, CV_32FC3).
This line of code will change the depth of each channel, so that temp_image has each channel of depth 16 bit (short). Specifically it's a signed short, since the type specifier has the S: CV_16SC3.
Note that if you are narrowing down the depth, as in the case from float to signed short, then saturate_cast will make sure that all the values in temp_image will be in the correct range, i.e. in [–32768, 32767] for signed short.
Why you need to change the depth of an image?
Some OpenCV functions require input images with a specific depth.
You need a matrix to contain a different range of values. E.g. if you need to sum (or subtract) some images CV_8UC3 (tipically BGR images), you'd better store the result in a CV_16SC3 or you'll probably get wrong results due to saturations, since the range for CV_8U images is in [0,255]
You read with imread, or want to store with imwrite images with 16bit depth. This are usually used (AFAIK) in medical or graphics application to allow a wider range of colors. However, most monitors do not support 16bit image visualization.
There may be other cases, let me know if I miss the one important to you.
An image is a matrix of pixel information (i.e. a 1080p image will be a 1,920 × 1,080 matrix where each entry contains rbg values for that pixel). All you are doing is reformatting that matrix (each pixel entry, iteratively) into a new type (CV_16SC3) so it can be read by different programs.
The temp_image is a new matrix of pixel information based off of image formatted into CV_16SC3.
The first one is a source, the second one - destination. So, it takes image, converts it into type CV_16SC3 and stores in temp_image.
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 float matrix of 1024x1024 and I want to keep sign of this matrix inside a file. For this purpose, I want to keep the sign matrix as Matrix of boolean which I fail to do.
Assume, my matrix is:
2.312, 0.232, -2,132
5.754, -4,34, -3.23
-4.34, -1.23, 7.9453
My output should be
1,1,0
1,0,0
0,0,1
Since float is 4Byte and my matrix size is 10^20(1M) the size is 4MB and boolean is 1bit and matrix size is 1M, I expect the bool mat to be around 1Mb=128KB however, when I use threshold method in opencv my output file is 1MB which means the file is saved as uchar(8bit).
I tried to use imwrite but it didn't work.
EDIT: I realized that I didn't mention speed is also another important factor for my tests. I'm loading approximately 10 million of 1K*1K matrix from disk.
Thanks in advance
In OpenCV you can write
Mat input;
Mat A = (input >= 0);
Now the problem is that OpenCV has no bitmap data type. So the best you can get is Mat1u (unsigned char).
If you want to save space in your storage, you need to do it on your own. For example, you can use libpng to write out a PNG file of bit depth 1. Unfortunately, imwrite does not support setting that bit depth (it can write PNGs with bit depths 8 and 16).
If you want to write a compressed PNG with bitdepth 8, you can use imwrite:
std::vector<int> flags;
flags.push_back(CV_IMWRITE_PNG_COMPRESSION);
flags.push_back(9); // [0-9] 9 being max compression, default is 3
cv::imwrite("output.png", A, flags);
This will result in the best compression effort. Now you can use Imagemagick to compare the filesize against the same image stored with bit depth 1:
convert output.png -type Bilevel -define "png:bit-depth=1" -define "png:compression-level=9" output-1b.png
I tested with a random example image (see below).
8 bit, compressed PNG: 24,732 bytes
1 bit, compressed PNG: 20,529 bytes
8 bit, uncompressed PGM: 270,015 bytes
1 bit, uncompressed PBM: 34,211 bytes
As you can see, a compressed 8bit storage still beats uncompressed 1bit storage in this example.