i have a pixel array containing the values from 0 to 255 ...
i have passed it to my c++ function ...
this pixel array i want to save it to jpeg image file...
how to do it with correct encoding ??
i have converted the array to binary string
and saved it into the file in the below code but it just saves an empty image of 4 byte size ...
FILE *file = fopen("/media/internal/wallpapers/04.jpeg", "w+");
fwrite(binaryStr , 1 , sizeof(binaryStr) ,file );
fclose(file);
thnks
Use libjpeg. Don't try to reimplement jpeg encoding yourself, there are too many ways it can go wrong.
I think you need a JPEG library, like libjpeg.
Independent JPEG Group: http://www.ijg.org/
Info: http://en.wikipedia.org/wiki/Libjpeg
From your description it looks like you have YUV-data that you need to convert to jpeg. Correct? Imagemagick is a very powerful tool that can handle this.
From wikipedias entry on YUV:
Y' values are conventionally shifted and scaled to the range [16, 235] rather than using the full range of [0, 255]. This confusing practice derives from the MPEG standards and explains why 16 is added to Y' and why the Y' coefficients in the basic transform sum to 220 instead of 255. U and V values, which may be positive or negative, are summed with 128 to make them always positive
I.e. 0-255 is not a valid range for YUV-data
It seems that sizeof(binaryStr) is 4. So, you'll need to get a length of the binaryStr, not the sizeof(pointer). And it's more simply to use something already cooked like libjpeg.
Related
is there a way to get the int value of a pixel returned with cimg? I'm in the process of building a basic ASCII art program that converts JPG's to character arrays, and I have the entire utility built out but I cann not find a way to get the unsigned char's converted into the range of ints I need (0-255, although the specifics don't matter so long as its a predictable interval).
Does anyone have any idea how to get a numerical pixel value from a JPG? (library suggestions or anything else are completely welcome)
Here is the pixel output:
\�_b��}�HaX�gNzԴ�����p��-�u�����lqu��Lߐ_"T������{�y�sricX[[TXgZ]`a~�t91960d�BpvJ0kY#uR!BpMWb\W?j"#���dCy2+4?ڽ�TT<Tght%P%y;mhͬ�����8#1�H��)����:4lu���CY|��u&<_��ī��������������ȿF�����LP:����N���-�Q�+�2;E3(�SdRO6��NI16j{#�0((
: pixel data
It's already been converted to black and white, so even accessing the numerical value of one color channel off the cimg would be fine. I just can't seem to get any kind of intelligible/manipulable output from the image, even though the image itself is exactly what i'm looking for.
cast it as an int using (int)img(x,y) and ignore the extra channels
I'm currenty trying to save a cv::Mat of type CV_64FC3 to my harddisk of float values between about 0 and 5. If you want to know why, read below.
I have been looking into saving it in hdr format, but if I just call imwrite("myname.hdr", myCV_64FC3);, which they mention in this tutorial http://www.programmersought.com/article/33831043697/, it outputs a format that if you imread it again, is a UINT8 type and does not contain the float values anymore.
How do you save this correctly?
Why do I want this?
I want to use this to save the standard deviations of my background, which have float values of about 0 to 5, can be a bit higher.
Now I can cast them to CV_8UC3 and be done with it, but it will round my values down. While I want to use it to threshold at x times the standard deviation, I want to keep this precise, since 3 times a standard deviation of 0.7, is something else than zero.
Does anyone know a way to save this?
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 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.
I have a really weird error,
so I'm trying to read a pgm image by loading its pixel values into an array, I was able to correctly read in its version, height, width, and maximum possible pixel value. However, when I start reading the pixel values, I always get 0. (I know it's not zero because I can read it using imread in matlab, but have to implement it in c++, plus I couldn't use the opencv library so..)
And besides, when I read the pgm file in like NotePad++, the first few lines are good representing the information about this image ,how ever, the actual pixel values are not readable. I'm wondering if I need some sort of parsing to read a pgm image? Its version is p5.
Thanks!
You must have an assignment to solve as there is no sane reason implementing a PGM reader otherwise.
There are two different PGM formats: ASCII and binary. You seem to expect an ASCII PGM but the one you have is binary.
Have a look at the specs: http://netpbm.sourceforge.net/doc/pgm.html
It says:
/1. A "magic number" for identifying the file type. A pgm image's
magic number is the two characters "P5".
[…]
/9. A raster of Height rows, in order from top to bottom. Each row
consists of Width gray values, in order from left to right. Each gray
value is a number from 0 through Maxval, with 0 being black and Maxval
being white. Each gray value is represented in pure binary by either
1 or 2 bytes. If the Maxval is less than 256, it is 1 byte.
Otherwise, it is 2 bytes. The most significant byte is first.
The format you are expecting is described further down below as the Plain PGM format. Its magic number is "P2".