I have a freshly compiled libjpeg version 9 and tried running jpegtran.exe in command line with the arguments:
.\jpegtran.exe -rotate 180 -outfile test_output1.jpg testimg.jpg
testimg.jpg: test_output1.jpg:
As you can see it does rotate the image but it clips it and it's not put together correctly. The usage.txt file that comes with the package isn't totally up to date because I had to use the -outfile switch instead of what it says:
jpegtran uses a command line syntax similar to cjpeg or djpeg. On
Unix-like systems, you say:
jpegtran [switches] [inputfile] >outputfile
On most non-Unix systems, you say:
jpegtran [switches] inputfile outputfile
where both the input and output files are JPEG
files.
To specify the coded JPEG representation used in the output file,
jpegtran accepts a subset of the switches recognized by cjpeg:
-optimize Perform optimization of entropy encoding parameters.
-progressive Create progressive JPEG file.
-arithmetic Use arithmetic coding.
-restart N Emit a JPEG restart marker every N MCU rows, or every N MCU blocks if "B" is attached to the number.
-scans file Use the scan script given in the specified text file.
See the previous discussion of cjpeg for more details about these
switches. If you specify none of these switches, you get a plain
baseline-JPEG output file. The quality setting and so forth are
determined by the input file.
The image can be losslessly transformed by giving one of these
switches:
-flip horizontal Mirror image horizontally (left-right).
-flip vertical Mirror image vertically (top-bottom).
-rotate 90 Rotate image 90 degrees clockwise.
-rotate 180 Rotate image 180 degrees.
-rotate 270 Rotate image 270 degrees clockwise (or 90 ccw).
-transpose Transpose image (across UL-to-LR axis).
-transverse Transverse transpose (across UR-to-LL axis).
Oddly enough (or maybe not), if I execute .\jpegtran.exe -rotate 180 -outfile test_output2.jpg test_output1.jpg I get the original image back without any clipping issues. It's flipping the clipped parts but just not lining it up right with the rest of the image.
test_output2.jpg:
I get the same result by executing jpegtran.exe -rotate 90 twice.
Also, I tried it on a larger .jpg file which resulted in the same issue but the file size was 18KB smaller for the output. I imagine the issue is related to this.
Edit - I also found this blurb which seems to describe the problem:
jpegtran's default behavior when transforming an odd-size image is
designed to preserve exact reversibility and mathematical consistency
of the transformation set. As stated, transpose is able to flip the
entire image area. Horizontal mirroring leaves any partial iMCU
column at the right edge untouched, but is able to flip all rows of
the image. Similarly, vertical mirroring leaves any partial iMCU row
at the bottom edge untouched, but is able to flip all columns. The
other transforms can be built up as sequences of transpose and flip
operations; for consistency, their actions on edge pixels are defined
to be the same as the end result of the corresponding
transpose-and-flip sequence.
The -trim switch works, if you can call it that, and trims out the disorganized data but the image is smaller and lost data.
test_output5.jpg:
Adding the -perfect switch which supposedly stops the above from happening results in this: transformation is not perfect for output and no image.
So is it not possible to losslessly rotate a .jpg? I could, myself, go into paint and reconstruct the original image by simply moving the edge lines into their correct place. Is there a method to do this within libjpeg?
A lossless rotation works with whole DCT blocks contained within the JPEG file. These blocks are always 8x8 or 16x16 pixels (depending on the compression downsampling settings). The file contains a width and height so the extra pixels can be thrown away when the image is decoded, but there's no way to move the clipping from the right/bottom edge to the left/top edge. The software is doing the best it can with an impossible problem.
As you've discovered the way around this problem is to make the width and height evenly divisible by 16. You'll find that images from cameras for example will have this property.
Related
I need some code that will ask names of 2 input files(windows explorer), then compare pixel by pixel. Files must have same specs. If value is same, move pixel to new output file. If value is >, move the > pixel over.
At end save output file as newfile.png
New file must have EXACTLY same specs as input files. These are gray scale images, no color.
Haven’t tried anything . Need help. I’m not coder
I have closely studied the MS documentation on EMF files and from the definitions for the 3 header types I can't see how to convert from logical coords (which the graphics records coords are stored as) to device coords. The header has a Frame part that specifies the page size surrounding (but not necessarily bounding) the composite image in 0.01mm units; and a Bounds part that specifies the actual bounds of the composite image in logical units. And finally there are the Device and Millimeters parts that specify the size of the recording device.
From these there seems no way that calculating the ratio to convert from logical coords to device coords is possible.
I must be missing something simple :-)
Think I sussed it: you use the records:
EMR_SETVIEWPORTEXTEX - device units
EMR_SETVIEWPORTORGEX - (ditto)
EMR_SETWINDOWEXTEX - logical units
EMR_SETWINDOWORGEX - (ditto)
EMR_SETWORLDTRANSFORM
Yes, the Bounds header property is specified as the actual bounds of the composite image (in logical units) but, on investigating Inkscape and Adobe Illustrator created emf's, I find that they do not adhere to this.
After creating your DC (createDC), use getdevicecaps to get the total number of dots (raster lines) available for your DC. Horzres for width, Vertres for height. The dots aren't square. Then after reading your EMF file with getenhmetafile, use getenhmetafileheader to get the header record. You then look at either rclbound or rclframe in the header record. The second rectangle is a multiple of the first rectangle. For emfs created by powerpoint, the top and left is zero in my experience, so you focus on the bottom and right. The ratio of the two is your aspect ratio. You use that ratio to calculate the rectangle in DC units that has the same aspect ratio as rclbound, but likely adds margins all around so your image doesn't go right to the edge of your device. That rectangle, with units that fall within the range provide by vertres and horzres is the third arguement to the playenhmetafile command where your finish up. In sum, you convert from the EMF logical units to the DC logical units by using vertres and horzres (from your DC) combined with the aspect ratio you calculate (from your EMF).
I am still a beginner in coding. I am currently working on a program in C/C++ that is determining pixel position of a defined mark (which is a black circle with white surroundings) in a photo.
I made a mask from the mark and a vector, which contains mask's every pixel value as it's elements (using Magick++ I summed values for Red, Green and Blue). Vector contains aprox. 10 000 values since the mask is 100x100px. I also used threshold functions for simplifying the image.
Than I made a grid, that is doing the same for the picture, where I want to find the coordinates of the mark. It is basically a loop, that is going throught the image and when the program knows pixel values in the grid it immediately compares them with the mask. Main idea is to find lowest difference between the mask and one of the grid positions.
The problem is however that this procedure of evaluating all grids position takes huge amount of time (e.g. the image has 1920x1080px so more than 2 million vectors containing 10 000 values). I decided to cycle the grid not every pixel but for example every 10th column and row, and than for the best corellation from this procedure I selected area where I used every pixel loop. But, this still takes lot of time.
I would like to ask you, if there is some way of improving this method for better (faster) results or this whole idea is not time efficient and I should use different approach.
Thanks for every advice!
Edit: The program will be used for processing multiple images and on all of them the size will be same. This is the picture after threshold, the mark is the big black dot.
Image
The idea that I find interesting is a pyramidal scheme - or progressive refinement: you find the spot at a lower size image then search only a small rectangle in the larger image.
If you reduce your image by 2 in each dimension then you would reduce the time by 4 plus some search effort in the larger image.
This has some problems: the reduction will affect accuracy I expect. You might miss the spot.
You have to cut the sample (template) by the same so you create a half-size template in this case. As you half half half... the template will get blurred into the surrounding objects so it will not be possible to have a valid template; for half size once I guess the dot has a couple of pixels around it.
As you haven't specified a tool or OS, I will choose ImageMagick which is installed on most Linux distros and is available for OSX and Windows. I am just using it at the command-line here but there are C, C++, Python, Perl, PHP, Ruby, Java and .Net bindings available.
I would use a "Connect Components Analysis" or "Blob Analysis" like this:
convert image.png -negate \
-define connected-components:area-threshold=1200 \
-define connected-components:verbose=true \
-connected-components 8 -auto-level result.png
I have inverted your image with -negate because in morphological operations, the foreground is usually white rather than black. I have excluded blobs smaller than 1200 pixels because your circles seem to have a radius of 22 pixels which makes for an area of 1520 pixels (Pi * 22^2).
That gives this output, which means 7 blobs - one per line - with the bounding box and area of each:
Objects (id: bounding-box centroid area mean-color):
0: 1358x1032+0+0 640.8,517.0 1296947 gray(0)
3: 341x350+1017+287 1206.5,468.9 90143 gray(255)
106: 64x424+848+608 892.2,829.3 6854 gray(255)
95: 38x101+44+565 61.5,619.1 2619 gray(255)
49: 17x145+1341+379 1350.3,446.7 2063 gray(0)
64: 43x43+843+443 864.2,464.1 1451 gray(255)
86: 225x11+358+546 484.7,551.9 1379 gray(255)
Note that, as your circle is 42x42 pixels you will be looking for a blob that is square-ish and close to that size - so I am looking at the second to last line. I can draw that in in red on your original image like this:
convert image.png -fill none -stroke red -draw "rectangle 843,443 886,486" result.png
Also, note that as you are looking for a circle, you would expect the area to be pi * r^2 or around 1500 pixels and you can check that in the penultimate column of the output.
That runs in 0.4 seconds on a reasonable spec iMac. Note that you could divide the image into 4 and run each quarter in parallel to speed things up. So, if you do something like this:
#!/bin/bash
# Split image into 4 (maybe should allow 23 pixels overlap)
convert image.png -crop 1x4# tile-%02d.mpc
# Do Blob Analysis on 4 strips in parallel
for f in tile-*mpc; do
convert $f -negate \
-define connected-components:area-threshold=1200 \
-define connected-components:verbose=true \
-connected-components 8 info: &
done
# Wait for all 4 to finish
wait
That runs in around 0.14 seconds.
I read dicom images with ITK using itk::ImageSeriesReader and itk::GDCMImageIO after reading i flip the images with itk::FlipImageFilter (to get right orientation of the images) and convert the itkImageData to vtkImageData using itk::ImageToVTKImageFilter. I visualization images with VTK using vtkResliceImageViewer in QVTKWidget2.
I set:
(vtkResliceImageViewer)m_imageViewer[i]->SetColorWindow(windowWidthTAGvalue[0028|1051]);
(vtkResliceImageViewer)m_imageViewer[i]->SetColorLevel(windowCenterTAGvalue[0028|1050]);
and i set following blac&white LookUpTable:
vtkLookupTable* lutbw = vtkLookupTable::New();
lutbw->SetTableRange(0,1000);
lutbw->SetSaturationRange(0,0);
lutbw->SetHueRange(0,0);
lutbw->SetValueRange(0,1);
lutbw->Build();
And images shown into my software compared with the same images shown into other software are much darker, i can not get the same effect as other DICOM viewers
My software images are right other software image is left also when i use some other LookUpTable in this example Flow i can not get the same effect (2nd row images) my image on right is much darker then other.
What i am missing why my images are darker what can i do? i was research a lot into dicom and ikt/vtk can not find good solution any help is appreciate.
Please check the values for Rescale Slope (0028,1053) and Rescale Intercept(0028,1052) and apply the Modality LUT transformation before applying the Window level.
Your dataset may have VOI LUT Function (0028,1056) attribute value of "SIGMOID" instead of "LINEAR".
I extracted the image data from one of your DICOM file (brain_009.dcm) and looked at the histogram of the image data. It looks like, the minimum value stored in the image is 0 and maximum value is 960 regardless of interpreting the data is signed or unsigned. Also, the Window Width (0028:1051) has an invalid value of “0” and you cannot use that for displaying the image.
So your default display could set the Window Width to 960 and Window Center to half the window width plus the minimum value.
I want to merge 2 images. How can i remove the same area between 2 images?
Can you tell me an algorithm to solve this problem. Thanks.
Two image are screenshoot image. They have the same width and image 1 always above image 2.
When two images have the same width and there is no X-offset at the left side this shouldn't be too difficult.
You should create two vectors of integer and store the CRC of each pixel row in the corresponding vector element. After doing this for both pictures you find the CRC of the first line of the lower image in the first vector. This is the offset in the upper picture. Then you check that all following CRCs from both pictures are identical. If not, you have to look up the next occurrence of the initial CRC in the upper image again.
After checking that the CRCs between both pictures are identical when you apply the offset you can use the bitblit function of your graphics format and build the composite picture.
I haven't come across something similar before but I think the following might work:
Convert both to grey-scale.
Enhance the contrast, the grey box might become white for example and the text would become more black. (This is just to increase the confidence in the next step)
Apply some threshold, converting the pictures to black and white.
afterwards, you could find the similar areas (and thus the offset of overlap) with a good degree of confidence. To find the similar parts, you could harper's method (which is good but I don't know how reliable it would be without the said filtering), or you could apply some DSP operation(s) like convolution.
Hope that helps.
If your images are same width and image 1 is always on top. I don't see how that hard could it be..
Just store the bytes of the last line of image 1.
from the first line to the last of the image 2, make this test :
If the current line of image 2 is not equal to the last line of image 1 -> continue
else -> break the loop
you have to define a new byte container for your new image :
Just store all the lines of image 1 + all the lines of image 2 that start at (the found line + 1).
What would make you sweat here is finding the libraries to manipulate all these data structures. But after a few linkage and documentation digging, you should be able to easily implement that.