coordtransform | Provides conversion between Baidu coordinates
kandi X-RAY | coordtransform Summary
kandi X-RAY | coordtransform Summary
Provides conversion between Baidu coordinates (BD09), National Bureau of Survey coordinates (Mars coordinates, GCJ02), and WGS84 coordinate systems
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QUESTION
I am making an Image viewer application for my own interest. I am using JPanel
to display a bufferedImage
.
Objective - I want to manipulate the image and give user the options to zoom, pan, flip or rotate images.
I am now able to zoom, pan and rotate. But the problem is that it is not the same as what you see when you open a .pdf file in Microsoft Edge.
I have added the code below:
...ANSWER
Answered 2019-Sep-23 at 18:09Here is a simple program that demonstrates what I suggested, namely a large image set as the icon of a JLabel
and the JLabel
is the scrollable client for a JScrollPane
. Note that I set the preferred size of the JScrollPane
to 800 x 600 pixels because the default behavior is to make the JScrollPane
the same size as its scrollable client. The image dimensions are 2312 x 1536 pixels which means it is larger than my computer screen. So if I don't set the preferred size, when you run the program the JFrame
overflows the screen. You don't see all of the JFrame
. This is on purpose so you can see how the image scrolls. The URL for the image is https://unsplash.com/photos/l68Z6eF2peA. And here is the code, which is a MCVE
By the way, I downloaded and saved the image in a file that I named worldmap.jpg
QUESTION
Description :- I am trying to rotate an image without using OpenCV functions in C++. The rotation center need not be the center of the image. It could be a different point (offset from the image center). So far I followed a variety of sources to do image interpolation and I am aware of a source which does the job perfectly in MATLAB. I tried to mimic the same in C++ without OpenCV functions. But I am not getting the expected rotated image. Instead my output appears like a small horizontal line on the screen.
...ANSWER
Answered 2019-Jul-11 at 15:14First, I have to admit I agree with generic_opto_guy:
The approach with the loop looks good, so we would need to check the math. On thing I noticed: if (oldRow > 0 && oldCol > 0 && oldRow <= oldHeight && oldCol <= oldWidth) implies you start indexing with 1. I belife that opencv starts indexing with 0.
For all that, I couldn't resist to answer. (May be, it's just an image phase of mine.)
Instead of fiddling with sin() and cos(), I would recommend to use matrix transformation. At the first glance, this might appear over-engineered but later you will recognize that it bears much more flexibility. With a transformation matrix, you can express a lot of transformations (translation, rotation, scaling, shearing, projection) as well as combining multiple transformations into one matrix.
(A teaser for what is possible: SO: How to paint / deform a QImage in 2D?)
In an image, the pixels may be addressed by 2d coordinates. Hence a 2×2 matrix comes into mind but a 2×2 matrix cannot express translations. For this, homogeneous coordinates has been introduced – a math trick to handle positions and directions in the same space by extending the dimension by one.
To make it short, a 2d position (x, y) has the homogeneous coordinates (x, y, 1).
A position transformed with a transformation matrix:
v´ = M · v.
This may or may not change the value of third component. To convert the homogeneous coordinate to 2D position again, x and y has to be divided by 3rd component.
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