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The difference is that with your method, if you imagine the histogram, you're simply going to stretch it to span from 0 to 255, but its shape will be preserved. Histogram equalization not only stretches your histogram, but also tries to make it flat, so that you get an approximately even distribution of pixels of every shade of gray.
In terms of why one is ...
7
For you questions:
1. After applying gaussian filter on a histogram, the pixel value of new histogram will be changed.
2. The sum of pixels in new histogram is almost impossible to remain unchanged.
Visually speaking, after your applying the gaussian filter (low pass), the histogram shall become more smooth than before. Thus, the new histogram is ...
4
To answer your question histogram equalization is called like this because its function is to produce an equalized histogram (that is an uniform probability density function).
There are different algorithms that may approach this function, and obviously there is a problem in the example that is shown:
In fact, the algorithm used there will always have ...
4
I'm not sure if this is the best answer to your question out there, but I would disagree that your 256-bin histogram isn't equalized. Think of it this way.
Each bin represents the count of pixels of a certain intensity. When you're performing histogram equalization, you're moving that bin to a new intensity value, according to the empirical distribution (i....
4
First you'll have to compute the histogram of one of the two images.
H = hist(img(:), bins);
Next find the cdf of the image:
cdf = [0, cumsum(Hmod)/sum(Hmod)];
Next you'll have to make the second image follow the exact same cdf of the first image. This is usually known as histogram specification.
Here is the algorithm:
Say you have a 3 bit image, that ...
3
If I understood right, you are stuck in matching a given histogram into a desired one and creating a new image from this matched histogram obtained by your filtering method.
I would first suggest you to get rid of all the unnecessary stuff (including the python code) above and isolate your problem as "histogram matching" in mathematical terms.
Now I will ...
3
From the paint.NET manual, the parameter in the middle is a gamma correction, which can be used to enhance the contrast in the dark tones or high tones. The gamma-curve is the simplest non-linear level transfer curve. A more sophisticated non-linear technique for automatic level adjustment is histogram equalization, which consists in applying a monotonic, ...
3
Are you looking for a specific function to perform this task? Otherwise this could probably be done with a simple for loop.
Determine your unique rows using the function
bins = unique(A, 'rows');
Then just loop through each row of A and count each instance of each bin you find in A.
This is probably good enough if you dont care about how long the ...
2
Human eye perception is a complicated thing, the human perception tend to fail in different situations even in large changes of gray tones, also is capable of note minimal changes in the gray levels but all depends on the shape of the figure rather than its histogram.
In some cases you can enhance an image by applying histogram equalization techniques but ...
2
In your example, $x_3[n]$ is the sum of two iid random variables, $x_1[n]$ and $x_1[n-1]$. Since each is taken from the uniform distribution, the probability density function (pdf) of each is:
$$
f_{x_1}(x) = \begin{cases}
\frac{1}{b-a}, & x \in [a,b] \\
0, & \text{otherwise}
\end{cases}
$$
This is just a rectangle function that is scaled and ...
2
In step 8 of the reference you provide, you obtain 4 possible values for a given pixel (based on the values obtained from its 4 neighbours, which I assume are the usual North-East-South_West neighbours).
Then, to avoid introducing tile artifacts, you can use any sufficiently regular interpolation method to obtain the value in the center of these 4 pixels.
...
2
If I hand you a uniform distribution spanning say, 0 to 1, then I can say that the probability of a variable with this uniform distribution taking on value of say, 0.3 is, equal to the probability that it takes on a value of 0.7, or 0.01, or 0.93, etc.
In other words, since the uniform distribution is flat, all probability values are equal.
Now forget ...
2
To make a signal with that distribution have a truly uniform distribution would probably require that the transformation used be time-varying: a value $10$ at time $t_0$ would be moved to, say, a value of $15$, but a value $10$ at time $t_1$ (later) would be moved to, for example, $20$.
So, what the equalizer tries to do is to make the cumulative ...
2
I don't think you have any choice other than to use the same number of bins for each observation. Otherwise not only will you not be able to average the histograms, you will also not be able to compare them.
And you definitely need to change the histogram slowly, i. e.
$$h = (1 - \alpha)h + \alpha h_{obs}$$ where $h$ is your "moving average" histogram, $h_{...
2
Yes it is true. A simple histogram equalization may fail for certain image intensity distributions. In your case there are some bins with excessive number of pixels. For other methods, try for example gamma and contrast corrections.
2
This is a wierd thing you are trying to do. What i the purpose behind this signal generation? What particular application needs this?
I think of two ways, but maybe one is not correct.
1 - Generate Uniform White noise, and then filter it to your desired bandwith of interest. But that will surely affect the probability distribution of the amplitudes and ...
2
I don't see any other way than to adapt one of the histograms such that both histograms have the same number of bins before performing histogram intersection. This will always involve some guesswork since the histograms themselves don't give you enough information to do this accurately. I would adapt the histogram with the larger number of bins to the number ...
2
As pointed out by Scott in comment, the quick'n dirty way of doing it is to just multiply by 255 and taking the integer part of the result.
The actual answer is (unfortunately, it's a very common answer to many Image Processing tasks...) "it depends":
float images are usually assumed to have values between 0 and 1, but maybe it's not the case of your ...
2
You should check the Envi documentation (a standard tool for working with multi/hyper spectral images) on the stretches it uses to fit multi/hyperspectral data into 8 bits for display.
Here is the page on some of the stretches used, which apply an affine function to the pixel values and clip the values to the lowest and highest displayable value. In ...
2
Assuming that in your question $T(r)$ denotes a map $:\mathbb{R}\to[0,1]$ and assuming that it is continuous and strictly increasing, then it is obviously an isomorphism among $r\in[0,1]$ onto its image.
As a consequence you get the following result: let $r$ be any value in $[0,1)$, and let $a=T(r)$, then for any $\rho\in[r,r+dr]$ it is $\alpha = T(\rho) \in ...
2
I think random sampling approach seems to be not effective, since the statistical population (pixel intensities) distribution in images is heavily localized. There might be more scientific approaches, but if you need only a rough near estimation I'd like to suggest a method. Here is how:
Blur the image with a low-pass mask (average out neighboring pixels) [...
2
What you are looking for called Otsu Method / Otsu Thresholding.
It solves an optimization problem to set the optimal threshold between two modal distribution.
2
The expectation of information is called entropy. The loss of information can hence be understood as difference in entropy between source and processed image, assuming no random effect was added.
Together with my answer on why contrast is not an appropriate measure of entropy, this gives us the simple answer:
The loss in information is simply the number ...
2
Brightness. You can find more detailed information in here:
Bi-Histogram Equalization with Brightness Preservation Using Contras
Enhancement
2
An easy way, especially one that is used in the detection and SNR estimation of PSK signals in a noisy signal, is based on stochastic moments of the received amplitude:
The kurtosis can be used as measure of how little "centered" a symmetric distribution is.
You could calculate a sample kurtosis frequency bin-wise in your spectrogram; those bins with high ...
2
If the test is just to test the hypothesis that there is noise vs. the hypothesis that there is noise plus some bimodal signal, then I would compare the mean absolute value of the signal, or some higher-order even moment (i.e. 4, 6 or 8) vs. its standard deviation (assuming it's zero mean).
For high SNR, that should give a pretty distinct prediction, ...
1
There are a few answers to a similar question over on Cross Validated.SE.
One suggested answer is to use Hartigan's dip test. Another is to use the mixtools package.
I've simulated some example data in R and used the diptest package and the mixtools package. The diagram below shows the raw data in the top to graphs, and the estimated underlying ...
1
First, I personally wouldn't recommend you using background subtraction in general for tracking applications.
There are way better methods like L1-minimzation of lasso problems, Kenalized Correlation Filters, Multiple Instance learnng, Struck, and whole bunch of stuff.
But if you must use background subtraction,, here are my thoughts:
You could obviously ...
1
Since the Histogram Equalization isn't Linear Operator the order does matter and the result of the doing one before the other changes the final result.
Sharpening would work best when all the Dynamic Range is used.
Hence I'd first maximize the Dynamic Range by Histogram Equalization and then would apply the sharpening.
You might also think of the noise when ...
1
If the noise is a constant background/remains in the same place, you could take various photos in the absence of any light to create an image for background subtraction. My image processing expertise is limited to grayscale but i would imagine the same could be done for RGB. Produce several background images, from that make an average/low pass filter. Then ...
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