I'm reading an image segmentation paper in which the problem is approached using the paradigm "signal separation", the idea that a signal (in this case, an image) is composed of several signals (objects in the image) as well as noise, and the task is to separate out the signals (segment the image).
The output of the algorithm is a matrix, which represents a segmentation of an image into M components. T is the total number of pixels in the image, is the value of the source component (/signal/object) i at pixel j
In the paper I'm reading, the authors wish to select a component m for which matches certain smoothness and entropy criteria. But I'm failing to understand what entropy is in this case.
Entropy is defined as the following:
and they say that '' are probabilities associated with the bins of the histogram of ''
The target component is a tumor and the paper reads: "the tumor related component with "almost" constant values is expected to have the lowest value of entropy."
But what does low entropy mean in this context? What does each bin represent? What does a vector with low entropy look like?
link to paper
They are talking about Shannon's entropy. One way to view entropy is to relate it to the amount of uncertainty about an event associated with a given probability distribution. Entropy can serve as a measure of 'disorder'. As the level of disorder rises, the entropy rises and events become less predictable.
Back to the definition of entropy in the paper:
H(s_m) is the entropy of the random variable s_m. Here is the probability that outcome s_m happens. m are all the possible outcomes. The probability density p_n is calculated using the gray level histogram, that is the reason why the sum runs from 1 to 256. The bins represent possible states.
So what does this mean? In image processing entropy might be used to classify textures, a certain texture might have a certain entropy as certain patterns repeat themselves in approximately certain ways. In the context of the paper low entropy (H(s_m) means low disorder, low variance within the component m. A component with low entropy is more homogenous than a component with high entropy, which they use in combination with the smoothness criterion to classify the components.
Another way of looking at entropy is to view it as the measure of information content. A vector with relatively 'low' entropy is a vector with relatively low information content. It might be [0 1 0 1 1 1 0]. A vector with relatively 'high' entropy is a vector with relatively high information content. It might be [0 242 124 222 149 13].
It's a fascinating and complex subject which really can't be summarised in one post.
Entropy was introduced by Shanon (1948), were the higher value of Entropy = more detailed information.
Entropy is a measure of image information content, which is interpreted as the average uncertainty of information source.
In Image, Entropy is defined as corresponding states of intensity level which individual pixels can adapt.
It is used in the quantitative analysis and evaluation image details, the entropy value is used as it provides better comparison of the image details.
Perhaps, another way to think about entropy and information content in an image is to consider how much an image can be compressed. Independent of the compression scheme (run length encoding being one of many), you can imagine a simple image having little information (low entropy) can be encoded with fewer bytes of data while completely random images (like white noise) cannot be compressed much, if at all.
Related
I have built a FCN for image segmentation. The object to be segmented is only very few pixels relatively to the image size (1024x1024). This results in that the accuracy is very high, even if I only train with 10 images instead of 18000 (my full training set).
My approach to solve this is to use some kind of weighted accuracy, so that the accuracy actually say something about the performance of identifying the small object (now it gets high accuracy since so many pixels are not the object and by not classifying anything the accuracy still gets high).
How do I decide the weight, anybody with some experience?
As you wrote, use a custom weight function which penalizes misclassification of underrepresented pixels more. You can get the weight by calculating the quotient between the number of object pixels versus all of the pixels in the image, or you can try it by hand - just make sure you follow the metrics which tell you the accuracy of object pixels. Hope it helps.
You can use infogain loss layer for a "weighted" loss.
The infogain loss is a generalization of the cross entropy loss commonly used. It is defined using a weight matrix H (of size L-by-L, where L is the number of classes):
L(p) = -H log(p)
Where p is a vector of class probabilities.
You can find more details on this loss here.
I have input (r,c) in range (0, 1] as the coordinate of a pixel of an image and its color 1 or 2 only.
I have about 6,400 pixels.
My attempt of fitting X=(r,c) and y=color was a failure the accuracy won't go higher than 70%.
Here's the image:
The first is the actual image, the 2nd is the image I use to train on, it has only 2 colors. The last is the image that the neural network generated with about 500 weights training with 50 iterations. Input Layer is 2, one hidden layer of size 100, and the output layer is 2. (for binary classification like this, I may need only one output layer but I am just preparing for multi-class classification)
The classifier failed to fit the training set, why is that? I tried generating high polynomial terms of those 2 features but it doesn't help. I tried using Gaussian kernel and random 20-100 landmarks on the picture to add more features, also got similar output. I tried using logistic regressions, doesn't help.
Please help me increase the accuracy.
Here's the input:input.txt (you can load it into Octave the variable is coordinate (r,c features) and idx (color)
You can try plotting it first to make sure that you understand the input then try training on it and tell me if you get better result.
Your problem is hard to model. You are trying to fit function from R^2 to R, which has lots of complexity - lots of "spikes", lots of discontinuous regions (pixels that are completely separated from the rest). This is not an easy problem, and not usefull one.. In order to overfit your network to such setting you will need plenty of hidden units. Thus, what are the options to do so?
General things that are missing in the question, and are important
Your output variable should be {0, 1} if you are fitting your network through cross entropy cost (log likelihood), which you should use for classification.
50 iteraions (if you are talking about some mini-batch iteraions) is orders of magnitude to small, unless you mean 50 epochs (iterations over whole training set).
Actual things, that will probably need to be done (at least one of the below):
I assume that you are using ReLU activations (or Tanh, hard to say looking at the output) - you can instead use RBF activations, and increase number of hidden neurons to ~5000,
If you do not want to go with RBFs, then you will need 1-2 additional hidden layers to fit function of this complexity. Try architecture of type 100-100-100 instaed.
If the above fails - increase number of hidden units, that's all you need - enough capacity.
In general: neural networks are not designed for working with low dimensional datasets. This is nice example from the web, that you can learn pix-pos to color mapping, but it is completely artificial and seems to actually harm people intuitions.
I am using Linear regression to predict data. But, I am getting totally contrasting results when I Normalize (Vs) Standardize variables.
Normalization = x -xmin/ xmax – xmin
Zero Score Standardization = x - xmean/ xstd
a) Also, when to Normalize (Vs) Standardize ?
b) How Normalization affects Linear Regression?
c) Is it okay if I don't normalize all the attributes/lables in the linear regression?
Thanks,
Santosh
Note that the results might not necessarily be so different. You might simply need different hyperparameters for the two options to give similar results.
The ideal thing is to test what works best for your problem. If you can't afford this for some reason, most algorithms will probably benefit from standardization more so than from normalization.
See here for some examples of when one should be preferred over the other:
For example, in clustering analyses, standardization may be especially crucial in order to compare similarities between features based on certain distance measures. Another prominent example is the Principal Component Analysis, where we usually prefer standardization over Min-Max scaling, since we are interested in the components that maximize the variance (depending on the question and if the PCA computes the components via the correlation matrix instead of the covariance matrix; but more about PCA in my previous article).
However, this doesn’t mean that Min-Max scaling is not useful at all! A popular application is image processing, where pixel intensities have to be normalized to fit within a certain range (i.e., 0 to 255 for the RGB color range). Also, typical neural network algorithm require data that on a 0-1 scale.
One disadvantage of normalization over standardization is that it loses some information in the data, especially about outliers.
Also on the linked page, there is this picture:
As you can see, scaling clusters all the data very close together, which may not be what you want. It might cause algorithms such as gradient descent to take longer to converge to the same solution they would on a standardized data set, or it might even make it impossible.
"Normalizing variables" doesn't really make sense. The correct terminology is "normalizing / scaling the features". If you're going to normalize or scale one feature, you should do the same for the rest.
That makes sense because normalization and standardization do different things.
Normalization transforms your data into a range between 0 and 1
Standardization transforms your data such that the resulting distribution has a mean of 0 and a standard deviation of 1
Normalization/standardization are designed to achieve a similar goal, which is to create features that have similar ranges to each other. We want that so we can be sure we are capturing the true information in a feature, and that we dont over weigh a particular feature just because its values are much larger than other features.
If all of your features are within a similar range of each other then theres no real need to standardize/normalize. If, however, some features naturally take on values that are much larger/smaller than others then normalization/standardization is called for
If you're going to be normalizing at least one variable/feature, I would do the same thing to all of the others as well
First question is why we need Normalisation/Standardisation?
=> We take a example of dataset where we have salary variable and age variable.
Age can take range from 0 to 90 where salary can be from 25thousand to 2.5lakh.
We compare difference for 2 person then age difference will be in range of below 100 where salary difference will in range of thousands.
So if we don't want one variable to dominate other then we use either Normalisation or Standardization. Now both age and salary will be in same scale
but when we use standardiztion or normalisation, we lose original values and it is transformed to some values. So loss of interpretation but extremely important when we want to draw inference from our data.
Normalization rescales the values into a range of [0,1]. also called min-max scaled.
Standardization rescales data to have a mean (μ) of 0 and standard deviation (σ) of 1.So it gives a normal graph.
Example below:
Another example:
In above image, you can see that our actual data(in green) is spread b/w 1 to 6, standardised data(in red) is spread around -1 to 3 whereas normalised data(in blue) is spread around 0 to 1.
Normally many algorithm required you to first standardise/normalise data before passing as parameter. Like in PCA, where we do dimension reduction by plotting our 3D data into 1D(say).Here we required standardisation.
But in Image processing, it is required to normalise pixels before processing.
But during normalisation, we lose outliers(extreme datapoints-either too low or too high) which is slight disadvantage.
So it depends on our preference what we chose but standardisation is most recommended as it gives a normal curve.
None of the mentioned transformations shall matter for linear regression as these are all affine transformations.
Found coefficients would change but explained variance will ultimately remain the same. So, from linear regression perspective, Outliers remain as outliers (leverage points).
And these transformations also will not change the distribution. Shape of the distribution remains the same.
lot of people use Normalisation and Standardisation interchangeably. The purpose remains the same is to bring features into the same scale. The approach is to subtract each value from min value or mean and divide by max value minus min value or SD respectively. The difference you can observe that when using min value u will get all value + ve and mean value u will get bot + ve and -ve values. This is also one of the factors to decide which approach to use.
I have some geographical trajectories sampled to analyze, and I calculated the histogram of data in spatial and temporal dimension, which yielded a time domain based feature for each spatial element. I want to perform a discrete FFT to transform the time domain based feature into frequency domain based feature (which I think maybe more robust), and then do some classification or clustering algorithms.
But I'm not sure using what descriptor as frequency domain based feature, since there are amplitude spectrum, power spectrum and phase spectrum of a signal and I've read some references but still got confused about the significance. And what distance (similarity) function should be used as measurement when performing learning algorithms on frequency domain based feature vector(Euclidean distance? Cosine distance? Gaussian function? Chi-kernel or something else?)
Hope someone give me a clue or some material that I can refer to, thanks~
Edit
Thanks to #DrKoch, I chose a spatial element with the largest L-1 norm and plotted its log power spectrum in python and it did show some prominent peaks, below is my code and the figure
import numpy as np
import matplotlib.pyplot as plt
sp = np.fft.fft(signal)
freq = np.fft.fftfreq(signal.shape[-1], d = 1.) # time sloth of histogram is 1 hour
plt.plot(freq, np.log10(np.abs(sp) ** 2))
plt.show()
And I have several trivial questions to ask to make sure I totally understand your suggestion:
In your second suggestion, you said "ignore all these values."
Do you mean the horizontal line represent the threshold and all values below it should be assigned to value zero?
"you may search for the two, three largest peaks and use their location and probably widths as 'Features' for further classification."
I'm a little bit confused about the meaning of "location" and "width", does "location" refer to the log value of power spectrum (y-axis) and "width" refer to the frequency (x-axis)? If so, how to combine them together as a feature vector and compare two feature vector of "a similar frequency and a similar widths" ?
Edit
I replaced np.fft.fft with np.fft.rfft to calculate the positive part and plot both power spectrum and log power spectrum.
code:
f, axarr = plt.subplot(2, sharex = True)
axarr[0].plot(freq, np.abs(sp) ** 2)
axarr[1].plot(freq, np.log10(np.abs(sp) ** 2))
plt.show()
figure:
Please correct me if I'm wrong:
I think I should keep the last four peaks in first figure with power = np.abs(sp) ** 2 and power[power < threshold] = 0 because the log power spectrum reduces the difference among each component. And then use the log spectrum of new power as feature vector to feed classifiers.
I also see some reference suggest applying a window function (e.g. Hamming window) before doing fft to avoid spectral leakage. My raw data is sampled every 5 ~ 15 seconds and I've applied a histogram on sampling time, is that method equivalent to apply a window function or I still need apply it on the histogram data?
Generally you should extract just a small number of "Features" out of the complete FFT spectrum.
First: Use the log power spec.
Complex numbers and Phase are useless in these circumstances, because they depend on where you start/stop your data acquisiton (among many other things)
Second: you will see a "Noise Level" e.g. most values are below a certain threshold, ignore all these values.
Third: If you are lucky, e.g. your data has some harmonic content (cycles, repetitions) you will see a few prominent Peaks.
If there are clear peaks, it is even easier to detect the noise: Everything between the peaks should be considered noise.
Now you may search for the two, three largest peaks and use their location and probably widths as "Features" for further classification.
Location is the x-value of the peak i.e. the 'frequency'. It says something how "fast" your cycles are in the input data.
If your cycles don't have constant frequency during the measuring intervall (or you use a window before caclculating the FFT), the peak will be broader than one bin. So this widths of the peak says something about the 'stability' of your cycles.
Based on this: Two patterns are similar if the biggest peaks of both hava a similar frequency and a similar widths, and so on.
EDIT
Very intersiting to see a logarithmic power spectrum of one of your examples.
Now its clear that your input contains a single harmonic (periodic, oscillating) component with a frequency (repetition rate, cycle-duration) of about f0=0.04.
(This is relative frquency, proprtional to the your sampling frequency, the inverse of the time beetween individual measurment points)
Its is not a pute sine-wave, but some "interesting" waveform. Such waveforms produce peaks at 1*f0, 2*f0, 3*f0 and so on.
(So using an FFT for further analysis turns out to be very good idea)
At this point you should produce spectra of several measurements and see what makes a similar measurement and how differ different measurements. What are the "important" features to distinguish your mesurements? Thinks to look out for:
Absolute amplitude: Height of the prominent (leftmost, highest) peaks.
Pitch (Main cycle rate, speed of changes): this is position of first peak, distance between consecutive peaks.
Exact Waveform: Relative amplitude of the first few peaks.
If your most important feature is absoulute amplitude, you're better off with calculating the RMS (root mean square) level of our input signal.
If pitch is important, you're better off with calculationg the ACF (auto-correlation function) of your input signal.
Don't focus on the leftmost peaks, these come from the high frequency components in your input and tend to vary as much as the noise floor.
Windows
For a high quality analyis it is importnat to apply a window to the input data before applying the FFT. This reduces the infulens of the "jump" between the end of your input vector ant the beginning of your input vector, because the FFT considers the input as a single cycle.
There are several popular windows which mark different choices of an unavoidable trade-off: Precision of a single peak vs. level of sidelobes:
You chose a "rectangular window" (equivalent to no window at all, just start/stop your measurement). This gives excellent precission of your peaks which now have a width of just one sample. Your sidelobes (the small peaks left and right of your main peaks) are at -21dB, very tolerable given your input data. In your case this is an excellent choice.
A Hanning window is a single cosine wave. It makes your peaks slightly broader but reduces side-lobe levels.
The Hammimg-Window (cosine-wave, slightly raised above 0.0) produces even broader peaks, but supresses side-lobes by -42 dB. This is a good choice if you expect further weak (but important) components between your main peaks or generally if you have complicated signals like speech, music and so on.
Edit: Scaling
Correct scaling of a spectrum is a complicated thing, because the values of the FFT lines depend on may things like sampling rate, lenght of FFT, window, and even implementation details of the FFT algorithm (there exist several different accepted conventions).
After all, the FFT should show the underlying conservation of energy. The RMS of the input signal should be the same as the RMS (Energy) of the spectrum.
On the other hand: if used for classification it is enough to maintain relative amplitudes. As long as the paramaters mentioned above do not change, the result can be used for classification without further scaling.
I've just run through the Wikipedia page about SVMs, and this line caught my eyes:
"If the kernel used is a Gaussian radial basis function, the corresponding feature space is a Hilbert space of infinite dimensions." http://en.wikipedia.org/wiki/Support_vector_machine#Nonlinear_classification
In my understanding, if I apply Gaussian kernel in SVM, the resulting feature space will be m-dimensional (where m is the number of training samples), as you choose your landmarks to be your training examples, and you're measuring the "similarity" between a specific example and all the examples with the Gaussian kernel. As a consequence, for a single example you'll have as many similarity values as training examples. These are going to be the new feature vectors which are going to m-dimensional vectors, and not infinite dimensionals.
Could somebody explain to me what do I miss?
Thanks,
Daniel
The dual formulation of the linear SVM depends only on scalar products of all training vectors. Scalar product essentially measures similarity of two vectors. We can then generalize it by replacing with any other "well-behaved" (it should be positive-definite, it's needed to preserve convexity, as well as enables Mercer's theorem) similarity measure. And RBF is just one of them.
If you take a look at the formula here you'll see that RBF is basically a scalar product in a certain infinitely dimensional space
Thus RBF is kind of a union of polynomial kernels of all possible degrees.
The other answers are correct but don't really tell the right story here. Importantly, you are correct. If you have m distinct training points then the gaussian radial basis kernel makes the SVM operate in an m dimensional space. We say that the radial basis kernel maps to a space of infinite dimension because you can make m as large as you want and the space it operates in keeps growing without bound.
However, other kernels, like the polynomial kernel do not have this property of the dimensionality scaling with the number of training samples. For example, if you have 1000 2D training samples and you use a polynomial kernel of <x,y>^2 then the SVM will operate in a 3 dimensional space, not a 1000 dimensional space.
The short answer is that this business about infinite dimensional spaces is only part of the theoretical justification, and of no practical importance. You never actually touch an infinite-dimensional space in any sense. It's part of the proof that the radial basis function works.
Basically, SVMs are proved to work the way they do by relying on properties of dot products over vector spaces. You can't just swap in the radial basis function and expect it necessarily works. To prove that it does, however, you show that the radial basis function is actually like a dot product over a different vector space, and it's as if we're doing regular SVMs in a transformed space, which works. And it happens that infinite dimensioal-ness is OK, and that the radial basis function does correspond to a dot product in such a space. So you can say SVMs still work when you use this particular kernel.