I have two questions:
I need to know that is there any difference between the terms Fast Hadamard Transform (FHT) and Fast Walsh-Hadamard Transform (FWHT) ?
Can i use these two terms interchangeably ?
By a Normalized FHT, do we mean dividing all the values of output matrix (after doing FHT on data) by the largest value in the output matrix ?
What is the physical significance of a Normalized Transform ? (for e.g., Normalized FHT)
regards.
According to Wikipedia it looks the same:
https://en.wikipedia.org/wiki/Hadamard_transform
https://en.wikipedia.org/wiki/Walsh_matrix
https://en.wikipedia.org/wiki/Walsh%E2%80%93Hadamard_transform
Up to the normalization factor.
And indeed there's an efficient implementation using the same trick in the FFT algorithm.
Related
I have a dataset where some data points are 0 and I'm trying to process it so that each data point is instead a percent change from the previous point. The problem is that some of these points have the value of 0, and so sometimes calculating percent change from the previous data point of 0 will lead the current data point to equal infinity.
Is there a better way to handle percent change or is it fine for a recurrent neural network to use infinity as some of its data points?
I am feeding this data into a recurrent neural network backed by Keras.
This is a classical problem in machine learning. In dealing with such problems you need to apply a so-called smoothing which is usually adding a small constant eps to denumerator. So you need to apply following transformation:
ration = next_step / (eps + small_step)
I advise you to set eps to be greater than 1e5 as 1e6 is a decimal point precision of float32 format used in keras.
I am using libSVM.
Say my feature values are in the following format:
instance1 : f11, f12, f13, f14
instance2 : f21, f22, f23, f24
instance3 : f31, f32, f33, f34
instance4 : f41, f42, f43, f44
..............................
instanceN : fN1, fN2, fN3, fN4
I think there are two scaling can be applied.
scale each instance vector such that each vector has zero mean and unit variance.
( (f11, f12, f13, f14) - mean((f11, f12, f13, f14) ). /std((f11, f12, f13, f14) )
scale each colum of the above matrix to a range. for example [-1, 1]
According to my experiments with RBF kernel (libSVM) I found that the second scaling (2) improves the results by about 10%. I did not understand the reason why (2) gives me a improved results.
Could anybody explain me what is the reason for applying scaling and why the second option gives me improved results?
The standard thing to do is to make each dimension (or attribute, or column (in your example)) have zero mean and unit variance.
This brings each dimension of the SVM into the same magnitude. From http://www.csie.ntu.edu.tw/~cjlin/papers/guide/guide.pdf:
The main advantage of scaling is to avoid attributes in greater numeric
ranges dominating those in smaller numeric ranges. Another advantage is to avoid
numerical diculties during the calculation. Because kernel values usually depend on
the inner products of feature vectors, e.g. the linear kernel and the polynomial ker-
nel, large attribute values might cause numerical problems. We recommend linearly
scaling each attribute to the range [-1,+1] or [0,1].
I believe that it comes down to your original data a lot.
If your original data has SOME extreme values for some columns, then in my opinion you lose some definition when scaling linearly, for example in the range [-1,1].
Let's say that you have a column where 90% of values are between 100-500 and in the remaining 10% the values are as low as -2000 and as high as +2500.
If you scale this data linearly, then you'll have:
-2000 -> -1 ## <- The min in your scaled data
+2500 -> +1 ## <- The max in your scaled data
100 -> -0.06666666666666665
234 -> -0.007111111111111068
500 -> 0.11111111111111116
You could argue that the discernibility between what was originally 100 and 500 is smaller in the scaled data in comparison to what it was in the original data.
At the end, I believe it very much comes down to the specifics of your data and I believe the 10% improved performance is very coincidental, you will certainly not see a difference of this magnitude in every dataset you try both scaling methods on.
At the same time, in the paper in the link listed in the other answer, you can clearly see that the authors recommend data to be scaled linearly.
I hope someone finds this useful!
The accepted answer speaks of "Standard Scaling", which is not efficient for high-dimensional data stored in sparse matrices (text data is a use-case); in such cases, you may resort to "Max Scaling" and its variants, which works with sparse matrices.
I have generated a cascade using Local Binary Patterns for object recongition. It seems that the tool to evaluate the detection rate is opencv_performance.exe but I found out that it only works with haar cascades ? Is there something wrong with my cascade ? Do I need to change the format ?
You can code your own evaluator and use the same metric that opencv_performance uses. You need to change the old function that loads the cascades and also the function that makes the detection (change to detectMultiScale),
A suggestion of another metric is Intersection over Union (IOU), that calculates the percent of overlapping of two rectangles (the groundtruth rect and the detected rect).
The pseudo-code would be:
IOU = area(intersection(rect1,rect2)) / area(union(rect1,rect2)) and compare with, for example, if it is greater than 0.5.
Take a look here: http://answers.opencv.org/question/117/performance-evaluation-for-detection/
I try to implement a people detecting system based on SVM and HOG using OpenCV2.3. But I got stucked.
I came this far:
I can compute HOG values from an image database and then I calculate with LIBSVM the SVM vectors, so I get e.g. 1419 SVM vectors with 3780 values each.
OpenCV just wants one feature vector in the method hog.setSVMDetector(). Therefore I have to calculate one feature vector from my 1419 SVM vectors, that LIBSVM has calculated.
I found one hint, how to calculate this single feature vector: link
“The detecting feature vector at component i (where i is in the range e.g. 0-3779) is built out of the sum of the support vectors at i * the alpha value of that support vector, e.g.
det[i] = sum_j (sv_j[i] * alpha[j]) , where j is the number of the support vector, i
is the number of the components of the support vector.”
According to this, my routine works this way:
I take the first element of my first SVM vector, multiply it with the alpha value and add it with the first element of the second SVM vector that has been multiplied with alpha value, …
But after summing up all 1419 elements I get quite high values:
16.0657, -0.351117, 2.73681, 17.5677, -8.10134,
11.0206, -13.4837, -2.84614, 16.796, 15.0564,
8.19778, -0.7101, 5.25691, -9.53694, 23.9357,
If you compare them, to the default vector in the OpenCV sample peopledetect.cpp (and hog.cpp in the OpenCV source)
0.05359386f, -0.14721455f, -0.05532170f, 0.05077307f,
0.11547081f, -0.04268804f, 0.04635834f, -0.05468199f, 0.08232084f,
0.10424068f, -0.02294518f, 0.01108519f, 0.01378693f, 0.11193510f,
0.01268418f, 0.08528346f, -0.06309239f, 0.13054633f, 0.08100729f,
-0.05209739f, -0.04315529f, 0.09341384f, 0.11035026f, -0.07596218f,
-0.05517511f, -0.04465296f, 0.02947334f, 0.04555536f,
you see, that the default vector values are in the boundaries between –1 and +1, but my values exceed them far.
I think, my single feature vector routine needs some adjustment, any ideas?
Regards,
Christoph
The aggregated vector's values do look high.
I used the loadSVMfromModelFile() located in http://lnx.mangaitalia.net/trainer/main.cpp
I had to remove svinstr.sync(); from the code since it caused losing parts of the lines and getting wrong results.
I don't know much about the rest of the file, I only used this function.
According to "Introduction to Neural Networks with Java By Jeff Heaton", the input to the Kohonen neural network must be the values between -1 and 1.
It is possible to normalize inputs where the range is known beforehand:
For instance RGB (125, 125, 125) where the range is know as values between 0 and 255:
1. Divide by 255: (125/255) = 0.5 >> (0.5,0.5,0.5)
2. Multiply by two and subtract one: ((0.5*2)-1)=0 >> (0,0,0)
The question is how can we normalize the input where the range is unknown like our height or weight.
Also, some other papers mention that the input must be normalized to the values between 0 and 1. Which is the proper way, "-1 and 1" or "0 and 1"?
You can always use a squashing function to map an infinite interval to a finite interval. E.g. you can use tanh.
You might want to use tanh(x * l) with a manually chosen l though, in order not to put too many objects in the same region. So if you have a good guess that the maximal values of your data are +/- 500, you might want to use tanh(x / 1000) as a mapping where x is the value of your object It might even make sense to subtract your guess of the mean from x, yielding tanh((x - mean) / max).
From what I know about Kohonen SOM, they specific normalization does not really matter.
Well, it might through specific choices for the value of parameters of the learning algorithm, but the most important thing is that the different dimensions of your input points have to be of the same magnitude.
Imagine that each data point is not a pixel with the three RGB components but a vector with statistical data for a country, e.g. area, population, ....
It is important for the convergence of the learning part that all these numbers are of the same magnitude.
Therefore, it does not really matter if you don't know the exact range, you just have to know approximately the characteristic amplitude of your data.
For weight and size, I'm sure that if you divide them respectively by 200kg and 3 meters all your data points will fall in the ]0 1] interval. You could even use 50kg and 1 meter the important thing is that all coordinates would be of order 1.
Finally, you could a consider running some linear analysis tools like POD on the data that would give you automatically a way to normalize your data and a subspace for the initialization of your map.
Hope this helps.