Related
I do data normalization as:
X = ( X - X.mean(axis=0) ) / X.std(axis=0)
But some features of X have 0 variance. It gives me Runtime error for ZeroDivision.
I know we can normalize using "StandardScalar" class from sklearn. But how can I normalize data by myself from scratch if std=0 ?
To quote sklearn documentation for StandardScaler:
Per feature relative scaling of the data to achieve zero mean and unit variance. Generally this is calculated using np.sqrt(var_). If a variance is zero, we can’t achieve unit variance, and the data is left as-is, giving a scaling factor of 1.
Therefore, like what the other answer said, you can omit the standard deviation term and just do X - X.mean(axis=0) when standard deviation is 0. However this only works if the whole of X has 0 standard deviation.
To make this work where you have a mix of values with some std dev and values that don't, use this instead:
std = X.std(axis=0)
std = np.where(std == 0, 1, std)
X = ( X - X.mean(axis=0) ) / std
This code checks if standard deviation is zero for each row of values in axis 0, and replaces them with 1 if true.
If standard deviation is 0 for a particular feature, than all of its values are identical. In this case X = X - X.mean(axis=0) should suffice. This would give you 0 mean and 0 standardeviation.
I am a student working with time-series data which we feed into a neural network for classification (my task is to build and train this NN).
We're told to use a band-pass filter of 10 Hz to 150 Hz since anything outside that is not interesting.
After applying the band-pass, I've also down-sampled the data to 300 samples per second (originally it was 768 Hz). My understanding of the Shannon Nyquist sampling theorem is that, after applying the band-pass, any information in the data will be perfectly preserved at this sample-rate.
However, I got into a discussion with my supervisor who claimed that 300 Hz might not be sufficient even if the signal was band-limited. She says that it is only the minimum sample rate, not necessarily the best sample rate.
My understanding of the sampling theorem makes me think the supervisor is obviously wrong, but I don't want to argue with my supervisor, especially in case I'm actually the one who has misunderstood.
Can anyone help to confirm my understanding or provide some clarification? And how should I take this up with my supervisor (if at all).
The Nyquist-Shannon theorem states that the sampling frequency should at-least be twice of bandwidth, i.e.,
fs > 2B
So, this is the minimal criteria. If the sampling frequency is less than 2B then there will be aliasing. There is no upper limit on sampling frequency, but more the sampling frequency, the better will be the reconstruction.
So, I think your supervisor is right in saying that it is the minimal condition and not the best one.
Actually, you and your supervisor are both wrong. The minimum sampling rate required to faithfully represent a real-valued time series whose spectrum lies between 10 Hz and 150 Hz is 140 Hz, not 300 Hz. I'll explain this, and then I'll explain some of the context that shows why you might want to "oversample", as it is referred to (spoiler alert: Bailian-Low Theorem). The supervisor is mixing folklore into the discussion, and when folklore is not properly-contexted, it tends to telephone tag into fakelore. (That's a common failing even in the peer-reviewed literature, by the way). And there's a lot of fakelore, here, that needs to be defogged.
For the following, I will use the following conventions.
There's no math layout on Stack Overflow (except what we already have with UTF-8), so ...
a^b denotes a raised to the power b.
∫_I (⋯x⋯) dx denotes an integral of (⋯x⋯) taken over all x ∈ I, with the default I = ℝ.
The support supp φ (or supp_x φ(x) to make the "x" explicit) of a function φ(x) is the smallest closed set containing all the x-es for which φ(x) ≠ 0. For regularly-behaving (e.g. continuously differentiable) functions that means a union of closed intervals and/or half-rays or the whole real line, itself. This figures centrally in the Shannon-Nyquist sampling theorem, as its main condition is that a spectrum have bounded support; i.e. a "finite bandwidth".
For the Fourier transform I will use the version that has the 2π up in the exponent, and for added convenience, I will use the convention 1^x = e^{2πix} = cos(2πx) + i sin(2πx) (which I refer to as the Ramanujan Convention, as it is the convention I frequently used in my previous life oops I mean which Ramanujan secretly used in his life to make the math a whole lot simpler).
The set ℤ = {⋯, -2, -1, 0, +1, +2, ⋯ } is the integers, and 1^{x+z} = 1^x for all z∈ℤ - making 1^x the archetype of a periodic function whose period is 1.
Thus, the Fourier transform f̂(ν) of a function f(t) and its inverse are given by:
f̂(ν) = ∫ f(t) 1^{-νt} dt, f(t) = ∫ f̂(ν) 1^{+νt} dν.
The spectrum of the time series given by the function f(t) is the function f̂(ν) of the cyclic frequency ν, which is what is measured in Hertz (Hz.); t, itself, being measured in seconds. A common convention is to use the angular frequency ω = 2πν, instead, but that muddies the picture.
The most important example, with respect to the issue at hand, is the Fourier transform χ̂_Ω of the interval function given by χ_Ω(t) = 1 if t ∈ [-½Ω,+½Ω] and χ_Ω(t) = 0 else:
χ̂_Ω(t) = ∫_[-½Ω,+½Ω] 1^ν dν
= {1^{+½Ω} - 1^{-½Ω}}/{2πi}
= {2i sin πΩ}/{2πi}
= Ω sinc πΩ
which is where the function sinc x = (sin πx)/(πx) comes into play.
The cardinal form of the sampling theorem is that a function f(t) can be sampled over an equally-spaced sampled domain T ≡ { kΔt: k ∈ ℤ }, if its spectrum is bounded by supp f̂ ⊆ [-½Ω,+½Ω] ⊆ [-1/(2Δt),+1/(2Δt)], with the sampling given as
f(t) = ∑_{t'∈T} f(t') Ω sinc(Ω(t - t')) Δt.
So, this generally applies to [over-]sampling with redundancy factors 1/(ΩΔt) ≥ 1. In the special case where the sampling is tight with ΩΔt = 1, then it reduces to the form
f(t) = ∑_{t'∈T} f(t') sinc({t - t'}/Δt).
In our case, supp f̂ = [10 Hz., 150 Hz.] so the tightest fits are with 1/Δt = Ω = 300 Hz.
This generalizes to equally-spaced sampled domains of the form T ≡ { t₀ + kΔt: k ∈ ℤ } without any modification.
But it also generalizes to frequency intervals supp f̂ = [ν₋,ν₊] of width Ω = ν₊ - ν₋ and center ν₀ = ½ (ν₋ + ν₊) to the following form:
f(t) = ∑_{t'∈T} f(t') 1^{ν₀(t - t')} Ω sinc(Ω(t - t')) Δt.
In your case, you have ν₋ = 10 Hz., ν₊ = 150 Hz., Ω = 140 Hz., ν₀ = 80 Hz. with the condition Δt ≤ 1/140 second, a sampling rate of at least 140 Hz. with
f(t) = (140 Δt) ∑_{t'∈T} f(t') 1^{80(t - t')} sinc(140(t - t')).
where t and Δt are in seconds.
There is a larger context to all of this. One of the main places where this can be used is for transforms devised from an overlapping set of windowed filters in the frequency domain - a typical case in point being transforms for the time-scale plane, like the S-transform or the continuous wavelet transform.
Since you want the filters to be smoothly-windowed functions, without sharp corners, then in order for them to provide a complete set that adds up to a finite non-zero value over all of the frequency spectrum (so that they can all be normalized, in tandem, by dividing out by this sum), then their respective supports have to overlap.
(Edit: Generalized this example to cover both equally-spaced and logarithmic-spaced intervals.)
One example of such a set would be filters that have end-point frequencies taken from the set
Π = { p₀ (α + 1)ⁿ + β {(α + 1)ⁿ - 1} / α: n ∈ {0,1,2,⋯} }
So, for interval n (counting from n = 0), you would have ν₋ = p_n and ν₊ = p_{n+1}, where the members of Π are enumerated
p_n = p₀ (α + 1)ⁿ + β {(α + 1)ⁿ - 1} / α,
Δp_n = p_{n+1} - p_n = α p_n + β = (α p₀ + β)(α + 1)ⁿ,
n ∈ {0,1,2,⋯}
The center frequency of interval n would then be ν₀ = p_n + ½ Δp₀ (α + 1)ⁿ and the width would be Ω = Δp₀ (α + 1)ⁿ, but the actual support for the filter would overlap into a good part of the neighboring intervals, so that when you add up the filters that cover a given frequency ν the sum doesn't drop down to 0 as ν approaches any of the boundary points. (In the limiting case α → 0, this produces an equally-spaced frequency domain, suitable for an equalizer, while in the case β → 0, it produces a logarithmic scale with base α + 1, where octaves are equally-spaced.)
The other main place where you may apply this is to time-frequency analysis and spectrograms. Here, the role of a function f and its Fourier transform f̂ are reversed and the role of the frequency bandwidth Ω is now played by the (reciprocal) time bandwidth 1/Ω. You want to break up a time series, given by a function f(t) into overlapping segments f̃(q,λ) = g(λ)* f(q + λ), with smooth windowing given by the functions g(λ) with bounded support supp g ⊆ [-½ 1/Ω, +½ 1/Ω], and with interval spacing Δq much larger than the time sampling Δt (the ratio Δq/Δt is called the "hop" factor). The analogous role of Δt is played, here, by the frequency interval in the spectrogram Δp = Ω, which is now constant.
Edit: (Fixed the numbers for the Audacity example)
The minimum sampling rate for both supp_λ g and supp_λ f(q,λ) is Δq = 1/Ω = 1/Δp, and the corresponding redundancy factor is 1/(ΔpΔq). Audacity, for instance, uses a redundancy factor of 2 for its spectrograms. A typical value for Δp might be 44100/2048 Hz., while the time-sampling rate is Δt = 1/(2×3×5×7)² second (corresponding to 1/Δt = 44100 Hz.). With a redundancy factor of 2, Δq would be 1024/44100 second and the hop factor would be Δq/Δt = 1024.
If you try to fit the sampling windows, in either case, to the actual support of the band-limited (or time-limited) function, then the windows won't overlap and the only way to keep their sum from dropping to 0 on the boundary points would be for the windowing functions to have sharp corners on the boundaries, which would wreak havoc on their corresponding Fourier transforms.
The Balian-Low Theorem makes the actual statement on the matter.
https://encyclopediaofmath.org/wiki/Balian-Low_theorem
And a shout-out to someone I've been talking with, recently, about DSP-related matters and his monograph, which provides an excellent introductory reference to a lot of the issues discussed here.
A Friendly Guide To Wavelets
Gerald Kaiser
Birkhauser 1994
He said it's part of a trilogy, another installment of which is forthcoming.
I'm trying to understand RNNs and I would like to find a simple example that actually shows the one hot vectors and the numerical operations. Preferably conceptual since actual code may make it even more confusing. Most examples I google just show boxes with loops coming out of them and its really difficult to understand what exactly is going on. In the rare case where they do show the vectors its still difficult to see how they are getting the values.
for example I don't know where the values are coming from in this picture https://i1.wp.com/karpathy.github.io/assets/rnn/charseq.jpeg
If the example could integrate LSTMs and other popular extensions that would be cool too.
In the simple RNN case, a network accepts an input sequence x and produces an output sequence y while a hidden sequence h stores the network's dynamic state, such that at timestep i: x(i) ∊ ℝM, h(i) ∊ ℝN, y(i) ∊ ℝP the real valued vectors of M/N/P dimensions corresponding to input, hidden and output values respectively. The RNN changes its state and omits output based on the state equations:
h(t) = tanh(Wxh ∗ [x(t); h(t-1)]), where Wxh a linear map: ℝM+N ↦ ℝN, * the matrix multiplication and ; the concatenation operation. Concretely, to obtain h(t) you concatenate x(t) with h(t-1), you apply matrix multiplication between Wxh (of shape (M+N, N)) and the concatenated vector (of shape M+N) , and you use a tanh non-linearity on each element of the resulting vector (of shape N).
y(t) = sigmoid(Why * h(t)), where Why a linear map: ℝN ↦ ℝP. Concretely, you apply matrix multiplication between Why (of shape (N, P)) and h(t) (of shape N) to obtain a P-dimensional output vector, on which the sigmoid function is applied.
In other words, obtaining the output at time t requires iterating through the above equations for i=0,1,...,t. Therefore, the hidden state acts as a finite memory for the system, allowing for context-dependent computation (i.e. h(t) fully depends on both the history of the computation and the current input, and so does y(t)).
In the case of gated RNNs (GRU or LSTM), the state equations get somewhat harder to follow, due to the gating mechanisms which essentially allow selection between the input and the memory, but the core concept remains the same.
Numeric Example
Let's follow your example; we have M = 4, N = 3, P = 4, so Wxh is of shape (7, 3) and Why of shape (3, 4). We of course do not know the values of either W matrix, so we cannot reproduce the same results; we can still follow the process though.
At timestep t<0, we have h(t) = [0, 0, 0].
At timestep t=0, we receive input x(0) = [1, 0, 0, 0]. Concatenating x(0) with h(0-), we get [x(t); h(t-1)] = [1, 0, 0 ..., 0] (let's call this vector u to ease notation). We apply u * Wxh (i.e. multiplying a 7-dimensional vector with a 7 by 3 matrix) and get a vector v = [v1, v2, v3], where vi = Σj uj Wji = u1 W1i + u2 W2i + ... + u7 W7i. Finally, we apply tanh on v, obtaining h(0) = [tanh(v1), tanh(v2), tanh(v3)] = [0.3, -0.1, 0.9]. From h(0) we can also get y(0) via the same process; multiply h(0) with Why (i.e. 3 dimensional vector with a 3 by 4 matrix), get a vector s = [s1, s2, s3, s4], apply sigmoid on s and get σ(s) = y(0).
At timestep t=1, we receive input x(1) = [0, 1, 0, 0]. We concatenate x(1) with h(0) to get a new u = [0, 1, 0, 0, 0.3, -0.1, 0.9]. u is again multiplied with Wxh, and tanh is again applied on the result, giving us h(1) = [1, 0.3, 1]. Similarly, h(1) is multiplied by Why, giving us a new s vector on which we apply the sigmoid to obtain σ(s) = y(1).
This process continues until the input sequence finishes, ending the computation.
Note: I have ignored bias terms in the above equations because they do not affect the core concept and they make notation impossible to follow
I understand well perceptron so put accent only on kernel but I am not familiar with matemathic expressions so please give me an numerical example and a guide on kernel.
For example:
My hyperplane of perceptron is x1*w1+x2*w2+x3*w3+b=0; The RBF kernel formula: k(x,z) = exp((-|x-z|^2)/2*variance^2) where takes action the radial basis function kernel here. Is x an input and what is z variable here?
Or of what I have to calculate variance if it is variance in the formula?
Somewhere I have understood so that I have to plug this formula in perceptron decision function x1*w1+x2*w2+x3*w3+b=0; but how does it look look like If I plug in?
I would like to ask a numerical example to avoid confusion.
Linear Perceptron
As you know linear perceptrons can be trained for binary classification. More precisely, if there is n features, x1, x2, ..., xn in n-dimensional space, Rn, and you want to label them in 2 categories, y1 & y2 (usually -1 and +1), you can use linear perceptron which defines a hyperplane w1*x1 + ... + wn*xn + b = 0 to do so.
w1*x1 + ... + wn*xn + b > 0 or W.X + b > 0 ==> class = y1
w1*x1 + ... + wn*xn + b < 0 or W.X + b < 0 ==> class = y2
Linear perceptron will work well, only if the problem is linearly separable in Rn. For example, in 2D space, this means that one line can separate the 2 sets of points.
Algorithm
One common algorithm to train the perceptron, i.e., find weights and bias, w's & b, based on N data points, X1, ..., XN, and their labels, Y1, ..., YN is the following:
Initialize: W = zeros(n,1); b = 0
For i=1 to N:
Calculate F(Xi) = W.Xi + b
If F(Xi)*Yi <= 0:
W <--- W + Xi*Yi
b <--- b + Yi
This will give the final value for W & b. Besides, based on the training, W will be a linear combination of training points, Xi's, more precisely, the ones that were misclassified. So W = a1*X1 + ... + ...aN*XN where a's are in {0,y1,y2}.
Now, if there is a new point, let's say Z, to label, we check the sign of F(Z) = W.Z + b = a1*(X1.Z) + ... + aN*(XN.Z) + b. It is interesting that only the inner product of new point and training points take part in it.
Kernel Perceptron
Now, if the problem is not linearly separable, one may try to go to a higher dimensional space in which a hyperplane can do the classification. As an example, consider a circle in 2D space. The points inside and outside of the circle can't be separated by a line. However, if you find a transformation that can take the points to 3D space such that the first 2 coordinates remain the same for all points, and the 3rd coordinate become +1 and -1 for the points inside and outside of the circle respectively, then a plane defined as 3rd coordinate = 0 can separate the points.
Finding such transformations can be difficult and computationally heavy, so the kernel trick is introduced. Notice that we only used the inner product of new points with the training points. Kernel trick employs this fact and defines the inner product of the transformed points without actually finding the transformation.
If the unknown transformation is P(X) then Kernel function will be:
K(Xi,Xj) = <P(Xi),P(Xj)>. So instead of finding P, kernel functions are defined which represent the scalar result of the inner product in high-dimensional space. There are also theorems about what functions can be kernel functions, i.e., correspond to inner product in another space.
After choosing a kernel function, the algorithm will be modified as follows:
Initialize: F(X) = 0
For i=1 to N:
Calculate F(Xi)
If F(Xi)*Yi <= 0:
F(.) <--- F(.) + K(.,Xi)*Yi + Yi
At the end, F(.) = a1*K(.,X1) + ... + ...aN*K(.,XN) + b where a's are in {0,y1,y2}.
RBF Kernel
Radial basis function is one type of kernel function that is actually computing the inner product in an infinite-dimensional space. It can be written as
K(Xi,Xj) = exp(- norm2(Xi-Xj)^2 / (2*sigma^2))
Sigma is some parameter that you can work with to find an optimum value for. For example, you can train the model with different values of sigma and then find the best value based on the performance. You can start with sigma = 1
After training the model to find F(.), for a new data Z, the sign of F(Z) = a1*K(Z,X1) + ... + ...aN*K(Z,XN) + b will determine the class.
Remarks:
Regarding to your question about variance, you don't need to find any variance.
About x and z in your question, in each iteration, you should find the kernel output for the current data point and all the previously added points (the points that were misclassified and hence were added to F).
I couldn't come up with a simple instructive numerical example.
References:
I borrowed some notation from
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0ahUKEwjVu-fXo8DOAhVDxCYKHQkcDDAQFggoMAE&url=http%3A%2F%2Falex.smola.org%2Fteaching%2Fpune2007%2Fpune_3.pdf&usg=AFQjCNHlxy9TnY8xNe2-QDERipN_GycSqQ&bvm=bv.129422649,d.eWE
I'm trying to develop an application using SOM in analyzing data. However, after finishing training, I cannot find a way to visualize the result. I know that U-Matrix is one of the method but I cannot understand it properly. Hence, I'm asking for a specific and detail example how to construct U-Matrix.
I also read an answer at U-matrix and self organizing maps but it only refers to 1 row map, how about 3x3 map? I know that for 3x3 map:
m(1) m(2) m(3)
m(4) m(5) m(6)
m(7) m(8) m(9)
a 5x5 matrix must me created:
u(1) u(1,2) u(2) u(2,3) u(3)
u(1,4) u(1,2,4,5) u(2,5) u(2,3,5,6) u(3,6)
u(4) u(4,5) u(5) u(5,6) u(6)
u(4,7) u(4,5,7,8) u(5,8) u(5,6,8,9) u(6,9)
u(7) u(7,8) u(8) u(8,9) u(9)
but I don't know how to calculate u-weight u(1,2,4,5), u(2,3,5,6), u(4,5,7,8) and u(5,6,8,9).
Finally, after constructing U-Matrix, is there any way to visualize it using color, e.g. heat map?
Thank you very much for your time.
Cheers
I don't know if you are still interested in this but I found this link
http://www.uni-marburg.de/fb12/datenbionik/pdf/pubs/1990/UltschSiemon90
which explains very speciffically how to calculate the U-matrix.
Hope it helps.
By the way, the site were I found the link has several resources referring to SOMs I leave it here in case anyone is interested:
http://www.ifs.tuwien.ac.at/dm/somtoolbox/visualisations.html
The essential idea of a Kohonen map is that the data points are mapped to a
lattice, which is often a 2D rectangular grid.
In the simplest implementations, the lattice is initialized by creating a 3D
array with these dimensions:
width * height * number_features
This is the U-matrix.
Width and height are chosen by the user; number_features is just the number
of features (columns or fields) in your data.
Intuitively this is just creating a 2D grid of dimensions w * h
(e.g., if w = 10 and h = 10 then your lattice has 100 cells), then
into each cell, placing a random 1D array (sometimes called "reference tuples")
whose size and values are constrained by your data.
The reference tuples are also referred to as weights.
How is the U-matrix rendered?
In my example below, the data is comprised of rgb tuples, so the reference tuples
have length of three and each of the three values must lie between 0 and 255).
It's with this 3D array ("lattice") that you begin the main iterative loop
The algorithm iteratively positions each data point so that it is closest to others similar to it.
If you plot it over time (iteration number) then you can visualize cluster
formation.
The plotting tool i use for this is the brilliant Python library, Matplotlib,
which plots the lattice directly, just by passing it into the imshow function.
Below are eight snapshots of the progress of a SOM algorithm, from initialization to 700 iterations. The newly initialized (iteration_count = 0) lattice is rendered in the top left panel; the result from the final iteration, in the bottom right panel.
Alternatively, you can use a lower-level imaging library (in Python, e.g., PIL) and transfer the reference tuples onto the 2D grid, one at a time:
for y in range(h):
for x in range(w):
img.putpixel( (x, y), (
SOM.Umatrix[y, x, 0],
SOM.Umatrix[y, x, 1],
SOM.Umatrix[y, x, 2])
)
Here img is an instance of PIL's Image class. Here the image is created by iterating over the grid one pixel at a time; for each pixel, putpixel is called on img three times, the three calls of course corresponding to the three values in an rgb tuple.
From the matrix that you create:
u(1) u(1,2) u(2) u(2,3) u(3)
u(1,4) u(1,2,4,5) u(2,5) u(2,3,5,6) u(3,6)
u(4) u(4,5) u(5) u(5,6) u(6)
u(4,7) u(4,5,7,8) u(5,8) u(5,6,8,9) u(6,9)
u(7) u(7,8) u(8) u(8,9) u(9)
The elements with single numbers like u(1), u(2), ..., u(9) as just the elements with more than two numbers like u(1,2,4,5), u(2,3,5,6), ... , u(5,6,8,9) are calculated using something like the mean, median, min or max of the values in the neighborhood.
It's a nice idea calculate the elements with two numbers first, one possible code for that is:
for i in range(self.h_u_matrix):
for j in range(self.w_u_matrix):
nb = (0,0)
if not (i % 2) and (j % 2):
nb = (0,1)
elif (i % 2) and not (j % 2):
nb = (1,0)
self.u_matrix[(i,j)] = np.linalg.norm(
self.weights[i //2, j //2] - self.weights[i //2 +nb[0], j // 2 + nb[1]],
axis = 0
)
In the code above the self.h_u_matrix = self.weights.shape[0]*2 - 1 and self.w_u_matrix = self.weights.shape[1]*2 - 1 are the dimensions of the U-Matrix. With that said, for calculate the others elements it's necessary obtain a list with they neighboors and apply a mean for example. The following code implements that's idea:
for i in range(self.h_u_matrix):
for j in range(self.w_u_matrix):
if not (i % 2) and not (j % 2):
nodelist = []
if i > 0:
nodelist.append((i-1,j))
if i < 4:
nodelist.append((i+1, j))
if j > 0:
nodelist.append((i,j -1))
if j < 4:
nodelist.append((i,j+1))
meanlist = [self.u_matrix[u_node] for u_node in nodelist]
self.u_matrix[(i,j)] = np.mean(meanlist)
elif (i % 2) and (j % 2):
meanlist = [
(i - 1, j),
(i + 1, j),
(i, j - 1),
(i, j + 1)]
self.u_matrix[(i,j)] = np.mean(meanlist)