Related
Say the two series are:
x = [4,4,4,4,6,8,10,8,6,4,4,4,4,4,4,4,4,4,4,4,4,4,4]
y = [4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,6,8,10,8,6,4,4]
Series x clearly lags y by 12 time periods.
However, using the following code as suggested in Python cross correlation:
import numpy as np
c = np.correlate(x, y, "full")
lag = np.argmax(c) - c.size/2
leads to an incorrect lag of -0.5.
What's wrong here?
If you want to do it the easy way you should simply use scipy correlation_lags
Also, remember to subtract the mean from the inputs.
import numpy as np
from scipy import signal
x = [4,4,4,4,6,8,10,8,6,4,4,4,4,4,4,4,4,4,4,4,4,4,4]
y = [4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,6,8,10,8,6,4,4]
correlation = signal.correlate(x-np.mean(x), y - np.mean(y), mode="full")
lags = signal.correlation_lags(len(x), len(y), mode="full")
lag = lags[np.argmax(abs(correlation))]
This gives lag=-12, that is the difference between the index of the first six in x and in y, if you swap inputs it gives +12
Edit
Why to subtract the mean
If the signals have non-zero mean the terms at the center of the correlation will become larger, because there you have a larger support sample to compute the correlation. Furthermore, for very large data, subtracting the mean makes the calculations more accurate.
Here I illustrate what would happen if the mean was not subtracted for this example.
plt.plot(abs(correlation))
plt.plot(abs(signal.correlate(x, y, mode="full")))
plt.plot(abs(signal.correlate(np.ones_like(x)*np.mean(x), np.ones_like(y)*np.mean(y))))
plt.legend(['subtracting mean', 'constant signal', 'keeping the mean'])
Notice that the maximum on the blue curve (at 10) does not coincide with the maximum of the orange curve.
I'm facing a problem while trying to implement the coupled differential equation below (also known as single-mode coupling equation) in Python 3.8.3. As for the solver, I am using Scipy's function scipy.integrate.solve_bvp, whose documentation can be read here. I want to solve the equations in the complex domain, for different values of the propagation axis (z) and different values of beta (beta_analysis).
The problem is that it is extremely slow (not manageable) compared with an equivalent implementation in Matlab using the functions bvp4c, bvpinit and bvpset. Evaluating the first few iterations of both executions, they return the same result, except for the resulting mesh which is a lot greater in the case of Scipy. The mesh sometimes even saturates to the maximum value.
The equation to be solved is shown here below, along with the boundary conditions function.
import h5py
import numpy as np
from scipy import integrate
def coupling_equation(z_mesh, a):
ka_z = k # Global
z_a = z # Global
a_p = np.empty_like(a).astype(complex)
for idx, z_i in enumerate(z_mesh):
beta_zf_i = np.interp(z_i, z_a, beta_zf) # Get beta at the desired point of the mesh
ka_z_i = np.interp(z_i, z_a, ka_z) # Get ka at the desired point of the mesh
coupling_matrix = np.empty((2, 2), complex)
coupling_matrix[0] = [-1j * beta_zf_i, ka_z_i]
coupling_matrix[1] = [ka_z_i, 1j * beta_zf_i]
a_p[:, idx] = np.matmul(coupling_matrix, a[:, idx]) # Solve the coupling matrix
return a_p
def boundary_conditions(a_a, a_b):
return np.hstack(((a_a[0]-1), a_b[1]))
Moreover, I couldn't find a way to pass k, z and beta_zf as arguments of the function coupling_equation, given that the fun argument of the solve_bpv function must be a callable with the parameters (x, y). My approach is to define some global variables, but I would appreciate any help on this too if there is a better solution.
The analysis function which I am trying to code is:
def analysis(k, z, beta_analysis, max_mesh):
s11_analysis = np.empty_like(beta_analysis, dtype=complex)
s21_analysis = np.empty_like(beta_analysis, dtype=complex)
initial_mesh = np.linspace(z[0], z[-1], 10) # Initial mesh of 10 samples along L
mesh = initial_mesh
# a_init must be complex in order to solve the problem in a complex domain
a_init = np.vstack((np.ones(np.size(initial_mesh)).astype(complex),
np.zeros(np.size(initial_mesh)).astype(complex)))
for idx, beta in enumerate(beta_analysis):
print(f"Iteration {idx}: beta_analysis = {beta}")
global beta_zf
beta_zf = beta * np.ones(len(z)) # Global variable so as to use it in coupling_equation(x, y)
a = integrate.solve_bvp(fun=coupling_equation,
bc=boundary_conditions,
x=mesh,
y=a_init,
max_nodes=max_mesh,
verbose=1)
# mesh = a.x # Mesh for the next iteration
# a_init = a.y # Initial guess for the next iteration, corresponding to the current solution
s11_analysis[idx] = a.y[1][0]
s21_analysis[idx] = a.y[0][-1]
return s11_analysis, s21_analysis
I suspect that the problem has something to do with the initial guess that is being passed to the different iterations (see commented lines inside the loop in the analysis function). I try to set the solution of an iteration as the initial guess for the following (which must reduce the time needed for the solver), but it is even slower, which I don't understand. Maybe I missed something, because it is my first time trying to solve differential equations.
The parameters used for the execution are the following:
f2 = h5py.File(r'path/to/file', 'r')
k = np.array(f2['k']).squeeze()
z = np.array(f2['z']).squeeze()
f2.close()
analysis_points = 501
max_mesh = 1e6
beta_0 = 3e2;
beta_low = 0; # Lower value of the frequency for the analysis
beta_up = beta_0; # Upper value of the frequency for the analysis
beta_analysis = np.linspace(beta_low, beta_up, analysis_points);
s11_analysis, s21_analysis = analysis(k, z, beta_analysis, max_mesh)
Any ideas on how to improve the performance of these functions? Thank you all in advance, and sorry if the question is not well-formulated, I accept any suggestions about this.
Edit: Added some information about performance and sizing of the problem.
In practice, I can't find a relation that determines de number of times coupling_equation is called. It must be a matter of the internal operation of the solver. I checked the number of callings in one iteration by printing a line, and it happened in 133 ocasions (this was one of the fastests). This must be multiplied by the number of iterations of beta. For the analyzed one, the solver returned this:
Solved in 11 iterations, number of nodes 529.
Maximum relative residual: 9.99e-04
Maximum boundary residual: 0.00e+00
The shapes of a and z_mesh are correlated, since z_mesh is a vector whose length corresponds with the size of the mesh, recalculated by the solver each time it calls coupling_equation. Given that a contains the amplitudes of the progressive and regressive waves at each point of z_mesh, the shape of a is (2, len(z_mesh)).
In terms of computation times, I only managed to achieve 19 iterations in about 2 hours with Python. In this case, the initial iterations were faster, but they start to take more time as their mesh grows, until the point that the mesh saturates to the maximum allowed value. I think this is because of the value of the input coupling coefficients in that point, because it also happens when no loop in beta_analysisis executed (just the solve_bvp function for the intermediate value of beta). Instead, Matlab managed to return a solution for the entire problem in just 6 minutes, aproximately. If I pass the result of the last iteration as initial_guess (commented lines in the analysis function, the mesh overflows even faster and it is impossible to get more than a couple iterations.
Based on semi-random inputs, we can see that max_mesh is sometimes reached. This means that coupling_equation can be called with a quite big z_mesh and a arrays. The problem is that coupling_equation contains a slow pure-Python loop iterating on each column of the arrays. You can speed the computation up a lot using Numpy vectorization. Here is an implementation:
def coupling_equation_fast(z_mesh, a):
ka_z = k # Global
z_a = z # Global
a_p = np.empty(a.shape, dtype=np.complex128)
beta_zf_i = np.interp(z_mesh, z_a, beta_zf) # Get beta at the desired point of the mesh
ka_z_i = np.interp(z_mesh, z_a, ka_z) # Get ka at the desired point of the mesh
# Fast manual matrix multiplication
a_p[0] = (-1j * beta_zf_i) * a[0] + ka_z_i * a[1]
a_p[1] = ka_z_i * a[0] + (1j * beta_zf_i) * a[1]
return a_p
This code provides a similar output with semi-random inputs compared to the original implementation but is roughly 20 times faster on my machine.
Furthermore, I do not know if max_mesh happens to be big with your inputs too and even if this is normal/intended. It may make sense to decrease the value of max_mesh in order to reduce the execution time even more.
so I'm trying to get the second derivative of the following formula using numpy.gradient, and I'm trying to differentiate it once by S[:,0] and then by S[:,1]
S = np.random.multivariate_normal(mean, covariance, N)
formula = (S[:,0]**2) * (S[:,1]**2)
But the thing is when I use spacings as the second argument of numpy.gradient
dx = np.diff(S[:,0])
dy = np.diff(S[:,1])
dfdx = np.gradient(formula,dx)
I get the error saying
ValueError: when 1d, distances must match the length of the corresponding dimension
And I get that's because the spacings vector length is one element less than the formula's, but I didn't know what to do to fix that.
I've read somewhere also that you can have coordinates of the point rather than the spacing as the second argument, but when I tried checking the result out of that by differentiating the formula by S[:,0] and then by S[:,1], and then trying to differentiate it this time by S[:,0] and then by S[:,1], and comparing the two results, which should be similar; there was a huge difference between those two results.
Can anybody explain to me what I'm doing wrong here?
When introducing the vector of coordinates of values of your function using Numpy's gradient, you have to be careful to either introduce it as a list with as many arrays as dimensions of your function, or to specify at which axis (as an argument of gradient) you want to calculate the gradient.
When you checked both ways of differentiation, I think the problem is that your formula isn't actually two-dimensional, but one-dimensional (even though you use data from two variables, note your f array has only one dimension).
Take a look at this little script in which we verify that, indeed, the order of differentiation doesn't alter the result (assuming your function is well-behaved).
import numpy as np
# Dummy arrays and function
x = np.linspace(0,1,50)
y = np.linspace(0,2,50)
f = np.sin(2*np.pi*x[:,None]) * np.cos(2*np.pi*y)
dfdx = np.gradient(f, x, axis = 0)
df2dy = np.gradient(dfdx, y, axis = 1)
dfdy = np.gradient(f, y, axis = 1)
df2dx = np.gradient(dfdy, x, axis = 0)
# Check how many values are essentially different
print(np.sum(~np.isclose(df2dx, df2dy)))
Does this apply to your problem?
I am having trouble understanding the output of my function to implement multiple-ridge regression. I am doing this from scratch in Python for the closed form of the method. This closed form is shown below:
I have a training set X that is 100 rows x 10 columns and a vector y that is 100x1.
My attempt is as follows:
def ridgeRegression(xMatrix, yVector, lambdaRange):
wList = []
for i in range(1, lambdaRange+1):
lambVal = i
# compute the inner values (X.T X + lambda I)
xTranspose = np.transpose(x)
xTx = xTranspose # x
lamb_I = lambVal * np.eye(xTx.shape[0])
# invert inner, e.g. (inner)**(-1)
inner_matInv = np.linalg.inv(xTx + lamb_I)
# compute outer (X.T y)
outer_xTy = np.dot(xTranspose, y)
# multiply together
w = inner_matInv # outer_xTy
wList.append(w)
print(wList)
For testing, I am running it with the first 5 lambda values.
wList becomes 5 numpy.arrays each of length 10 (I'm assuming for the 10 coefficients).
Here is the first of those 5 arrays:
array([ 0.29686755, 1.48420319, 0.36388528, 0.70324668, -0.51604451,
2.39045735, 1.45295857, 2.21437745, 0.98222546, 0.86124358])
My question, and clarification:
Shouldn't there be 11 coefficients, (1 for the y-intercept + 10 slopes)?
How do I get the Minimum Square Error from this computation?
What comes next if I wanted to plot this line?
I think I am just really confused as to what I'm looking at, since I'm still working on my linear-algebra.
Thanks!
First, I would modify your ridge regression to look like the following:
import numpy as np
def ridgeRegression(X, y, lambdaRange):
wList = []
# Get normal form of `X`
A = X.T # X
# Get Identity matrix
I = np.eye(A.shape[0])
# Get right hand side
c = X.T # y
for lambVal in range(1, lambdaRange+1):
# Set up equations Bw = c
lamb_I = lambVal * I
B = A + lamb_I
# Solve for w
w = np.linalg.solve(B,c)
wList.append(w)
return wList
Notice that I replaced your inv call to compute the matrix inverse with an implicit solve. This is much more numerically stable, which is an important consideration for these types of problems especially.
I've also taken the A=X.T#X computation, identity matrix I generation, and right hand side vector c=X.T#y computation out of the loop--these don't change within the loop and are relatively expensive to compute.
As was pointed out by #qwr, the number of columns of X will determine the number of coefficients you have. You have not described your model, so it's not clear how the underlying domain, x, is structured into X.
Traditionally, one might use polynomial regression, in which case X is the Vandermonde Matrix. In that case, the first coefficient would be associated with the y-intercept. However, based on the context of your question, you seem to be interested in multivariate linear regression. In any case, the model needs to be clearly defined. Once it is, then the returned weights may be used to further analyze your data.
Typically to make notation more compact, the matrix X contains a column of ones for an intercept, so if you have p predictors, the matrix is dimensions n by p+1. See Wikipedia article on linear regression for an example.
To compute in-sample MSE, use the definition for MSE: the average of squared residuals. To compute generalization error, you need cross-validation.
Also, you shouldn't take lambVal as an integer. It can be small (close to 0) if the aim is just to avoid numerical error when xTx is ill-conditionned.
I would advise you to use a logarithmic range instead of a linear one, starting from 0.001 and going up to 100 or more if you want to. For instance you can change your code to that:
powerMin = -3
powerMax = 3
for i in range(powerMin, powerMax):
lambVal = 10**i
print(lambVal)
And then you can try a smaller range or a linear range once you figure out what is the correct order of lambVal with your data from cross-validation.
So say I'm trying to create a 100-sample dataset that follows a certain line, maybe 2x+2. And I want the values on my X-axis to range from 0-1000. To do this, I use the following.
X = np.random.random(100,1) * 1000
Y = (2*X) + 2
data = np.hstack(X,Y)
The hstack gives me the array with corresponding x and y values. That part works. But if I want to inject noise into it in order to scatter the datapoints further away from that 2x+2 line...that's what I can't figure out.
Say for example, I want that Y array to have a standard deviation of 20. How would I inject that noise into the y values?
Maybe I'm missing something, but have you tried adding numpy.random.normal(scale=20,size=100) to Y? You can even write
Y=numpy.random.normal(2*X+2,20)
and do it all at once (and without repeating the array size).
To simulate noise use a normally distributed random number generator like np.random.randn.
Is this what you are trying to do:
X = np.linspace(0, 1000, 100)
Y = (2*X) + 2 + 20*np.random.randn(100)
data = np.hstack((X.reshape(100,1),Y.reshape(100,1)))