Alternative to Modulo operator - opencv

I am using OpenCV library. It doesn't have a modulo operator that can be applied to WHOLE matrix.
It has a Multiplication, Subtraction operators for a matrix.
Is there any way I can combine these operations to have an alternative to modulo operation?
I am looking for modulo of 2*pi.
If there are multiple of 2*pi in a row of matrix, the product of that row will be zero. It will ease my life and save considerable performance compare to looping every element of matrix.
Thanks.

If you've also got Division and Floor, you can use:
mod(a,b) = a - floor(a/b) * b

Related

Trying to do PCA analysis on interest rate swaps data (multivariate time series)

I have a data set with 20 non-overlapping different swap rates (spot1y, 1y1y, 2y1y, 3y1y, 4y1y, 5y2y, 7y3y, 10y2y, 12y3y...) over the past year.
I want to use PCA / multiregression and look at residuals in order to determine which sectors on the curve are cheap/rich. Has anyone had experience with this? I've done PCA but not for time series. I'd ideally like to model something similar to the first figure here but in USD.
https://plus.credit-suisse.com/rpc4/ravDocView?docid=kv66a7
Thanks!
Here are some broad strokes that can help answer your question. Also, that's a neat analysis from CS :)
Let's be pythonistas and use NumPy. You can imagine your dataset as a 20x261 array of floats. The first place to start is creating the array. Suppose you have a CSV file storing the raw data persistently. Then a reasonable first step to load the data would be something as simple as:
import numpy
x = numpy.loadtxt("path/to/my/file")
The object x is our raw time series matrix, and we verify the truthness of x.shape == (20, 261). The next step is to transform this array into it's covariance matrix. Whether it has been done on the raw data already, or it still has to be done, the first step is centering each time series on it's mean, like this:
x_centered = x - x.mean(axis=1, keepdims=True)
The purpose of this step is to help simplify any necessary rescaling, and is a very good habit that usually shouldn't be skipped. The call to x.mean uses the parameters axis and keepdims to make sure each row (e.g. the time series for spot1yr, ...) is centered with it's mean value.
The next steps are to square and scale x to produce a swap rate covariance array. With 2-dimensional arrays like x, there are two ways to square it-- one that leads to a 261x261 array and another that leads to a 20x20 array. It's the second array we are interested in, and the squaring procedure that will work for our purposes is:
x_centered_squared = numpy.matmul(x_centered, x_centered.transpose())
Then, to scale one can chose between 1/261 or 1/(261-1) depending on the statistical context, which looks like this:
x_covariance = x_centered_squared * (1/261)
The array x_covariance has an entry for how each swap rate changes with itself, and changes with any one of the other swap rates. In linear-algebraic terms, it is a symmetric operator that characterizes the spread of each swap rate.
Linear algebra also tells us that this array can be decomposed into it's associated eigen-spectrum, with elements in this spectrum being scalar-vector pairs, or eigenvalue-eigenvector pairs. In the analysis you shared, x_covariance's eigenvalues are plotted in exhibit two as percent variance explained. To produce the data for a plot like exhibit two (which you will always want to furnish to the readers of your PCA), you simply divide each eigenvalue by the sum of all of them, then multiply each by 100.0. Due to the convenient properties of x_covariance, a suitable way to compute it's spectrum is like this:
vals, vects = numpy.linalg.eig(x_covariance)
We are now in a position to talk about residuals! Here is their definition (with our namespace): residuals_ij = x_ij − reconstructed_ij; i = 1:20; j = 1:261. Thus for every datum in x, there is a corresponding residual, and to find them, we need to recover the reconstructed_ij array. We can do this column-by-column, operating on each x_i with a change of basis operator to produce each reconstructed_i, each of which can be viewed as coordinates in a proper subspace of the original or raw basis. The analysis describes a modified Gram-Schmidt approach to compute the change of basis operator we need, which ensures this proper subspace's basis is an orthogonal set.
What we are going to do in the approach is take the eigenvectors corresponding to the three largest eigenvalues, and transform them into three mutually orthogonal vectors, x, y, z. Research the web for active discussions and questions geared toward developing the Gram-Schmidt process for all sorts of practical applications, but for simplicity let's follow the analysis by hand:
x = vects[0] - sum([])
xx = numpy.dot(x, x)
y = vects[1] - sum(
(numpy.dot(x, vects[1]) / xx) * x
)
yy = numpy.dot(y, y)
z = vects[2] - sum(
(numpy.dot(x, vects[2]) / xx) * x,
(numpy.dot(y, vects[2]) / yy) * y
)
It's reasonable to implement normalization before or after this step, which should be informed by the data of course.
Now with the raw data, we implicitly made the assumption that the basis is standard, we need a map between {e1, e2, ..., e20} and {x,y,z}, which is given by
ch_of_basis = numpy.array([x,y,z]).transpose()
This can be used to compute each reconstructed_i, like this:
reconstructed = []
for measurement in x.transpose().tolist():
reconstructed.append(numpy.dot(ch_of_basis, measurement))
reconstructed = numpy.array(reconstructed).transpose()
And then you get the residuals by subtraction:
residuals = x - reconstructed
This flow obviously might need further tuning, but it's the gist of how to do compute all the residuals. To get that periodic bar plot, take the average of each row in residuals.

Shape function for B21 (Timoshenko beam) element in Abaqus

I am wondering what are the appropriate shape/interpolation functions for the B21 element since it has 3 DoF per node, but is stated as linear interpolation element.
Update: (as per duffymo's comment)
I know there is a distinction between nodal DoF and interpolation order and I am looking for the relation between the two.
For example the standard Euler-Bernoulli beam element (B23) has a 3rd order polynomial interpolation and uses the four nodal DoF (2 displacements and 2 rotations) two determine the displacement field. This interpolation is still linear in the coefficients, but cubic in length. How is the interpolation kept linear in length for B21? Does it have separate first order polynomials for each DoF?
My end goal here is to calculate stresses from displacements, obtained by my own solver.
Any help is appreciated.

Find negative fractional power of a matrix in Armadillo

In Matlab I do A ^ -0.5 to find the negative fractional power of matrix A. What is the equivalent in Armadillo C++ library? The pow() function performs element wise operation.
You can do
expmat(-0.5 * logmat(A))
Use the powmat() function, like so:
mat A(5,5,fill::randu);
cx_mat B = powmat(A, -0.5);
Or use a combination of inv() and sqrtmat().

Wouldn't setting the first derivative of Cost function J to 0 gives the exact Theta values that minimize the cost?

I am currently doing Andrew NG's ML course. From my calculus knowledge, the first derivative test of a function gives critical points if there are any. And considering the convex nature of Linear / Logistic Regression cost function, it is a given that there will be a global / local optima. If that is the case, rather than going a long route of taking a miniscule baby step at a time to reach the global minimum, why don't we use the first derivative test to get the values of Theta that minimize the cost function J in a single attempt , and have a happy ending?
That being said, I do know that there is a Gradient Descent alternative called Normal Equation that does just that in one successful step unlike the former.
On a second thought, I am thinking if it is mainly because of multiple unknown variables involved in the equation (which is why the Partial Derivative comes into play?) .
Let's take an example:
Gradient simple regression cost function:
Δ[RSS(w) = [(y-Hw)T(y-Hw)]
y : output
H : feature vector
w : weights
RSS: residual sum of squares
Equating this to 0 for getting the closed form solution will give:
w = (H T H)-1 HT y
Now assuming there are D features, the time complexity for calculating transpose of matrix is around O(D3). If there are a million features, it is computationally impossible to do within reasonable amount of time.
We use these gradient descent methods since they give solutions with reasonably acceptable solutions within much less time.

Accelerate framework ios: Fastest pythagoren calculation

So I have 2 matrices: each is 100x100.
I am looking to calculate a 3rd matrix such that: M3[i]=sqrt(M1[i]^2 + M2[i]^2).
I can obviously do ForLoops but I am sure there is something faster.
I digged into the accelerate framework and got lost in Linpack world
Any help to get me on the right track...
Thanks
The Accelerate framework is a good idea.
You could use a function like vDSP_vsq to operate on one column of the matrix at a time, placing the result in the corresponding column of M3. You might have to make two resultant matrices (one which will be M1^2, the other M2^2), and then add them to form the correct M3 result matrix by adding the columns with a call to vDSP_vadd which, again, will be able to operate one column at a time.
There is sample code (showing how to add two vectors, etc) here at the Apple developer page.
I think the fastest way is to use
vDSP_vpythg
Vector Pythagoras; single precision.
Subtracts vector C from A and squares the differences, subtracts vector D from B and squares the differences, adds the two sets of squared differences, and then writes the square roots of the sums to vector E.
Obviously pass ZERO vector for C and D .

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