Started learning octave recently. How do I generate a matrix from another matrix by applying a function to each element?
eg:
Apply 2x+1 or 2x/(x^2+1) or 1/x+3 to a 3x5 matrix A.
The result should be a 3x5 matrix with the values now 2x+1
if A(1,1)=1 then after the operation with output matrix B then
B(1,1) = 2.1+1 = 3
My main concern is a function that uses the value of x like that of finding the inverse or something as indicated above.
regards.
You can try
B = A.*2 + 1
The operator . means application of the following operation * to each element of the matrix.
You will find a lot of documentation for Octave in the distribution package and on the Web. Even better, you can usually also use the extensive documentation on Matlab.
ADDED. For more complex operations you can use arrayfun(), e.g.
B = arrayfun(#(x) 2*x/(x^2+1), A)
Related
i have been following cs231n lectures of Stanford and trying to complete assignments on my own and sharing these solutions both on github and my blog. But i'm having a hard time on understanding how to modelize backpropagation. I mean i can code modular forward and backward passes but what bothers me is that if i have the model below : Two Layered Neural Network
Lets assume that our loss function here is a softmax loss function. In my modular softmax_loss() function i am calculating loss and gradient with respect to scores (dSoft = dL/dY). After that, when i'am following backwards lets say for b2, db2 would be equal to dSoft*1 or dW2 would be equal to dSoft*dX2(outputs of relu gate). What's the chain rule here ? Why isnt dSoft equal to 1 ? Because dL/dL would be 1 ?
The softmax function is outputs a number given an input x.
What dSoft means is that you're computing the derivative of the function softmax(x) with respect to the input x. Then to calculate the derivative with respect to W of the last layer you use the chain rule i.e. dL/dW = dsoftmax/dx * dx/dW. Note that x = W*x_prev + b where x_prev is the input to the last node. Therefore dx/dW is just x and dx/db is just 1, which means that dL/dW or simply dW is dsoftmax/dx * x_prev and dL/db or simply db is dsoftmax/dx * 1. Note that here dsoftmax/dx is dSoft we defined earlier.
I've been trying to display in my console an exponential equation like the following one:
y(t) = a*e^t + b*e^t + c*e^t
I would write it as a string, however the coefficients a,b and c, are numbers in a vector V = [a b c]. So I was trying to concatenate the numbers with strings "e^t", but I failed to do it. I know scilab displays polynomial equations, but I don't know it is possible to display exponential one. Anyone can help?
Usually this kind of thing is done with mprintf command, which places given numerical arguments into a string with formatting instructions.
V = [3 5 -7]
mprintf("y(t) = %f*e^t + %f*e^t + %f*e^t", V)
The output is
y(t) = 3.000000*e^t + 5.000000*e^t + -7.000000*e^t
which isn't ideal, and can be improved in some ways by tweaking the formatters, but is readable regardless.
Notice we don't have to list every entry V(1), V(2), ... individually; the vector V gets "unpacked" automatically.
If you wanted to have 2D output like what we get for polynomials,
then no, this kind of thing is what Scilab does for polynomials and rational functions only, not for general expressions.
There is also prettyprint but its output is LaTeX syntax, like $1+s+s^{2}-s^{123}$. It works for a few things: polynomials, rational functions, matrices... but again, Scilab is not meant for symbolic manipulations, and does not really support symbolic expressions.
So, I have a vector that corresponds to a given feature (same dimensionality). Is there a package in Julia that would provide a mathematical function that fits these data points, in relation to the original feature? In other words, I have x and y (both vectors) and need to find a decent mapping between the two, even if it's a highly complex one. The output of this process should be a symbolic formula that connects x and y, e.g. (:x)^3 + log(:x) - 4.2454. It's fine if it's just a polynomial approximation.
I imagine this is a walk in the park if you employ Genetic Programming, but I'd rather opt for a simpler (and faster) approach, if it's available. Thanks
Turns out the Polynomials.jl package includes the function polyfit which does Lagrange interpolation. A usage example would go:
using Polynomials # install with Pkg.add("Polynomials")
x = [1,2,3] # demo x
y = [10,12,4] # demo y
polyfit(x,y)
The last line returns:
Poly(-2.0 + 17.0x - 5.0x^2)`
which evaluates to the correct values.
The polyfit function accepts a maximal degree for the output polynomial, but defaults to using the length of the input vectors x and y minus 1. This is the same degree as the polynomial from the Lagrange formula, and since polynomials of such degree agree on the inputs only if they are identical (this is a basic theorem) - it can be certain this is the same Lagrange polynomial and in fact the only one of such a degree to have this property.
Thanks to the developers of Polynomial.jl for leaving me just to google my way to an Answer.
Take a look to MARS regression. Multi adaptive regression splines.
First, excuse me for not providing a minimal working example, it is that I just can't think of one, really. I'll just give some pieces of code and ask my question "in principle".
I'm doing thermophysical properties calculation with a real gas model (Peng-Robinson) and here I am having problems when translating a model, where I use pressure p and specific enthalpy h as inputs to calculate all other properties. When it comes to calculating the temperature T, it is linked to the enthalpy h via an equation called departure function, which is itself a function of T. In Modelica it looks like this:
Dh_real = R_m*T*(Z - 1) + (T*dadT - a)/(sqrt(8)*b)*log((Z + (1 + sqrt(2))*B)/(Z + (1 - sqrt(2))*B));
Here a, dadT and Z are also temperature-dependent scalars and partly calculated using matrix operations (dadT) or polynomial-root-calculation (Z) in functions, b and B are parameters.
Calculating the enthalpy from an input temperature (in another model) is straightforward and working fine, the solver can solve the departure function analytically. The other direction has to be solved numerically and this is, I think, why Dymola gives me this error, when translating.
Cannot find differentiation function:
DadT_Unique2([some parameters and T])
with respect to time
Failed to differentiate the equation
dadT = DadT_Unique2([some parameters and T]);
in order to reduce the DAE index.
Failed to reduce the DAE index.
Now DadT is a function within the model, where I use some simple matrix operations to calculate dadT from some parameters and the temperature T. Obviously, Dymola is in need of the derivative of some internal _Unique2-function.
I couldn't find anything in the specification nor in the web about this. Can I provide a derivative of the functions somehow? I tried the smoothOrder-annotation, but without effect. How can I deal with this?
This is not a full answer, but a list of interesting links that you should read:
Michael Tiller on annotation(derivative=dxyz) and other annotations:
http://book.xogeny.com/behavior/functions/func_annos/#derivative
Claytex on numerical Jacobians and flag Hidden.PrintFailureToDifferentiate:
http://www.claytex.com/blog/how-can-i-make-my-models-run-faster/
Two related questions here on StackOverflow:
Dymola solving stationary equation systems for Media-Model
Two-Phase Modelica Media example
Some related Modelica conference papers:
https://modelica.org/events/Conference2005/online_proceedings/Session1/Session1c2.pdf
http://dx.doi.org/10.3384/ecp15118647
http://dx.doi.org/10.3384/ecp15118653
Cubic equation of state, generalized form (table 4.2)
https://books.google.de/books?id=_Op6DQAAQBAJ&pg=PA187
Solving cubic equations of state:
http://dx.doi.org/10.1002/aic.690480421
https://books.google.com/books?id=dd410GGw8wUC&pg=PA48
https://books.google.com/books?id=1rOA5I6kQ7gC&pg=PA620 (Appendix C)
Rewriting partial derivatives:
https://scholar.google.com/scholar?cluster=3379879976574799663
I am using Matlab to do one of my projects. I am stuck at one basic thing.
I have 2 matricies - A and B and a vector V. What I want to do is this:
A(i, j) = V(B(i,j)) for all i, j.
I tried doing this in the most obvious way - nested loops. For some reason, A is not getting populated. Am I missing something? Is there a more efficient (in-built function) way of doing this.
Thanks,
Anil.
If all entries in B are integers larger than zero, and if the maximum of B is not larger than the number of elements in V, then you can simply write
A = V(B);