python,optimization,scipy,minimization

Passing arguments to the objects is done with parameter args. Optimizing rosen(x,2): import numpy as np from scipy.optimize import minimize def rosen(x, y): """The Rosenbrock function""" return sum(100.0*(x[1:]-x[:-1]**2.0)**y + (1-x[:-1])**2.0) x0 = np.array([1.3, 0.7, 0.8, 1.9, 1.2]) res = minimize(rosen, x0, args=(2,), method='nelder-mead', options={'xtol': 1e-8, 'disp': True}) Note that the...

First of all, the easiest way of optimizing a function using scipy.optimize is to construct the target function such that the first argument is a list of the parameters need to be optimized and the following arguments specify other things, such as data and fixed parameters. Second, it will be...

r,mathematical-optimization,minimization,quadratic-programming

Given the similarities of the optimization problem here to your previous question, I will borrow some language directly from my answer to that question. However they are quite a bit different (the previous problem was a linear programming problem and this is a quadratic programming problem, and the constraints differ),...

I recommend looking into Grunt, a JavaScript task runner. It has a lot of fantastic features and plugins which can automate mundane tasks for you. Uglify, specifically, sounds like what you're looking for. Uglify compresses multiple JavaScript files into one, while also adding the extra benefits of optimizing your code...

matlab,optimization,minimization

You can try using Unconstrained Optimization method such as fminsearch, for example: [email protected](x) x^2; [email protected](x) x^3; k1=2; k2=4; inital_guess=3; f = @(x) sum(abs([h(x)+k1; g(x)+k2])); [x,fval] = fminsearch(f,inital_guess) Note that I represent both eq in matrix form, and the minimization is by looking at the sum of their absolute values. For...

python,arrays,loops,minimization

There is already a handy formula for least squares fitting. I came up with two different ways to solve your problem. For the first one, consider the matrix K: L = len(X) K = np.identity(L) - np.ones((L, L)) / L In your case, A and B are defined as: A...

Specify finish=None: scipy.optimize.brute(lambda x:x, (slice(1,100,1),), finish=None) The default behavior is to pass the output through fmin to improve it. The ranges aren't passed to fmin; they seem to be considered hints, rather than bounds. Specifying finish=None makes brute give you the brute-force solution directly. If you want fmin to improve...

r,statistics,mathematical-optimization,minimization

On this occasion optim will not work obviously because you have equality constraints. constrOptim will not work either for the same reason (I tried converting the equality to two inequalities i.e. greater and less than 15 but this didn't work with constrOptim). However, there is a package dedicated to this...

python,scipy,ellipse,astronomy,minimization

First, you have a typo in your objective function that prevents optimization of one of the variables: dummy_ellipse = generate_ellipse(...,dz,dy,dz) should be dummy_ellipse = generate_ellipse(...,dx,dy,dz) Also, taking sqrt out and minimizing the sum of squared euclidean distances makes it numerically somewhat easier for the optimizer. Your objective function is also...

python,algorithm,graph,shortest-path,minimization

Think of each unoccupied square on the board as a node in a graph, with edges connecting squares that a knight can move between in a single turn. Then you can use the search algorithm of your choice to find the shortest path in this graph.

python,algorithm,numpy,minimization

Use following code p[residuals.index(min(residuals))] ...

python,scipy,params,minimization

Use the callback keyword argument. scipy.optimize.minimize can take a keyword argument callback. This should be a function that accepts, as input, the current vector of parameters. This function is called after every iteration. For instance, from scipy.optimize import minimize def objective_function(xs): """ Function to optimize. """ x, y = xs...

r,regression,mathematical-optimization,linear-programming,minimization

Because Quw and ruw are both data, all constraints as well as the objective are linear in the yuw decision variables. As a result, this is a linear programming problem that can be approached by the lpSolve package. To abstract this out a bit, let's assume R=20 and C=10 describe...

This looks like a modified multi-agent path planning problem. Normally agents are just restricted not to collide or cross paths, but in your version they cannot share vertices anywhere. It wouldn't be hard to modify the CBS algorithm to solve this problem. I'm going to call your k paths agents...