So currently I'm working with code looking like:
Q,R = np.linalg.qr(matrix)
Qb = np.dot(Q.T, new_mu[b][n])
x_qr = np.linalg.solve(R, Qb)
mu.append(x_qr)
The code works fine as long as my matrix is square, but as soon as it's not, the system is not solvable and I got errors. If I've understood it right I can't use linalg.solve on non-full rank matrices, but is there a way for me to get across this obstacle without using a lstsquare solution?
No, this is not possible, as specified in the np.linalg.solve docs.
The issue is that given Ax = b, if A is not square, then your equation is either over-determined or under-determined, assuming that all rows in A are linearly independent. This means that there does not exist a single x that solves this equation.
Intuitively, the idea is that if you have n (length of x) variables that you are trying to solve for, then you need exactly n equations to find a unique solution for x, assuming that these equations are not "redundant". In this case, "redundant" means linearly dependent: one equation is equal to the linear combination of one or more of the other equations.
In this scenario, one possibly useful thing to do is to find the x that minimizes norm(b - Ax)^2 (i.e. linear least squares solution):
x, _, _, _ = np.linalg.lsq(A, b)
Related
I need to solve a set of simultaneous equations of the form Ax = B for x. I've used the numpy.linalg.solve function, inputting A and B, but I get the error 'LinAlgError: Last 2 dimensions of the array must be square'. How do I fix this?
Here's my code:
A = matrix([[v1x, v2x], [v1y, v2y], [v1z, v2z]])
print A
B = [(p2x-p1x-nmag[0]), (p2y-p1y-nmag[1]), (p2z-p1z-nmag[2])]
print B
x = numpy.linalg.solve(A, B)
The values of the matrix/vector are calculated earlier in the code and this works fine, but the values are:
A =
(-0.56666301, -0.52472909)
(0.44034147, 0.46768087)
(0.69641397, 0.71129036)
B =
(-0.38038602567630364, -24.092279373295057, 0.0)
x should have the form (x1,x2,0)
In case you still haven't found an answer, or in case someone in the future has this question.
To solve Ax=b:
numpy.linalg.solve uses LAPACK gesv. As mentioned in the documentation of LAPACK, gesv requires A to be square:
LA_GESV computes the solution to a real or complex linear system of equations AX = B, where A is a square matrix and X and B are rectangular matrices or vectors. Gaussian elimination with row interchanges is used to factor A as A = PL*U , where P is a permutation matrix, L is unit lower triangular, and U is upper triangular. The factored form of A is then used to solve the above system.
If A matrix is not square, it means that you either have more variables than your equations or the other way around. In these situations, you can have the cases of no solution or infinite number of solutions. What determines the solution space is the rank of the matrix compared to the number of columns. Therefore, you first have to check the rank of the matrix.
That being said, you can use another method to solve your system of linear equations. I suggest having a look at factorization methods like LU or QR or even SVD. In LAPACK you can use getrs, in Python you can different things:
first do the factorization like QR and then feed the resulting matrices to a method like scipy.linalg.solve_triangular
solve the least-squares using numpy.linalg.lstsq
Also have a look here where a simple example is formulated and solved.
A square matrix is a matrix with the same number of rows and columns. The matrix you are doing is a 3 by 2. Add a column of zeroes to fix this problem.
I'm working on two functions. I have two data sets, eg [[x(1), y(1)], ..., [x(n), y(n)]], dataSet and testData.
createMatrix(D, S) which returns a data matrix, where D is the degree and S is a vector of real numbers [s(1), s(2), ..., s(n)].
I know numpy has a function called polyfit. But polyfit takes in three variables, any advice on how I'd create the matrix?
polyFit(D), which takes in the polynomial of degree D and fits it to the data sets using linear least squares. I'm trying to return the weight vector and errors. I also know that there is lstsq in numpy.linag that I found in this question: Fitting polynomials to data
Is it possible to use that question to recreate what I'm trying?
This is what I have so far, but it isn't working.
def createMatrix(D, S):
x = []
y = []
for i in dataSet:
x.append(i[0])
y.append(i[1])
polyfit(x, y, D)
What I don't get here is what does S, the vector of real numbers, have to do with this?
def polyFit(D)
I'm basing a lot of this on the question posted above. I'm unsure about how to get just w though, the weight vector. I'll be coding the errors, so that's fine I was just wondering if you have any advice on getting the weight vectors themselves.
It looks like all createMatrix is doing is creating the two vectors required by polyfit. What you have will work, but, the more pythonic way to do it is
def createMatrix(dataSet, D):
D = 3 # set this to whatever degree you're trying
x, y = zip(*dataSet)
return polyfit(x, y, D)
(This S/O link provides a detailed explanation of the zip(*dataSet) idiom.)
This will return a vector of coefficients that you can then pass to something like poly1d to generate results. (Further explanation of both polyfit and poly1d can be found here.)
Obviously, you'll need to decide what value you want for D. The simple answer to that is 1, 2, or 3. Polynomials of higher order than cubic tend to be rather unstable and the intrinsic errors make their output rather meaningless.
It sounds like you might be trying to do some sort of correlation analysis (i.e., does y vary with x and, if so, to what extent?) You'll almost certainly want to just use linear (D = 1) regression for this type of analysis. You can try to do a least squares quadratic fit (D = 2) but, again, the error bounds are probably wider than your assumptions (e.g. normality of distribution) will tolerate.
Given a 2D point p, I'm trying to calculate the smallest distance between that point and a functional curve, i.e., find the point on the curve which gives me the smallest distance to p, and then calculate that distance. The example function that I'm using is
f(x) = 2*sin(x)
My distance function for the distance between some point p and a provided function is
def dist(p, x, func):
x = np.append(x, func(x))
return sum([[i - j]**2 for i,j in zip(x,p)])
It takes as input, the point p, a position x on the function, and the function handle func. Note this is a squared Euclidean distance (since minimizing in Euclidean space is the same as minimizing in squared Euclidean space).
The crucial part of this is that I want to be able to provide bounds for my function so really I'm finding the closest distance to a function segment. For this example my bounds are
bounds = [0, 2*np.pi]
I'm using the scipy.optimize.minimize function to minimize my distance function, using the bounds. A result of the above process is shown in the graph below.
This is a contour plot showing distance from the sin function. Notice how there appears to be a discontinuity in the contours. For convenience, I've plotted a few points around that discontinuity and the "closet" points on the curve that they map to.
What's actually happening here is that the scipy function is finding a local minimum (given some initial guess), but not a global one and that is causing the discontinuity. I know finding the global minimum of any function is impossible, but I'm looking for a more reliable way to find the global minimum.
Possible methods for finding a global minimum would be
Choose a smart initial guess, but this amounts to knowing approximately where the global minimum is to begin with, which is using the solution of the problem to solve it.
Use a multiple initial guesses and choose the answer which gets to the best minimum. This however seems like a poor choice, especially when my functions get more complicated (and higher dimensional).
Find the minimum, then perturb the solution and find the minimum again, hoping that I may have knocked it into a better minimum. I'm hoping that maybe there is some way to do this simply without evoking some complicated MCMC algorithm or something like that. Speed counts for this process.
Any suggestions about the best way to go about this, or possibly directions to useful functions that may tackle this problem would be great!
As suggest in a comment, you could try a global optimization algorithm such as scipy.optimize.differential_evolution. However, in this case, where you have a well-defined and analytically tractable objective function, you could employ a semi-analytical approach, taking advantage of the first-order necessary conditions for a minimum.
In the following, the first function is the distance metric and the second function is (the numerator of) its derivative w.r.t. x, that should be zero if a minimum occurs at some 0<x<2*np.pi.
import numpy as np
def d(x, p):
return np.sum((p-np.array([x,2*np.sin(x)]))**2)
def diff_d(x, p):
return -2 * p[0] + 2 * x - 4 * p[1] * np.cos(x) + 4 * np.sin(2*x)
Now, given a point p, the only potential minimizers of d(x,p) are the roots of diff_d(x,p) (if any), as well as the boundary points x=0 and x=2*np.pi. It turns out that diff_d may have more than one root. Noting that the derivative is a continuous function, the pychebfun library offers a very efficient method for finding all the roots, avoiding cumbersome approaches based on the scipy root-finding algorithms.
The following function provides the minimum of d(x, p) for a given point p:
import pychebfun
def min_dist(p):
f_cheb = pychebfun.Chebfun.from_function(lambda x: diff_d(x, p), domain = (0,2*np.pi))
potential_minimizers = np.r_[0, f_cheb.roots(), 2*np.pi]
return np.min([d(x, p) for x in potential_minimizers])
Here is the result:
I found this chunk of code on http://rosettacode.org/wiki/Multiple_regression#Python, which does a multiple linear regression in python. Print b in the following code gives you the coefficients of x1, ..., xN. However, this code is fitting the line through the origin (i.e. the resulting model does not include a constant).
All I'd like to do is the exact same thing except I do not want to fit the line through the origin, I need the constant in my resulting model.
Any idea if it's a small modification to do this? I've searched and found numerous documents on multiple regressions in python, except they are lengthy and overly complicated for what I need. This code works perfect, except I just need a model that fits through the intercept not the origin.
import numpy as np
from numpy.random import random
n=100
k=10
y = np.mat(random((1,n)))
X = np.mat(random((k,n)))
b = y * X.T * np.linalg.inv(X*X.T)
print(b)
Any help would be appreciated. Thanks.
you only need to add a row to X that is all 1.
Maybe a more stable approach would be to use a least squares algorithm anyway. This can also be done in numpy in a few lines. Read the documentation about numpy.linalg.lstsq.
Here you can find an example implementation:
http://glowingpython.blogspot.de/2012/03/linear-regression-with-numpy.html
What you have written out, b = y * X.T * np.linalg.inv(X * X.T), is the solution to the normal equations, which gives the least squares fit with a multi-linear model. swang's response is correct (and EMS's elaboration)---you need to add a row of 1's to X. If you want some idea of why it works theoretically, keep in mind that you are finding b_i such that
y_j = sum_i b_i x_{ij}.
By adding a row of 1's, you are are setting x_{(k+1)j} = 1 for all j, which means that you are finding b_i such that:
y_j = (sum_i b_i x_{ij}) + b_{k+1}
because the k+1st x_ij term is always equal to one. Thus, b_{k+1} is your intercept term.
I'd like to linearly fit the data that were NOT sampled independently. I came across generalized least square method:
b=(X'*V^(-1)*X)^(-1)*X'*V^(-1)*Y
The equation is Matlab format; X and Y are coordinates of the data points, and V is a "variance matrix".
The problem is that due to its size (1000 rows and columns), the V matrix becomes singular, thus un-invertable. Any suggestions for how to get around this problem? Maybe using a way of solving generalized linear regression problem other than GLS? The tools that I have available and am (slightly) familiar with are Numpy/Scipy, R, and Matlab.
Instead of:
b=(X'*V^(-1)*X)^(-1)*X'*V^(-1)*Y
Use
b= (X'/V *X)\X'/V*Y
That is, replace all instances of X*(Y^-1) with X/Y. Matlab will skip calculating the inverse (which is hard, and error prone) and compute the divide directly.
Edit: Even with the best matrix manipulation, some operations are not possible (for example leading to errors like you describe).
An example of that which may be relevant to your problem is if try to solve least squares problem under the constraint the multiple measurements are perfectly, 100% correlated. Except in rare, degenerate cases this cannot be accomplished, either in math or physically. You need some independence in the measurements to account for measurement noise or modeling errors. For example, if you have two measurements, each with a variance of 1, and perfectly correlated, then your V matrix would look like this:
V = [1 1; ...
1 1];
And you would never be able to fit to the data. (This generally means you need to reformulate your basis functions, but that's a longer essay.)
However, if you adjust your measurement variance to allow for some small amount of independence between the measurements, then it would work without a problem. For example, 95% correlated measurements would look like this
V = [1 0.95; ...
0.95 1 ];
You can use singular value decomposition as your solver. It'll do the best that can be done.
I usually think about least squares another way. You can read my thoughts here:
http://www.scribd.com/doc/21983425/Least-Squares-Fit
See if that works better for you.
I don't understand how the size is an issue. If you have N (x, y) pairs you still only have to solve for (M+1) coefficients in an M-order polynomial:
y = a0 + a1*x + a2*x^2 + ... + am*x^m