I have been getting seemingly unacceptably high inaccuracies when computing matrix inverses (solving a linear system) in numpy.
Is this a normal level of inaccuracy?
How can I improve the accuracy of this computation?
Also, is there a way to solve this system more efficiently in numpy or scipy (scipy.linalg.cho_solve seemed promising but does not do what I want)?
In the code below, cholM is a 128 x 128 matrix. The matrix data is too large to include here but is located on pastebin: cholM.txt.
Also, the original vector, ovec, is being randomly selected, so for different ovec's the accuracy varies, but, for most cases, the error still seems unacceptably high.
Edit Solving the system using the singular value decomposition produces significantly lower error than the other methods.
import numpy.random as rnd
import numpy.linalg as lin
import numpy as np
cholM=np.loadtxt('cholM.txt')
dims=len(cholM)
print 'Dimensions',dims
ovec=rnd.normal(size=dims)
rvec=np.dot(cholM.T,ovec)
invCholM=lin.inv(cholM.T)
svec=np.dot(invCholM,rvec)
svec1=lin.solve(cholM.T,rvec)
def back_substitute(M,v):
r=np.zeros(len(v))
k=len(v)-1
r[k]=v[k]/M[k,k]
for k in xrange(len(v)-2,-1,-1):
r[k]=(v[k]-np.dot(M[k,k+1:],r[k+1:]))/M[k,k]
return r
svec2=back_substitute(cholM.T,rvec)
u,s,v=lin.svd(cholM)
svec3=np.dot(u,np.dot(np.diag(1./s),np.dot(v,rvec)))
for k in xrange(dims):
print '%20.3f%20.3f%20.3f%20.3f'%(ovec[k]-svec[k],ovec[k]-svec1[k],ovec[k]-svec2[k],ovec[k]-svec3[k])
assert np.all( np.abs(ovec-svec)<1e-5 )
assert np.all( np.abs(ovec-svec1)<1e-5 )
As noted by #Craig J Copi and #pv, the condition number of the cholM matrix is high, around 10^16, indicating that to achieve higher accuracy in the inverse, much greater numerical precision may be required.
Condition number can be determined by the ratio of maximum singular value to minimum singular value. In this instance, this ratio is not the same as the ratio of eigenvalues.
http://docs.scipy.org/doc/scipy/reference/tutorial/linalg.html
We could find the solution vector using a matrix inverse:
...
However, it is better to use the linalg.solve command which can be faster and more numerically stable
edit - from Steve Lord at MATLAB
http://www.mathworks.com/matlabcentral/newsreader/view_thread/63130
Why are you inverting? If you're inverting to solve a system, don't --
generally you would want to use backslash instead.
However, for a system with a condition number around 1e17 (condition numbers
must be greater than or equal to 1, so I assume that the 1e-17 figure in
your post is the reciprocal condition number from RCOND) you're not going to
get a very accurate result in any case.
Related
I'm trying to understand the difference between numpy fft and rfft. I've read the doc, and it only says rfft is meant for real inputs.
I've tested their performance on a large real array and found out rfft is faster than fft by about a third. My question is: why is rfft fast? Thanks!
An RFFT has half the degrees of freedom on the input, and half the number of complex outputs, compared to an FFT. Thus the FFT computation tree can be pruned to remove those adds and multiplies not needed for the non-existent inputs and/or those unnecessary since there are a lesser number of independant output values that need to be computed.
This is because an FFT of a strictly real input (e.g. all the input value imaginary components zero) produces a complex conjugate mirrored result, where each half can be trivially derived from the other half.
at the risk of using the wrong SE...
My question concerns the rcond parameter in numpy.linalg.lstsq(a, b, rcond). I know it's used to define the cutoff for small singular values in the singular value decomposition when numpy computed the pseudo-inverse of a.
But why is it advantageous to set "small" singular values (values below the cutoff) to zero rather than just keeping them as small numbers?
PS: Admittedly, I don't know exactly how the pseudo-inverse is computed and exactly what role SVD plays in that.
Thanks!
To determine the rank of a system, you could need compare against zero, but as always with floating point arithmetic we should compare against some finite range. Hitting exactly 0 never happens when adding and subtracting messy numbers.
The (reciprocal) condition number allows you to specify such a limit in a way that is independent of units and scale in your matrix.
You may also opt of of this feature in lstsq by specifying rcond=None if you know for sure your problem can't run the risk of being under-determined.
The future default sounds like a wise choice from numpy
FutureWarning: rcond parameter will change to the default of machine precision times max(M, N) where M and N are the input matrix dimensions
because if you hit this limit, you are just fitting against rounding errors of floating point arithmetic instead of the "true" mathematical system that you hope to model.
I'm trying to use linalg to find $P^{500}$ where $ P$ is a 9x9 matrix but Python displays the following:
Matrix full of inf
I think this is too much for this method so my question is, there is annother library to find $P^{500}$? Must I surrender?
Thank you all in advance
Use the eigendecomposition and then exponentiate the matrix of eigenvalues. Like this. You end up getting an inf up in the first column. Unless you control the type of matrix by their eigenvalues this won't happen I believe. In other words, your eigenvalues have to be bounded. You can generate a random matrix by the Schur decomposition putting the eigenvalues along the diagonal. This is a post I have about generating a matrix with given eigenvalues. This should be the way that method works anyways.
% Generate random 9x9 matrix
n=9;
A = randn(n);
[V,D] = eig(A);
p = 500;
Dp = D^p;
Ap = V^(-1)*Dp*V;
Ap1 = mpower(A,p);
NumPy arrays have homogeneous data types and float datatype maximum is
>>> np.finfo('d').max
1.7976931348623157e+308
>>> _**0.002
4.135322944991858
>>> np.array(4.135)**500
1.7288485271474026e+308
>>> np.array(4.136)**500
__main__:1: RuntimeWarning: overflow encountered in power
inf
So if there is an inner product that results higher than approx. 4.135 it is going to blowup and once it blows up, the next product will be multiplied with infinities and more entries will get infinities until everything becomes infinities.
Metahominid's suggestion certainly helps but it will not solve the issue if your eigenvalues are larger than this value. In general, you need to use specialized high-precision tools to get correct results.
I have a real signal in time given by:
And I am simply trying to compute its power spectrum, which is the Fourier transform of the autocorrelation of the signal, and is also a purely real and positive quantity in this case. To do this, I simply write:
import numpy as np
from scipy.fftpack import fft, arange, rfftfreq, rfft
from pylab import *
lags1, c1, line1, b1 = acorr(((Y_DATA)), usevlines=False, normed=True, maxlags=3998, lw=2)
Power_spectrum = (fft(np.real(c1)))
freqs = np.fft.fftfreq(len(c1), dx)
plt.plot(freqs,Power_spectrum)
plt.xlabel('f (Hz)')
plt.xlim([-20000,20000])
plt.show()
But the output gives:
which has negative-valued output. Although if I simply take the absolute value of the data on the y-axis and plot it (i.e. np.abs(Power_spectrum)), then the output is:
which is exactly what I expect. Although why is this only fixed by taking the absolute value of my power spectrum? I checked my autocorrelation and plotted it—it seems to be working as expected and matches what others have computed.
Although what appears odd is the next step when I take the FFT. The FFT function outputs negative values which is contrary to the theory discussed in the link above and I don't quite understand why. Any thoughts on what is going wrong?
The power spectrum is the FFT of the autocorrelation, but that's not an efficient way to calculate it.
The autocorrelation is probably calculated with an FFT and iFFT, anyway.
The power spectrum is also just the squared magnitude of the FFT coefficients.
Do that instead so that the total work will be one FFT instead of 3.
An fft produces a complex result (real and imaginary components to represent both magnitude and phase of the spectrum). You have to take the (squared) magnitude of the complex vector to get the power spectrum.
I'm trying to solve a large eigenvalue problem with Scipy where the matrix A is dense but I can compute its action on a vector without having to assemble A explicitly. So in order to avoid memory issues when the matrix A gets big I'd like to use the sparse solver scipy.sparse.linalg.eigs with a LinearOperator that implemements this action.
Applying eigs to an explicit numpy array A works fine. However, if I apply eigs to a LinearOperator instead then the iterative solver fails to converge. This is true even if the matvec method of the LinearOperator is simply matrix-vector multiplication with the given matrix A.
A minimal example illustrating the failure is attached below (I'm using shift-invert mode because I am interested in the smallest few eigenvalues). This computes the eigenvalues of a random matrix A just fine, but fails when applied to a LinearOperator that is directly converted from A. I tried to fiddle with the parameters for the iterative solver (v0, ncv, maxiter) but to no avail.
Am I missing something obvious? Is there a way to make this work? Any suggestions would be highly appreciated. Many thanks!
Edit: I should clarify what I mean by "make this work" (thanks, Dietrich). The example below uses a random matrix for illustration. However, in my application I know that the eigenvalues are almost purely imaginary (or almost purely real if I multiply the matrix by 1j). I'm interested in the 10-20 smallest-magnitude eigenvalues, but the algorithm doesn't behave well (i.e., never stops even for small-ish matrix sizes) if I specify which='SM'. Therefore I'm using shift-invert mode by passing the parameters sigma=0.0, which='LM'. I'm happy to try a different approach so long as it allows me to compute a bunch of smallest-magnitude eigenvalues.
from scipy.sparse.linalg import eigs, LinearOperator, aslinearoperator
import numpy as np
# Set a seed for reproducibility
np.random.seed(0)
# Size of the matrix
N = 100
# Generate a random matrix of size N x N
# and compute its eigenvalues
A = np.random.random_sample((N, N))
eigvals = eigs(A, sigma=0.0, which='LM', return_eigenvectors=False)
print eigvals
# Convert the matrix to a LinearOperator
A_op = aslinearoperator(A)
# Try to solve the same eigenproblem again.
# This time it produces an error:
#
# ValueError: Error in inverting M: function gmres did not converge (info = 1000).
eigvals2 = eigs(A_op, sigma=0.0, which='LM', return_eigenvectors=False)
I tried running your code, but not passing the sigma parameter to eigs() and it ran without problems (read eigs() docs for its meaning). I didn't see the benefit of it in your example.
Eigs can already find the smallest eigenvalues first. Set which = 'SM'