I am trying to perform integration of a function in order to find the mean position. However when I perform the integrating with quad I get problems of dimensions not matching. When I run the function on its own it works without a problem. However, when the function is used by the quad integration scheme it gives the error of dimenions mismatch. I will post the completely functional code below and the error message and I hope someone can tell me whats going wrong and how I can fix it.
Please let me know if anything is unclear so I can add more information.
import numpy as np
import time
from scipy.sparse.linalg import eigsh
from scipy.sparse.linalg import spsolve
from scipy.sparse import diags
from scipy.sparse import identity
import scipy.sparse as sp
from scipy.integrate import quad
x0 = 15
v = 16
sigma2 = 5
tmax = 4
def my_gaussian(x, mu=x0, var=5):
return np.exp(-((x - mu)**2 / (2*var))+(1j*v*x/2))
L = 100
N = 3200
dx = 1/32
x = np.linspace(0, L, N)
func = lambda x: my_gaussian(x)*my_gaussian(x).conjugate()
C,e = quad(func, 0, L)
def task3(x):
psi_0 = (C**-(1/2))*my_gaussian(x)
H = (dx**-2)*diags([-1, 2, -1], [-1, 0, 1], shape=(N, N))
H = sp.lil_matrix(H)
H[0,0]=0.5*H[0,0]
H[N-1,N-1]=0.5*H[N-1,N-1]
lam, phi = eigsh(H, 400, which="SM")
a = phi.T.dot(psi_0)
psi = phi.dot(a*np.exp(-1j*lam*tmax))
return psi*psi.conjugate()
func = lambda x: task3(x)*x
N1,e = quad(func, 0, L)
Initially this seems like a pretty straightforward shape mismatch. If you multiply a 2-d array by a 1-d array, numpy will only broadcast them automatically if the last dimension of the 2-d array is the same size as the 1-d array. Otherwise you have to explicitly reshape the 1-d array to be broadcastable with the 2-d array.
To fix the problem we need to know which arrays have the mismatched shapes. I added this to your code to see:
try:
psi = phi.dot(a * np.exp(-1j*lam*tmax).reshape(-1, 1))
except ValueError:
print('phi.shape:', phi.shape)
print('a.shape:', a.shape)
print('lam.shape:', lam.shape)
raise
The result was
phi.shape: (3200, 400)
a.shape: (400, 3200)
lam.shape: (400,)
So we need to reshape the lam term to be a column vector:
np.exp(-1j*lam*tmax).reshape(-1, 1))
This fixes the shape problem. But it doesn't fix the whole problem, because... the output of task3 is then a (3200,3200) array! This is, of course, not useful for an integration routine that expects a function that returns a single scalar.
This last problem is something you'll have to work out on your own, since I have no way of knowing what your goal is.
Related
When running y = multivariate_normal(np.zeros(d), np.eye(d)).rvs() we obtain a sample of dimension (d, ). However, when d=1 we obtain a scalar, which makes sense since it's 1 dimensional. Unfortunately, I have some piece of code that must work for any number of dimensions, including d=1, and basically takes the dot product of a d dimensional vector x with y. This breaks for d=1. How can I fix it?
import numpy as np
from scipy.stats import multivariate_normal as MVN
def mwe_function(d, x):
"""Minimal Working Example"""
y = MVN(np.zeros(d), np.eye(d)).rvs()
return x # y
mwe_function(2, np.ones(2)) # This works
mwe_function(1, np.ones(1)) # This doesn't
IMPORTANT: I want to avoid if statements. One could simply use scipy.stats.norm in that case, but I want to avoid if statements as they would slow down the code.
You can use np.reshape to fix the shape of your sample. By using -1 to specify the length of the first dimension, you will always get a 1-dimensional array and no scalar.
import numpy as np
from scipy.stats import multivariate_normal as MVN
def mwe_function(d, x):
"""Minimal Working Example"""
y = MVN(np.zeros(d), np.eye(d)).rvs().reshape([-1])
return x # y
v0 = mwe_function(2, np.ones(2)) # This works
print(v0) # -0.5718013906409207
v1 = mwe_function(1, np.ones(1)) # This works as well :-)
print(v1) # -0.20196038784485093
where .reshape([-1]) does the job.
Personally, I prefer reshaping over using np.atleast_1d, since the effect is directly visible - but in the end it is a matter of taste.
Given data with shape = (t,m,n), I need to find a vector variable of shape (n,) that minimizes a convex function of the data and vector. I've used cvxopt (and cvxpy) to perform convex optimizations using 2D input, but it seems like they don't support 3D arrays. Is there a way to implement this convex optimization using these or other similar packages?
Given data with shape (t,m,n) and (t,m) and var with shape (n,), here's a simplification of the type of function I need to minimize:
import numpy as np
obj_func(var,data1,data2):
#data1.shape = (t,m,n)
#data2.shape = (t,m)
#var.shape = (n,)
score = np.sum(data1*var,axis=2) #dot product along axis 2
time_series = np.sum(score*data2,axis=1) #weighted sum along axis 1
return np.sum(time_series)-np.sum(time_series**2) #some function
This seems like it should be a simple convex optimization, but unfortunately these functions aren't supported on N-dimensional arrays in cvxopt/cvxpy. Is there a way to implement this?
I think if you simply reshape data1 to be 2d temporarily you'll be fine, e.g.
import numpy as np
import cvxpy as cp
t, m, n = 10, 8, 6
data1 = np.ones((t, m, n))
data2 = np.ones((t, m))
x = cp.Variable(n)
score = cp.reshape(data1.reshape(-1, n) * x, (t, m))
time_series = cp.sum(cp.multiply(score, data2), axis=1)
expr = cp.sum(time_series) - cp.sum(time_series ** 2)
print(repr(expr))
Outputs:
Expression(CONCAVE, UNKNOWN, ())
My code is running fine for first iteration but after that it outputs the following error:
ValueError: matrix must be 2-dimensional
To the best of my knowledge (which is not much in python), my code is correct. but I don't know, why it is not running correctly for all given iterations. Could anyone help me in this problem.
from __future__ import division
import numpy as np
import math
import matplotlib.pylab as plt
import sympy as sp
from numpy.linalg import inv
#initial guesses
x = -2
y = -2.5
i1 = 0
while i1<5:
F= np.matrix([[(x**2)+(x*y**3)-9],[(3*y*x**2)-(y**3)-4]])
theta = np.sum(F)
J = np.matrix([[(2*x)+y**3, 3*x*y**2],[6*x*y, (3*x**2)-(3*y**2)]])
Jinv = inv(J)
xn = np.array([[x],[y]])
xn_1 = xn - (Jinv*F)
x = xn_1[0]
y = xn_1[1]
#~ print theta
print xn
i1 = i1+1
I believe xn_1 is a 2D matrix. Try printing it you and you will see [[something], [something]]
Therefore to get the x and y, you need to use multidimensional indexing. Here is what I did
x = xn_1[0,0]
y = xn_1[1,0]
This works because within the 2D matrix xn_1 are two single element arrays. Therefore we need to further index 0 to get that single element.
Edit: To clarify, xn_1[1,0] means to index 1 and then take that subarray and index 0 on that. And although according to Scipy it may seem that it should be functionally equivalent to xn_1[1][0], that only applies to the general np.array type and not the np.matrix type. Here is an excellent thread on SO that explains this.
So you should use the xn_1[1,0] way to get the element you want.
xn_1 is a numpy matrix, so it's elements are accessed with the item() method, not like an array. (with []s)
So just change
x = xn_1[0]
y = xn_1[1]
to
x = xn_1.item(0)
y = xn_1.item(1)
I found some examples online showing how to find the null space of a regular matrix in Python, but I couldn't find any examples for a sparse matrix (scipy.sparse.csr_matrix).
By null space I mean x such that M·x = 0, where '·' is matrix multiplication. Does anybody know how to do this?
Furthermore, in my case I know that the null space will consist of a single vector. Can this information be used to improve the efficiency of the method?
This isn't a complete answer yet, but hopefully it will be a starting point towards one. You should be able to compute the null space using a variant on the SVD-based approach shown for dense matrices in this question:
import numpy as np
from scipy import sparse
import scipy.sparse.linalg
def rand_rank_k(n, k, **kwargs):
"generate a random (n, n) sparse matrix of rank <= k"
a = sparse.rand(n, k, **kwargs)
b = sparse.rand(k, n, **kwargs)
return a.dot(b)
# I couldn't think of a simple way to generate a random sparse matrix with known
# rank, so I'm currently using a dense matrix for proof of concept
n = 100
M = rand_rank_k(n, n - 1, density=1)
# # this seems like it ought to work, but it doesn't
# u, s, vh = sparse.linalg.svds(M, k=1, which='SM')
# this works OK, but obviously converting your matrix to dense and computing all
# of the singular values/vectors is probably not feasible for large sparse matrices
u, s, vh = np.linalg.svd(M.todense(), full_matrices=False)
tol = np.finfo(M.dtype).eps * M.nnz
null_space = vh.compress(s <= tol, axis=0).conj().T
print(null_space.shape)
# (100, 1)
print(np.allclose(M.dot(null_space), 0))
# True
If you know that x is a single row vector then in principle you would only need to compute the smallest singular value/vector of M. It ought to be possible to do this using scipy.sparse.linalg.svds, i.e.:
u, s, vh = sparse.linalg.svds(M, k=1, which='SM')
null_space = vh.conj().ravel()
Unfortunately, scipy's svds seems to be badly behaved when finding small singular values of singular or near-singular matrices and usually either returns NaNs or throws an ArpackNoConvergence error.
I'm not currently aware of an alternative implementation of truncated SVD with Python bindings that will work on sparse matrices and can selectively find the smallest singular values - perhaps someone else knows of one?
Edit
As a side note, the second approach seems to work reasonably well using MATLAB or Octave's svds function:
>> M = rand(100, 99) * rand(99, 100);
% svds converges much more reliably if you set sigma to something small but nonzero
>> [U, S, V] = svds(M, 1, 1E-9);
>> max(abs(M * V))
ans = 1.5293e-10
I have been trying to find a solution to the same problem. Using Scipy's svds function provides unreliable results for small singular values. Therefore i have been using QR decomposition instead. The sparseqr https://github.com/yig/PySPQR provides a wrapper for Matlabs SuiteSparseQR method, and works reasonably well. Using this the null space can be calculated as:
from sparseqr import qr
Q, _, _,r = qr( M.transpose() )
N = Q.tocsr()[:,r:]
What's the easiest way to get the DFT matrix for 2-d DFT in python? I could not find such function in numpy.fft. Thanks!
The easiest and most likely the fastest method would be using fft from SciPy.
import scipy as sp
def dftmtx(N):
return sp.fft(sp.eye(N))
If you know even faster way (might be more complicated) I'd appreciate your input.
Just to make it more relevant to the main question - you can also do it with numpy:
import numpy as np
dftmtx = np.fft.fft(np.eye(N))
When I had benchmarked both of them I have an impression scipy one was marginally faster but I
have not done it thoroughly and it was sometime ago so don't take my word for it.
Here's pretty good source on FFT implementations in python:
http://nbviewer.ipython.org/url/jakevdp.github.io/downloads/notebooks/UnderstandingTheFFT.ipynb
It's rather from speed perspective, but in this case we can actually see that sometimes it comes with simplicity too.
I don't think this is built in. However, direct calculation is straightforward:
import numpy as np
def DFT_matrix(N):
i, j = np.meshgrid(np.arange(N), np.arange(N))
omega = np.exp( - 2 * pi * 1J / N )
W = np.power( omega, i * j ) / sqrt(N)
return W
EDIT For a 2D FFT matrix, you can use the following:
x = np.zeros(N, N) # x is any input data with those dimensions
W = DFT_matrix(N)
dft_of_x = W.dot(x).dot(W)
As of scipy 0.14 there is a built-in scipy.linalg.dft:
Example with 16 point DFT matrix:
>>> import scipy.linalg
>>> import numpy as np
>>> m = scipy.linalg.dft(16)
Validate unitary property, note matrix is unscaled thus 16*np.eye(16):
>>> np.allclose(np.abs(np.dot( m.conj().T, m )), 16*np.eye(16))
True
For 2D DFT matrix, it's just a issue of tensor product, or specially, Kronecker Product in this case, as we are dealing with matrix algebra.
>>> m2 = np.kron(m, m) # 256x256 matrix, flattened from (16,16,16,16) tensor
Now we can give it a tiled visualization, it's done by rearranging each row into a square block
>>> import matplotlib.pyplot as plt
>>> m2tiled = m2.reshape((16,)*4).transpose(0,2,1,3).reshape((256,256))
>>> plt.subplot(121)
>>> plt.imshow(np.real(m2tiled), cmap='gray', interpolation='nearest')
>>> plt.subplot(122)
>>> plt.imshow(np.imag(m2tiled), cmap='gray', interpolation='nearest')
>>> plt.show()
Result (real and imag part separately):
As you can see they are 2D DFT basis functions
Link to documentation
#Alex| is basically correct, I add here the version I used for 2-d DFT:
def DFT_matrix_2d(N):
i, j = np.meshgrid(np.arange(N), np.arange(N))
A=np.multiply.outer(i.flatten(), i.flatten())
B=np.multiply.outer(j.flatten(), j.flatten())
omega = np.exp(-2*np.pi*1J/N)
W = np.power(omega, A+B)/N
return W
Lambda functions work too:
dftmtx = lambda N: np.fft.fft(np.eye(N))
You can call it by using dftmtx(N). Example:
In [62]: dftmtx(2)
Out[62]:
array([[ 1.+0.j, 1.+0.j],
[ 1.+0.j, -1.+0.j]])
If you wish to compute the 2D DFT as a single matrix operation, it is necessary to unravel the matrix X on which you wish to compute the DFT into a vector, as each output of the DFT has a sum over every index in the input, and a single square matrix multiplication does not have this ability. Taking care to be sure we are handling the indices correctly, I find the following works:
M = 16
N = 16
X = np.random.random((M,N)) + 1j*np.random.random((M,N))
Y = np.fft.fft2(X)
W = np.zeros((M*N,M*N),dtype=np.complex)
hold = []
for m in range(M):
for n in range(N):
hold.append((m,n))
for j in range(M*N):
for i in range(M*N):
k,l = hold[j]
m,n = hold[i]
W[j,i] = np.exp(-2*np.pi*1j*(m*k/M + n*l/N))
np.allclose(np.dot(W,X.ravel()),Y.ravel())
True
If you wish to change the normalization to orthogonal, you can divide by 1/sqrt(MN) or if you wish to have the inverse transformation, just change the sign in the exponent.
This might be a little late, but there is a better alternative for creating the DFT matrix, that performs faster, using NumPy's vander
also, this implementation does not use loops (explicitly)
def dft_matrix(signal):
N = signal.shape[0] # num of samples
w = np.exp((-2 * np.pi * 1j) / N) # remove the '-' for inverse fourier
r = np.arange(N)
w_matrix = np.vander(w ** r, increasing=True) # faster than meshgrid
return w_matrix
if I'm not mistaken, the main improvement is that this method generates the elements of the power from the (already calculated) previous elements
you can read about vander in the documentation:
numpy.vander