Polynomial interpolation in python - sympy

I am studying function approximation and while trying to understand/implement polynomial interpolation I've found an example here. I find the code below a good example to understand what is actually going on instead of using ready functions, however it doesn't run:
Defining the interpolation algorithm. Essentially, we are trying to come up with a representation of true f as a linear combination of basis functions (psi-s).
import sympy as sym
def interpolation(f, psi, points):
N = len(psi) - 1 #order of the approximant polynomial
A = sym.zeros((N+1, N+1)) # initiating the square matrix, whose regular element is psi evaluated at each nodes
b = sym.zeros((N+1, 1)) # original function f evaluated at the selected nodes
psi_sym = psi # save symbolic expression
# Turn psi and f into Python functions
x = sym.Symbol('x')
psi = [sym.lambdify([x], psi[i]) for i in range(N+1)]
f = sym.lambdify([x], f)
for i in range(N+1):
for j in range(N+1):
A[i,j] = psi[j](points[i])
b[i,0] = f(points[i])
c = A.LUsolve(b) #finding the accurate weights for each psi
# c is a sympy Matrix object, turn to list
c = [sym.simplify(c[i,0]) for i in range(c.shape[0])]
u = sym.simplify(sum(c[i,0]*psi_sym[i] for i in range(N+1)))
return u, c
True function f:= 10(x-1)^2 -1, nodes: x0:= 1 + 1/3 and x1 = 1 + 2/3. Interval: [1,2].
x = sym.Symbol('x')
f = 10*(x-1)**2 - 1
psi = [1, x] # approximant polynomial of order 1 (linear approximation)
Omega = [1, 2] #interval
points = [1 + sym.Rational(1,3), 1 + sym.Rational(2,3)]
u, c = interpolation(f, psi, points)
comparison_plot(f, u, Omega)
The code doesn't run. The error occurs in line
A = sym.zeros((N+1, N+1))
Error message: ValueError: (2, 2) is not an integer
But A isn't supposed to be an integer, it is a square matrix whose each element is psi evaluated at each node. f = A*c.
Thank you!!!

Related

Component reconstruction for multivariate lagged time series

I am trying to write a multivariate Singular Spectrum Analysis with Monte Carlo test. To this extent I am working on a code piece that can reconstruct the input series using the lagged trajectory matrix and projection base (ST-PCs) that result from the pca/ssa decomposition of the input series. The attached code piece works for a lagged univariate (that is, single) time series, but I am struggling to make this reconstruction for a lagged multivariate time series. I don't quite get the procedure mathematically and - not surprisingly - I also did not manage to program it. Useful links are attached to the function descriptions of the accompanying code. Input data should be of the form (time * number of series), so say 288x3 implying 3 time series of 288 time levels.
I hope you can help me out!
import numpy as np
def lagged_covariance_matrix(data, M):
""" Computes the lagged covariance matrix using the Broomhead & King method
Background: Plaut, G., & Vautard, R. (1994). Spells of low-frequency oscillations and
weather regimes in the Northern Hemisphere. Journal of the atmospheric sciences, 51(2), 210-236.
Arguments:
data : pxn time series, where p denotes the length of the time series and n the number of channels
M : window length """
# explicitely 'add' spatial dimension if input is a single time series
if np.ndim(data) == 1:
data = np.reshape(data,(len(data),1))
T = data.shape[0]
L = data.shape[1]
N = T - M + 1
X = np.zeros((T, L, M))
for i in range(M):
X[:,:,i] = np.roll(data, -i, axis = 0)
X = X[:N]
# X constitutes the trajectory matrix and is a stacked hankel matrix
X = np.reshape(X, (N, M*L), order = 'C') # https://www.jstatsoft.org/article/viewFile/v067i02/v67i02.pdf
# choose the smallest projection basis for computation of the covariance matrix
if M*L >= N:
return 1/(M*L) * X.dot(X.T), X
else:
return 1/N * X.T.dot(X), X
def sort_by_eigenvalues(eigenvalues, PCs):
""" Sorts the PCs and eigenvalues by descending size of the eigenvalues """
desc = np.argsort(-eigenvalues)
return eigenvalues[desc], PCs[:,desc]
def Reconstruction(M, E, X):
""" Reconstructs the series as the sum of M subseries.
See: https://en.wikipedia.org/wiki/Singular_spectrum_analysis, 'Basic SSA' &
the work of Vivien Sainte Fare Garnot on univariate time series (https://github.com/VSainteuf/mcssa)
Arguments:
M : window length
E : eigenvector basis
X : trajectory matrix """
time = len(X) + M - 1
RC = np.zeros((time, M))
# step 3: grouping
for i in range(M):
d = np.zeros(M)
d[i] = 1
I = np.diag(d)
Q = np.flipud(X # E # I # E.T)
# step 4: diagonal averaging
for k in range(time):
RC[k, i] = np.diagonal(Q, offset = -(time - M - k)).mean()
return RC
#=====================================================================================================
#=====================================================================================================
#=====================================================================================================
# input data
data = None
# number of lags a.k.a. window length
M = 45 # M = 1 means no lag
covmat, X = lagged_covariance_matrix(data, M)
# get the eigenvalues and vectors of the covariance matrix
vals, vecs = np.linalg.eig(covmat)
eig_data, eigvec_data = sort_by_eigenvalues(vals, vecs)
# component reconstruction
recons_data = Reconstruction(M, eigvec_data, X)
The following works but does not make direct use of the projection base (ST-PCs). Hence the original question still stands, but this already helps a great lot and solves the problem for me. This code piece makes use of the similarity between the ST-PCs projection base and the u & vt matrices obtained from the single value decomposition of the lagged trajectory matrix. I think it gives back the same answer as one would obtain using the ST-PCs projection base?
def lag_reconstruction(data, X, M, pairs = None):
""" Reconstructs the series as the sum of M subseries using the lagged trajectory matrix.
Based on equation 2.9 of Plaut, G., & Vautard, R. (1994). Spells of low-frequency oscillations and weather regimes in the Northern Hemisphere. Journal of Atmospheric Sciences, 51(2), 210-236.
Inspired by work of R. van Westen and C. Wieners """
time = data.shape[0] # number of time levels of the original series
L = data.shape[1] # number of input series
N = time - M + 1
u, s, vt = np.linalg.svd(X, full_matrices = False)
rc = np.zeros((time, L, M))
for t in range(time):
counter = 0
for i in range(M):
if t-i >= 0 and t-i < N:
counter += 1
if pairs:
for k in pairs:
rc[t,:,i] += u[t-i, k] * s[k] * vt[k, i*L : i*L + L]
else:
for k in range(len(s)):
rc[t,:,i] += u[t-i, k] * s[k] * vt[k, i*L : i*L + L]
rc[t] = rc[t]/counter
return rc

Solving a matrix equation containing MatrixSymbols of symbolic size in Sympy?

As an introduction i want to point out that if one has a matrix A consisting of 4 submatrices in a 2x2 pattern, where the diagonal matrices are square, then if we denote its inverse as X, the submatrix X22 = (A22-A21(A11^-1)A12)^-1, which is quite easy to show by hand.
I was trying to do the same for a matrix of 4x4 submatrices, but its quite tedious by hand. So I thought Sympy would be of some help. But I cannot figure out how (I have started by just trying to reproduce the 2x2 result).
I've tried:
import sympy as s
def blockmatrix(name, sizes, names=None):
if names is None:
names = sizes
ll = []
for i, (s1, n1) in enumerate(zip(sizes, names)):
l = []
for j, (s2, n2) in enumerate(zip(sizes, names)):
l.append(s.MatrixSymbol(name+str(n1)+str(n2), s1, s2))
ll.append(l)
return ll
def eyes(*sizes):
ll = []
for i, s1 in enumerate(sizes):
l = []
for j, s2 in enumerate(sizes):
if i==j:
l.append(s.Identity(s1))
continue
l.append(s.ZeroMatrix(s1, s2))
ll.append(l)
return ll
n1, n2 = s.symbols("n1, n2", integer=True, positive=True, nonzero=True)
M = s.Matrix(blockmatrix("m", (n1, n2)))
X = s.Matrix(blockmatrix("x", (n1, n2)))
I = s.Matrix(eyes(n1, n2))
s.solve(M*X[:, 1:]-I[:, 1:], X[:, 1:])
but it just returns an empty list instead of the result.
I have also tried:
Using M*X==I but that just returns False (boolean, not an Expression)
Entering a list of equations
Using 'ordinary' symbols with commutative=False instead of MatrixSymbols -- this gives an exception with GeneratorsError: non-commutative generators: (x12, x22)
but all without luck.
Can you show how to derive a result with Sympy similar to the one I gave as an example for X22?
The most similar other questions on solving with MatrixSymbols seem to have been solved by working around doing exactly that, by using an array of the inner symbols or some such instead. But since I am dealing with symbolically sized MatrixSymbols, that is not an option for me.
Is this what you mean by a matrix of 2x2 matrices?
>>> a = [MatrixSymbol(i,2,2) for i in symbols('a1:5')]
>>> A = Matrix(2,2,a)
>>> X = A.inv()
>>> print(X[1,1]) # [1,1] instead of [2,2] because indexing starts at 0
a1*(a1*a3 - a3*a1)**(-1)
[You indicated not and pointed out that the above is not correct -- that appears to be an issue that should be resolved.]
I am not sure why this isn't implemented, but we can do the solving manually as follows:
>>> n = 2
>>> v = symbols('b:%s'%n**2,commutative=False)
>>> A = Matrix(n,n,symbols('a:%s'%n**2,commutative=False))
>>> B = Matrix(n,n,v)
>>> eqs = list(A*B - eye(n))
>>> for i in range(n**2):
... s = solve(eqs[i],v[i])[0]
... eqs[i+1:] = [e.subs(v[i],s) for e in eqs[i+1:]]
...
>>> s # solution for v[3] which is B22
(-a2*a0**(-1)*a1 + a3)**(-1)
You can change n to 3 and see a modestly complicated expression. Change it to 4 and check the result by hand to give a new definition to the word "tedious" ;-)
The special structure of the equations to be solved can allow for a faster solution, too: the variable of interest is the last factor in each term containing it:
>>> for i in range(n**2):
... c,d = eqs[i].expand().as_independent(v[i])
... assert all(j.args[-1]==v[i] for j in Add.make_args(d))
... s = 1/d.subs(v[i], 1)*-c
... eqs[i+1:] = [e.subs(v[i], s) for e in eqs[i+1:]]

Tensor contraction with Kronecker deltas in sympy

I'm trying to use sympy to do some index gymnastics for me. I'm trying to calculate the derivatives of a cost function that looks like
cost = sumi (Mii)2
where M is given by a rotation
Mij = U*ki M0kl Ulj
I've written up a parametrization for the rotation matrix, from which I get the derivatives as products of Kronecker deltas. What I've got so far is
def Uder(p,q,r,s):
return KroneckerDelta(p,r)*KroneckerDelta(q,s) - KroneckerDelta(p,s)*KroneckerDelta(q,r)
from sympy import *
# Matrix size
n = symbols('n')
p = symbols('p');
i = Dummy('i')
k = Dummy('k')
l = Dummy('l')
# Matrix elements
M0 = IndexedBase('M')
U = IndexedBase('U')
# Indices
r, s = map(tensor.Idx, ['r', 's'])
# Derivative
cost_x = Sum(Sum(Sum(M0[i,i]*(Uder(k,i,r,s)*M0[k,l]*U[l,i] + U[k,i]*M0[k,l]*Uder(l,i,r,s)),(k,1,n)),(l,1,n)),(i,1,n))
print cost_x
but sympy is not evaluating the contractions for me, which should reduce to simple sums in terms of r and s, which are the rotation indices. Instead, what I get is
Sum(((-KroneckerDelta(_i, r)*KroneckerDelta(_k, s) + KroneckerDelta(_i, s)*KroneckerDelta(_k, r))*M[_k, _l]*U[_l, _i] + (-KroneckerDelta(_i, r)*KroneckerDelta(_l, s) + KroneckerDelta(_i, s)*KroneckerDelta(_l, r))*M[_k, _l]*U[_k, _i])*M[_i, _i], (_k, 1, n), (_l, 1, n), (_i, 1, n))
I'm using the latest git snapshot 4633fd5713c434c3286e3412a2399bd40fbd9569 of sympy.

Solving Differential Equation Sympy

I haven't been able to find particular solutions to this differential equation.
from sympy import *
m = float(raw_input('Mass:\n> '))
g = 9.8
k = float(raw_input('Drag Coefficient:\n> '))
v = Function('v')
f1 = g * m
t = Symbol('t')
v = Function('v')
equation = dsolve(f1 - k * v(t) - m * Derivative(v(t)), 0)
print equation
for m = 1000 and k = .2 it returns
Eq(f(t), C1*exp(-0.0002*t) + 49000.0)
which is correct but I want the equation solved for when v(0) = 0 which should return
Eq(f(t), 49000*(1-exp(-0.0002*t))
I believe Sympy is not yet able to take into account initial conditions. Although dsolve has the option ics for entering initial conditions (see the documentation), it appears to be of limited use.
Therefore, you need to apply the initial conditions manually. For example:
C1 = Symbol('C1')
C1_ic = solve(equation.rhs.subs({t:0}),C1)[0]
print equation.subs({C1:C1_ic})
Eq(v(t), 49000.0 - 49000.0*exp(-0.0002*t))

Matrix multiplication with Python

I have a numerical analysis assignment and I need to find some coefficients by multiplying matrices. We were given an example in Mathcad, but now we have to do it in another programming language so I chose Python.
The problem is, that I get different results by multiplying matrices in respective environments. Here's the function in Python:
from numpy import *
def matrica(C, n):
N = len(C) - 1
m = N - n
A = [[0] * (N + 1) for i in range(N+1)]
A[0][0] = 1
for i in range(0, n + 1):
A[i][i] = 1
for j in range(1, m + 1):
for i in range(0, N + 1):
if i + j <= N:
A[i+j][n+j] = A[i+j][n+j] - C[i]/2
A[int(abs(i - j))][n+j] = A[int(abs(i - j))][n+j] - C[i]/2
M = matrix(A)
x = matrix([[x] for x in C])
return [float(y) for y in M.I * x]
As you can see I am using numpy library. This function is consistent with its analog in Mathcad until return statement, the part where matrices are multiplied, to be more specific. One more observation: this function returns correct matrix if N = 1.
I'm not sure I understand exactly what your code do. Could you explain a little more, like what math stuff you're actually computing. But if you want a plain regular product and if you use a numpy.matrix, why don't you use the already written matrix product?
a = numpy.matrix(...)
b = numpy.matrix(...)
p = a * b #matrix product