Say I have a NxN numpy matrix. I am looking for the fastest way of extracting all square chunks (sub-matrices) from this matrix. Meaning all CxC parts of the original matrix for 0 < C < N+1. The sub-matrices should correspond to contiguous rows/columns indexes of the original matrix. I want to achieve this in as little time as possible.
You could use Numpy slicing,
import numpy as np
n = 20
x = np.random.rand(n, n)
slice_list = [slice(k, l) for k in range(0, n) for l in range(k, n)]
results = [x[sl,sl] for sl in slice_list]
avoiding loops in Numpy, is not a goal by itself. As long as you are being mindful about it, there shouldn't be much overhead.
Tricky enough, but here is an example of extracting all MxM submatrices in a NxN matrix.
import numpy as NP
import numpy.random as RNG
P = N - M + 1
x = NP.arange(P).repeat(M)
y = NP.tile(NP.arange(M), P) + x
a = RNG.randn(N, N)
b = a[NP.newaxis].repeat(P, axis=0)
c = b[x, y]
d = c.reshape(P, M, N)
e = d[:, NP.newaxis].repeat(P, axis=1)
f = e[:, x, :, y]
g = f.reshape(P, M, P, M)
h = g.transpose(2, 0, 3, 1)
for i in range(0, P):
for j in range(0, P):
assert NP.equal(h[i, j], a[i:i+M, j:j+M]).all()
Related
I have two numpy arrays: A of shape (b, i) and B of shape (b, o). I would like to compute an array R of shape (b, i, o) where every line l of R contains the outer product of the row l of A and the row l of B. So far what i have is:
import numpy as np
A = np.ones((10, 2))
B = np.ones((10, 6))
R = np.asarray([np.outer(a, b) for a, b in zip(A, B)])
assert R.shape == (10, 2, 6)
I think this method is too slow, because of the zip and the final transformation into a numpy array.
Is there a more efficient way to do it ?
That is possible with numpy.matmul, which can do multiplication of "matrix stacks". In this case we want to multiply a stack of column vectors with a stack of row vectors. First bring matrix A to shape (b, i, 1) and B to shape (b, 1, o). Then use matmul to perform b times the outer product:
import numpy as np
i, b, o = 3, 4, 5
A = np.ones((b, i))
B = np.ones((b, o))
print(np.matmul(A[:, :, np.newaxis], B[:, np.newaxis, :]).shape) # (4, 3, 5)
An alternative could be to use numpy.einsum, which can directly represent your index notation:
np.einsum('bi,bo->bio', A, B)
Why not simply
A[:, :, None] * B[:, None, :]
Depending on your convention and your dtype, you might need to throw in another np.conj somewhere. Note that np.newaxis is simply None
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.
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
I encounter a problem when defining a kernel by myself in scikit-learn.
I define by myself the gaussian kernel and was able to fit the SVM but not to use it to make a prediction.
More precisely I have the following code
from sklearn.datasets import load_digits
from sklearn.svm import SVC
from sklearn.utils import shuffle
import scipy.sparse as sparse
import numpy as np
digits = load_digits(2)
X, y = shuffle(digits.data, digits.target)
gamma = 1.0
X_train, X_test = X[:100, :], X[100:, :]
y_train, y_test = y[:100], y[100:]
m1 = SVC(kernel='rbf',gamma=1)
m1.fit(X_train, y_train)
m1.predict(X_test)
def my_kernel(x,y):
d = x - y
c = np.dot(d,d.T)
return np.exp(-gamma*c)
m2 = SVC(kernel=my_kernel)
m2.fit(X_train, y_train)
m2.predict(X_test)
m1 and m2 should be the same, but m2.predict(X_test) return the error :
operands could not be broadcast together with shapes (260,64) (100,64)
I don't understand the problem.
Furthermore if x is one data point, the m1.predict(x) gives a +1/-1 result, as expexcted, but m2.predict(x) gives an array of +1/-1...
No idea why.
The error is at the x - y line. You cannot subtract the two like that, because the first dimensions of both may not be equal. Here is how the rbf kernel is implemented in scikit-learn, taken from here (only keeping the essentials):
def row_norms(X, squared=False):
if issparse(X):
norms = csr_row_norms(X)
else:
norms = np.einsum('ij,ij->i', X, X)
if not squared:
np.sqrt(norms, norms)
return norms
def euclidean_distances(X, Y=None, Y_norm_squared=None, squared=False):
"""
Considering the rows of X (and Y=X) as vectors, compute the
distance matrix between each pair of vectors.
[...]
Returns
-------
distances : {array, sparse matrix}, shape (n_samples_1, n_samples_2)
"""
X, Y = check_pairwise_arrays(X, Y)
if Y_norm_squared is not None:
YY = check_array(Y_norm_squared)
if YY.shape != (1, Y.shape[0]):
raise ValueError(
"Incompatible dimensions for Y and Y_norm_squared")
else:
YY = row_norms(Y, squared=True)[np.newaxis, :]
if X is Y: # shortcut in the common case euclidean_distances(X, X)
XX = YY.T
else:
XX = row_norms(X, squared=True)[:, np.newaxis]
distances = safe_sparse_dot(X, Y.T, dense_output=True)
distances *= -2
distances += XX
distances += YY
np.maximum(distances, 0, out=distances)
if X is Y:
# Ensure that distances between vectors and themselves are set to 0.0.
# This may not be the case due to floating point rounding errors.
distances.flat[::distances.shape[0] + 1] = 0.0
return distances if squared else np.sqrt(distances, out=distances)
def rbf_kernel(X, Y=None, gamma=None):
X, Y = check_pairwise_arrays(X, Y)
if gamma is None:
gamma = 1.0 / X.shape[1]
K = euclidean_distances(X, Y, squared=True)
K *= -gamma
np.exp(K, K) # exponentiate K in-place
return K
You might want to dig deeper into the code, but look at the comments for the euclidean_distances function. A naive implementation of what you're trying to achieve would be this:
def my_kernel(x,y):
d = np.zeros((x.shape[0], y.shape[0]))
for i, row_x in enumerate(x):
for j, row_y in enumerate(y):
d[i, j] = np.exp(-gamma * np.linalg.norm(row_x - row_y))
return d
I am trying to integrate a second order differential equation using 'scipy.integrate.odeint'. My eqution is as follows
m*x[i]''+x[i]'= K/N*sum(j=0 to N)of sin(x[j]-x[i])
which I have converted into two first order ODEs as followed. In the below code, yinit is array of the initial values x(0) and x'(0). My question is what should be the values of x(0) and x'(0) ?
x'[i]=y[i]
y'[i]=(-y[i]+K/N*sum(j=0 to N)of sin(x[j]-x[i]))/m
from numpy import *
from scipy.integrate import odeint
N = 50
def f(theta, t):
global N
x, y = theta
m = 0.95
K = 1.0
fx = zeros(N, float)
for i in range(N):
s = 0.0
for j in range(i+1,N):
s = s + sin(x[j] - x[i])
fx[i] = (-y[i] + (K*s)/N)/m
return array([y, fx])
t = linspace(0, 10, 100, endpoint=False)
Uniformly generating random number
theta = random.uniform(-180, 180, N)
Integrating function f using odeint
yinit = array([x(0), x'(0)])
y = odeint(f, yinit, t)[:,0]
print (y)
You can choose as initial condition whatever you want.
In your case, you decided to use a random initial condition for x for all the oscillators. You can use a random initial condition for 'y' as well I guess, as I did below.
There were a few errors in the above code, mostly on how to unpack x,y from theta and how to repack them at the end (see concatenate below in the corrected code). See also the concatenate for yinit.
The rest are stylish/minor changes.
from numpy import concatenate, linspace, random, mod, zeros, sin
from scipy.integrate import odeint
Nosc = 20
assert mod(Nosc, 2) == 0
def f(theta, _):
N = theta.size / 2
x, y = theta[:N], theta[N:]
m = 0.95
K = 1.0
fx = zeros(N, float)
for i in range(N):
s = 0.0
for j in range(i + 1, N):
s = s + sin(x[j] - x[i])
fx[i] = (-y[i] + (K * s) / N) / m
return concatenate(([y, fx]))
t = linspace(0, 10, 50, endpoint=False)
theta = random.uniform(-180, 180, Nosc)
theta2 = random.uniform(-180, 180, Nosc) #added initial condition for the velocities of the oscillators
yinit = concatenate((theta, theta2))
res = odeint(f, yinit, t)
X = res[:, :Nosc].T
Y = res[:, Nosc:].T
To plot the time evolution of the system, you can use something like
import matplotlib.pylab as plt
fig, ax = plt.subplots()
for displacement in X:
ax.plot(t, displacement)
ax.set_xlabel('t')
ax.set_ylabel('x')
fig.show()
What are you modelling? At first the eq. looked a bit like kuramoto oscillators, but then I noticed you also have a x[i]'' term.
Notice how in your model, as you do not have a spring term in the equation, like a term x(t) at the LHS, the value of x converges to an arbitrary value: