A bit of background:
I want to calculate the array factor of a MxN antenna array, which is given by the following equation:
Where w_i are the complex weight of the i-th element, (x_i,y_i,z_i) is the position of the i-th element, k is the wave number, theta and phi are the elevation and azimuth respectively, and i ranges from 0 to MxN-1.
In the code I have:
-theta and phi are np.mgrid with shape (200,200) each,
-w_i, and (x,y,z)_i are np.array with shape (NxM,) each
so AF is a np.array with shape (200,200) (sum over i).There is no problem so far, and I can get AF easily doing:
af = zeros([theta.shape[0],phi.shape[0]])
for i in range(self.size[0]*self.size[1]):
af = af + ( w[i]*e**(-1j*(k * x_pos[i]*sin(theta)*cos(phi) + k * y_pos[i]* sin(theta)*sin(phi)+ k * z_pos[i] * cos(theta))) )
Now, each w_i depends on frequency, so AF too, and now I have w_i with shape (NxM,1000) (I have 1000 samples of each w_i in frequency). I tried to use the above code changing
af = zeros([1000,theta.shape[0],phi.shape[0]])
but I get 'operands could not be broadcast together'. I can solve this by using a for loop through the 1000 values, but it is slow and is a bit ugly. So, what is the correct way to do the summation, or the correct way to properly define w_i and AF ?
Any help would be appreciated. Thanks.
edit
The code with the new dimension I'm trying to add is the next:
from numpy import *
class AntennaArray:
def __init__(self,f,asize=None,tipo=None,dx=None,dy=None):
self.Lambda = 299792458 / f
self.k = 2*pi/self.Lambda
self.size = asize
self.type = tipo
self._AF_DATA_SIZE = 200
self.theta,self.phi = mgrid[0 : pi : self._AF_DATA_SIZE*1j,0 : 2*pi : self._AF_DATA_SIZE*1j]
self.element_pos = None
self.element_amp = None
self.element_pha = None
if dx == None:
self.dx = self.Lambda/2
else:
self.dx = dx
if dy == None:
self.dy = self.Lambda/2
else:
self.dy = dy
self.generate_array()
def generate_array(self):
M = self.size[0]
N = self.size[1]
dx = self.dx
dy = self.dy
x_pos = arange(0,dx*N,dx)
y_pos = arange(0,dy*M,dy)
z_pos = 0
ele = zeros([N*M,3])
for i in range(M):
ele[i*N:(i+1)*N,0] = x_pos[:]
for i in range(M):
ele[i*N:(i+1)*N,1] = y_pos[i]
self.element_pos = ele
#self.array_factor = self.calculate_array_factor()
def calculate_array_factor(self):
theta,phi = self.theta,self.phi
k = self.k
x_pos = self.element_pos[:,0]
y_pos = self.element_pos[:,1]
z_pos = self.element_pos[:,2]
w = self.element_amp*exp(1j*self.element_pha)
if len(self.element_pha.shape) > 1:
#I have f_size samples of w_i(f)
f_size = self.element_pha.shape[1]
af = zeros([f_size,theta.shape[0],phi.shape[0]])
else:
#I only have w_i
af = zeros([theta.shape[0],phi.shape[0]])
for i in range(self.size[0]*self.size[1]):
**strong text**#This for loop does the summation over i
af = af + ( w[i]*e**(-1j*(k * x_pos[i]*sin(theta)*cos(phi) + k * y_pos[i]* sin(theta)*sin(phi)+ k * z_pos[i] * cos(theta))) )
return af
I tried to test it with the next main
from numpy import *
f_points = 10
M = 2
N = 2
a = AntennaArray(5.8e9,[M,N])
a.element_amp = ones([M*N,f_points])
a.element_pha = zeros([M*N,f_points])
af = a.calculate_array_factor()
But I get
ValueError: 'operands could not be broadcast together with shapes (10,) (200,200) '
Note that if I set
a.element_amp = ones([M*N])
a.element_pha = zeros([M*N])
This works well.
Thanks.
I had a look at the code, and I think this for loop:
af = zeros([theta.shape[0],phi.shape[0]])
for i in range(self.size[0]*self.size[1]):
af = af + ( w[i]*e**(-1j*(k * x_pos[i]*sin(theta)*cos(phi) + k * y_pos[i]* sin(theta)*sin(phi)+ k * z_pos[i] * cos(theta))) )
is wrong in many ways. You are mixing dimensions, you cannot loop that way.
And by the way, to make full use of numpy efficiency, never loop over the arrays. It slows down the execution significantly.
I tried to rework that part.
First, I advice you to not use from numpy import *, it's bad practice (see here). Use import numpy as np. I reintroduced the np abbreviation, so you can understand what comes from numpy.
Frequency independent case
This first snippet assumes that w is a 1D array of length 4: I am neglecting the frequency dependency of w, to show you how you can get what you already obtained without the for loop and using instead the power of numpy.
af_points = w[:,np.newaxis,np.newaxis]*np.e**(-1j*
(k * x_pos[:,np.newaxis,np.newaxis]*np.sin(theta)*np.cos(phi) +
k * y_pos[:,np.newaxis,np.newaxis]*np.sin(theta)*np.sin(phi) +
k * z_pos[:,np.newaxis,np.newaxis]*np.cos(theta)
))
af = np.sum(af_points, axis=0)
I am using numpy broadcasting to obtain a 3D array named af_points, whose shape is (4, 200, 200). To do it, I use np.newaxis to extend the number of axis of an array in order to use broadcasting correctly. More here on np.newaxis.
So, w[:,np.newaxis,np.newaxis] is an array of shape (4, 1, 1). Similarly for x_pos[:,np.newaxis,np.newaxis], y_pos[:,np.newaxis,np.newaxis] and z_pos[:,np.newaxis,np.newaxis]. Since the angles have shape (200, 200), broadcasting can be done, and af_points has shape (4, 200, 200).
Finally the sum is done by np.sum, summing over the first axis to obtain a (200, 200) array.
Frequency dependent case
Now w has shape (4, 10), where 10 are the frequency points. The idea is the same, just consider that the frequency is an additional dimension in your numpy arrays: now af_points will be an array of shape (4, 10, 200, 200) where 10 are the f_points you have defined.
To keep it understandable, I've split the calculation:
#exp_point is only the exponent, frequency independent. Will be a (4, 200, 200) array.
exp_points = np.e**(-1j*
(k * x_pos[:,np.newaxis,np.newaxis]*np.sin(theta)*np.cos(phi) +
k * y_pos[:,np.newaxis,np.newaxis]*np.sin(theta)*np.sin(phi) +
k * z_pos[:,np.newaxis,np.newaxis]*np.cos(theta)
))
af_points = w[:,:,np.newaxis,np.newaxis] * exp_points[:,np.newaxis,:,:]
af = np.sum(af_points, axis=0)
And now af has shape (10, 200, 200).
Related
I'm trying to apply the method for baselinining vibrational spectra, which is announced as an improvement over asymmetric and iterative re-weighted least-squares algorithms in the 2015 paper (doi:10.1039/c4an01061b), where the following matlab code was provided:
function z = baseline(y, lambda, ratio)
% Estimate baseline with arPLS in Matlab
N = length(y);
D = diff(speye(N), 2);
H = lambda*D'*D;
w = ones(N, 1);
while true
W = spdiags(w, 0, N, N);
% Cholesky decomposition
C = chol(W + H);
z = C \ (C' \ (w.*y) );
d = y - z;
% make d-, and get w^t with m and s
dn = d(d<0);
m = mean(d);
s = std(d);
wt = 1./ (1 + exp( 2* (d-(2*s-m))/s ) );
% check exit condition and backup
if norm(w-wt)/norm(w) < ratio, break; end
end
that I rewrote into python:
def baseline_arPLS(y, lam, ratio):
# Estimate baseline with arPLS
N = len(y)
k = [numpy.ones(N), -2*numpy.ones(N-1), numpy.ones(N-2)]
offset = [0, 1, 2]
D = diags(k, offset).toarray()
H = lam * numpy.matmul(D.T, D)
w_ = numpy.ones(N)
while True:
W = spdiags(w_, 0, N, N, format='csr')
# Cholesky decomposition
C = cholesky(W + H)
z_ = spsolve(C.T, w_ * y)
z = spsolve(C, z_)
d = y - z
# make d- and get w^t with m and s
dn = d[d<0]
m = numpy.mean(dn)
s = numpy.std(dn)
wt = 1. / (1 + numpy.exp(2 * (d - (2*s-m)) / s))
# check exit condition and backup
norm_wt, norm_w = norm(w_-wt), norm(w_)
if (norm_wt / norm_w) < ratio:
break
w_ = wt
return(z)
Except for the input vector y the method requires parameters lam and ratio and it runs ok for values lam<1.e+07 and ratio>1.e-01, but outputs poor results. When values are changed outside this range, for example lam=1e+07, ratio=1e-02 the CPU starts heating up and job never finishes (I interrupted it after 1min). Also in both cases the following warning shows up:
/usr/local/lib/python3.9/site-packages/scipy/sparse/linalg/dsolve/linsolve.py: 144: SparseEfficencyWarning: spsolve requires A to be CSC or CSR matrix format warn('spsolve requires A to be CSC or CSR format',
although I added the recommended format='csr' option to the spdiags call.
And here's some synthetic data (similar to one in the paper) for testing purposes. The noise was added along with a 3rd degree polynomial baseline The method works well for parameters bl_1 and fails to converge for bl_2:
import numpy
from matplotlib import pyplot
from scipy.sparse import spdiags, diags, identity
from scipy.sparse.linalg import spsolve
from numpy.linalg import cholesky, norm
import sys
x = numpy.arange(0, 1000)
noise = numpy.random.uniform(low=0, high = 10, size=len(x))
poly_3rd_degree = numpy.poly1d([1.2e-06, -1.23e-03, .36, -4.e-04])
poly_baseline = poly_3rd_degree(x)
y = 100 * numpy.exp(-((x-300)/15)**2)+\
200 * numpy.exp(-((x-750)/30)**2)+ \
100 * numpy.exp(-((x-800)/15)**2) + noise + poly_baseline
bl_1 = baseline_arPLS(y, 1e+07, 1e-01)
bl_2 = baseline_arPLS(y, 1e+07, 1e-02)
pyplot.figure(1)
pyplot.plot(x, y, 'C0')
pyplot.plot(x, poly_baseline, 'C1')
pyplot.plot(x, bl_1, 'k')
pyplot.show()
sys.exit(0)
All this is telling me that I'm doing something very non-optimal in my python implementation. Since I'm not knowledgeable enough about the intricacies of scipy computations I'm kindly asking for suggestions on how to achieve convergence in this calculations.
(I encountered an issue in running the "straight" matlab version of the code because the line D = diff(speye(N), 2); truncates the last two rows of the matrix, creating dimension mismatch later in the function. Following the description of matrix D's appearance I substituted this line by directly creating a tridiagonal matrix using the diags function.)
Guided by the comment #hpaulj made, and suspecting that the loop exit wasn't coded properly, I re-visited the paper and found out that the authors actually implemented an exit condition that was not featured in their matlab script. Changing the while loop condition provides an exit for any set of parameters; my understanding is that algorithm is not guaranteed to converge in all cases, which is why this condition is necessary but was omitted by error. Here's the edited version of my python code:
def baseline_arPLS(y, lam, ratio):
# Estimate baseline with arPLS
N = len(y)
k = [numpy.ones(N), -2*numpy.ones(N-1), numpy.ones(N-2)]
offset = [0, 1, 2]
D = diags(k, offset).toarray()
H = lam * numpy.matmul(D.T, D)
w_ = numpy.ones(N)
i = 0
N_iterations = 100
while i < N_iterations:
W = spdiags(w_, 0, N, N, format='csr')
# Cholesky decomposition
C = cholesky(W + H)
z_ = spsolve(C.T, w_ * y)
z = spsolve(C, z_)
d = y - z
# make d- and get w^t with m and s
dn = d[d<0]
m = numpy.mean(dn)
s = numpy.std(dn)
wt = 1. / (1 + numpy.exp(2 * (d - (2*s-m)) / s))
# check exit condition and backup
norm_wt, norm_w = norm(w_-wt), norm(w_)
if (norm_wt / norm_w) < ratio:
break
w_ = wt
i += 1
return(z)
I have a code that works perfectly well but I wish to speed up the time it takes to converge. A snippet of the code is shown below:
def myfunction(x, i):
y = x + (min(0, target[i] - data[i, :]x))*data[i]/(norm(data[i])**2))
return y
rows, columns = data.shape
start = time.time()
iterate = 0
iterate_count = []
norm_count = []
res = 5
x_not = np.ones(columns)
norm_count.append(norm(x_not))
iterate_count.append(0)
while res > 1e-8:
for row in range(rows):
y = myfunction(x_not, row)
x_not = y
iterate += 1
iterate_count.append(iterate)
norm_count.append(norm(x_not))
res = abs(norm_count[-1] - norm_count[-2])
print('Converge at {} iterations'.format(iterate))
print('Duration: {:.4f} seconds'.format(time.time() - start))
I am relatively new in Python. I will appreciate any hint/assistance.
Ax=b is the problem we wish to solve. Here, 'A' is the 'data' and 'b' is the 'target'
Ugh! After spending a while on this I don't think it can be done the way you've set up your problem. In each iteration over the row, you modify x_not and then pass the updated result to get the solution for the next row. This kind of setup can't be vectorized easily. You can learn the thought process of vectorization from the failed attempt, so I'm including it in the answer. I'm also including a different iterative method to solve linear systems of equations. I've included a vectorized version -- where the solution is updated using matrix multiplication and vector addition, and a loopy version -- where the solution is updated using a for loop to demonstrate what you can expect to gain.
1. The failed attempt
Let's take a look at what you're doing here.
def myfunction(x, i):
y = x + (min(0, target[i] - data[i, :] # x)) * (data[i] / (norm(data[i])**2))
return y
You subtract
the dot product of (the ith row of data and x_not)
from the ith row of target,
limited at zero.
You multiply this result with the ith row of data divided my the norm of that row squared. Let's call this part2
Then you add this to the ith element of x_not
Now let's look at the shapes of the matrices.
data is (M, N).
target is (M, ).
x_not is (N, )
Instead of doing these operations rowwise, you can operate on the entire matrix!
1.1. Simplifying the dot product.
Instead of doing data[i, :] # x, you can do data # x_not and this gives an array with the ith element giving the dot product of the ith row with x_not. So now we have data # x_not with shape (M, )
Then, you can subtract this from the entire target array, so target - (data # x_not) has shape (M, ).
So far, we have
part1 = target - (data # x_not)
Next, if anything is greater than zero, set it to zero.
part1[part1 > 0] = 0
1.2. Finding rowwise norms.
Finally, you want to multiply this by the row of data, and divide by the square of the L2-norm of that row. To get the norm of each row of a matrix, you do
rownorms = np.linalg.norm(data, axis=1)
This is a (M, ) array, so we need to convert it to a (M, 1) array so we can divide each row. rownorms[:, None] does this. Then divide data by this.
part2 = data / (rownorms[:, None]**2)
1.3. Add to x_not
Finally, we're adding each row of part1 * part2 to the original x_not and returning the result
result = x_not + (part1 * part2).sum(axis=0)
Here's where we get stuck. In your approach, each call to myfunction() gives a value of part1 that depends on target[i], which was changed in the last call to myfunction().
2. Why vectorize?
Using numpy's inbuilt methods instead of looping allows it to offload the calculation to its C backend, so it runs faster. If your numpy is linked to a BLAS backend, you can extract even more speed by using your processor's SIMD registers
The conjugate gradient method is a simple iterative method to solve certain systems of equations. There are other more complex algorithms that can solve general systems well, but this should do for the purposes of our demo. Again, the purpose is not to have an iterative algorithm that will perfectly solve any linear system of equations, but to show what kind of speedup you can expect if you vectorize your code.
Given your system
data # x_not = target
Let's define some variables:
A = data.T # data
b = data.T # target
And we'll solve the system A # x = b
x = np.zeros((columns,)) # Initial guess. Can be anything
resid = b - A # x
p = resid
while (np.abs(resid) > tolerance).any():
Ap = A # p
alpha = (resid.T # resid) / (p.T # Ap)
x = x + alpha * p
resid_new = resid - alpha * Ap
beta = (resid_new.T # resid_new) / (resid.T # resid)
p = resid_new + beta * p
resid = resid_new + 0
To contrast the fully vectorized approach with one that uses iterations to update the rows of x and resid_new, let's define another implementation of the CG solver that does this.
def solve_loopy(data, target, itermax = 100, tolerance = 1e-8):
A = data.T # data
b = data.T # target
rows, columns = data.shape
x = np.zeros((columns,)) # Initial guess. Can be anything
resid = b - A # x
resid_new = b - A # x
p = resid
niter = 0
while (np.abs(resid) > tolerance).any() and niter < itermax:
Ap = A # p
alpha = (resid.T # resid) / (p.T # Ap)
for i in range(len(x)):
x[i] = x[i] + alpha * p[i]
resid_new[i] = resid[i] - alpha * Ap[i]
# resid_new = resid - alpha * A # p
beta = (resid_new.T # resid_new) / (resid.T # resid)
p = resid_new + beta * p
resid = resid_new + 0
niter += 1
return x
And our original vector method:
def solve_vect(data, target, itermax = 100, tolerance = 1e-8):
A = data.T # data
b = data.T # target
rows, columns = data.shape
x = np.zeros((columns,)) # Initial guess. Can be anything
resid = b - A # x
resid_new = b - A # x
p = resid
niter = 0
while (np.abs(resid) > tolerance).any() and niter < itermax:
Ap = A # p
alpha = (resid.T # resid) / (p.T # Ap)
x = x + alpha * p
resid_new = resid - alpha * Ap
beta = (resid_new.T # resid_new) / (resid.T # resid)
p = resid_new + beta * p
resid = resid_new + 0
niter += 1
return x
Let's solve a simple system to see if this works first:
2x1 + x2 = -5
−x1 + x2 = -2
should give a solution of [-1, -3]
data = np.array([[ 2, 1],
[-1, 1]])
target = np.array([-5, -2])
print(solve_loopy(data, target))
print(solve_vect(data, target))
Both give the correct solution [-1, -3], yay! Now on to bigger things:
data = np.random.random((100, 100))
target = np.random.random((100, ))
Let's ensure the solution is still correct:
sol1 = solve_loopy(data, target)
np.allclose(data # sol1, target)
# Output: False
sol2 = solve_vect(data, target)
np.allclose(data # sol2, target)
# Output: False
Hmm, looks like the CG method doesn't work for badly conditioned random matrices we created. Well, at least both give the same result.
np.allclose(sol1, sol2)
# Output: True
But let's not get discouraged! We don't really care if it works perfectly, the point of this is to demonstrate how amazing vectorization is. So let's time this:
import timeit
timeit.timeit('solve_loopy(data, target)', number=10, setup='from __main__ import solve_loopy, data, target')
# Output: 0.25586539999994784
timeit.timeit('solve_vect(data, target)', number=10, setup='from __main__ import solve_vect, data, target')
# Output: 0.12008900000000722
Nice! A ~2x speedup simply by avoiding a loop while updating our solution!
For larger systems, this will be even better.
for N in [10, 50, 100, 500, 1000]:
data = np.random.random((N, N))
target = np.random.random((N, ))
t_loopy = timeit.timeit('solve_loopy(data, target)', number=10, setup='from __main__ import solve_loopy, data, target')
t_vect = timeit.timeit('solve_vect(data, target)', number=10, setup='from __main__ import solve_vect, data, target')
print(N, t_loopy, t_vect, t_loopy/t_vect)
This gives us:
N t_loopy t_vect speedup
00010 0.002823 0.002099 1.345390
00050 0.051209 0.014486 3.535048
00100 0.260348 0.114601 2.271773
00500 0.980453 0.240151 4.082644
01000 1.769959 0.508197 3.482822
For reasons, I need to implement the Runge-Kutta4 method in PyTorch (so no, I'm not going to use scipy.odeint). I tried and I get weird results on the simplest test case, solving x'=x with x(0)=1 (analytical solution: x=exp(t)). Basically, as I reduce the time step, I cannot get the numerical error to go down. I'm able to do it with a simpler Euler method, but not with the Runge-Kutta 4 method, which makes me suspect some floating point issue here (maybe I'm missing some hidden conversion from double precision to single)?
import torch
import numpy as np
import matplotlib.pyplot as plt
def Euler(f, IC, time_grid):
y0 = torch.tensor([IC])
time_grid = time_grid.to(y0[0])
values = y0
for i in range(0, time_grid.shape[0] - 1):
t_i = time_grid[i]
t_next = time_grid[i+1]
y_i = values[i]
dt = t_next - t_i
dy = f(t_i, y_i) * dt
y_next = y_i + dy
y_next = y_next.unsqueeze(0)
values = torch.cat((values, y_next), dim=0)
return values
def RungeKutta4(f, IC, time_grid):
y0 = torch.tensor([IC])
time_grid = time_grid.to(y0[0])
values = y0
for i in range(0, time_grid.shape[0] - 1):
t_i = time_grid[i]
t_next = time_grid[i+1]
y_i = values[i]
dt = t_next - t_i
dtd2 = 0.5 * dt
f1 = f(t_i, y_i)
f2 = f(t_i + dtd2, y_i + dtd2 * f1)
f3 = f(t_i + dtd2, y_i + dtd2 * f2)
f4 = f(t_next, y_i + dt * f3)
dy = 1/6 * dt * (f1 + 2 * (f2 + f3) +f4)
y_next = y_i + dy
y_next = y_next.unsqueeze(0)
values = torch.cat((values, y_next), dim=0)
return values
# differential equation
def f(T, X):
return X
# initial condition
IC = 1.
# integration interval
def integration_interval(steps, ND=1):
return torch.linspace(0, ND, steps)
# analytical solution
def analytical_solution(t_range):
return np.exp(t_range)
# test a numerical method
def test_method(method, t_range, analytical_solution):
numerical_solution = method(f, IC, t_range)
L_inf_err = torch.dist(numerical_solution, analytical_solution, float('inf'))
return L_inf_err
if __name__ == '__main__':
Euler_error = np.array([0.,0.,0.])
RungeKutta4_error = np.array([0.,0.,0.])
indices = np.arange(1, Euler_error.shape[0]+1)
n_steps = np.power(10, indices)
for i, n in np.ndenumerate(n_steps):
t_range = integration_interval(steps=n)
solution = analytical_solution(t_range)
Euler_error[i] = test_method(Euler, t_range, solution).numpy()
RungeKutta4_error[i] = test_method(RungeKutta4, t_range, solution).numpy()
plots_path = "./plots"
a = plt.figure()
plt.xscale('log')
plt.yscale('log')
plt.plot(n_steps, Euler_error, label="Euler error", linestyle='-')
plt.plot(n_steps, RungeKutta4_error, label="RungeKutta 4 error", linestyle='-.')
plt.legend()
plt.savefig(plots_path + "/errors.png")
The result:
As you can see, the Euler method converges (slowly, as expected of a first order method). However, the Runge-Kutta4 method does not converge as the time step gets smaller and smaller. The error goes down initially, and then up again. What's the issue here?
The reason is indeed a floating point precision issue. torch defaults to single precision, so once the truncation error becomes small enough, the total error is basically determined by the roundoff error, and reducing the truncation error further by increasing the number of steps <=> decreasing the time step doesn't lead to any decrease in the total error.
To fix this, we need to enforce double precision 64bit floats for all floating point torch tensors and numpy arrays. Note that the right way to do this is to use respectively torch.float64 and np.float64 rather than, e.g., torch.double and np.double, because the former are fixed-sized float values, (always 64bit) while the latter depend on the machine and/or compiler. Here's the fixed code:
import torch
import numpy as np
import matplotlib.pyplot as plt
def Euler(f, IC, time_grid):
y0 = torch.tensor([IC], dtype=torch.float64)
time_grid = time_grid.to(y0[0])
values = y0
for i in range(0, time_grid.shape[0] - 1):
t_i = time_grid[i]
t_next = time_grid[i+1]
y_i = values[i]
dt = t_next - t_i
dy = f(t_i, y_i) * dt
y_next = y_i + dy
y_next = y_next.unsqueeze(0)
values = torch.cat((values, y_next), dim=0)
return values
def RungeKutta4(f, IC, time_grid):
y0 = torch.tensor([IC], dtype=torch.float64)
time_grid = time_grid.to(y0[0])
values = y0
for i in range(0, time_grid.shape[0] - 1):
t_i = time_grid[i]
t_next = time_grid[i+1]
y_i = values[i]
dt = t_next - t_i
dtd2 = 0.5 * dt
f1 = f(t_i, y_i)
f2 = f(t_i + dtd2, y_i + dtd2 * f1)
f3 = f(t_i + dtd2, y_i + dtd2 * f2)
f4 = f(t_next, y_i + dt * f3)
dy = 1/6 * dt * (f1 + 2 * (f2 + f3) +f4)
y_next = y_i + dy
y_next = y_next.unsqueeze(0)
values = torch.cat((values, y_next), dim=0)
return values
# differential equation
def f(T, X):
return X
# initial condition
IC = 1.
# integration interval
def integration_interval(steps, ND=1):
return torch.linspace(0, ND, steps, dtype=torch.float64)
# analytical solution
def analytical_solution(t_range):
return np.exp(t_range, dtype=np.float64)
# test a numerical method
def test_method(method, t_range, analytical_solution):
numerical_solution = method(f, IC, t_range)
L_inf_err = torch.dist(numerical_solution, analytical_solution, float('inf'))
return L_inf_err
if __name__ == '__main__':
Euler_error = np.array([0.,0.,0.], dtype=np.float64)
RungeKutta4_error = np.array([0.,0.,0.], dtype=np.float64)
indices = np.arange(1, Euler_error.shape[0]+1)
n_steps = np.power(10, indices)
for i, n in np.ndenumerate(n_steps):
t_range = integration_interval(steps=n)
solution = analytical_solution(t_range)
Euler_error[i] = test_method(Euler, t_range, solution).numpy()
RungeKutta4_error[i] = test_method(RungeKutta4, t_range, solution).numpy()
plots_path = "./plots"
a = plt.figure()
plt.xscale('log')
plt.yscale('log')
plt.plot(n_steps, Euler_error, label="Euler error", linestyle='-')
plt.plot(n_steps, RungeKutta4_error, label="RungeKutta 4 error", linestyle='-.')
plt.legend()
plt.savefig(plots_path + "/errors.png")
Result:
Now, as we decrease the time step, the error of the RungeKutta4 approximation decreases with the correct rate.
So I have defined a fucntion to generate a random number in a given circle:
def rand_gen(R,c):
# random angle
alpha = 2 * math.pi * random.random()
# random radius
r = R * math.sqrt(random.random())
# calculating coordinates
x = r * math.cos(alpha) + c[0]
y = r * math.sin(alpha) + c[1]
return (x,y)
Now I want to add a parameter n (rand_gen(R,c,n)) such that we can get n such numbers instead of one
You can achieve that goal also using generator function:
import random
import math
def rand_gen(R,c,n):
while n:
# random angle
alpha = 2 * math.pi * random.random()
# random radius
r = R * math.sqrt(random.random())
# calculating coordinates
x = r * math.cos(alpha) + c[0]
y = r * math.sin(alpha) + c[1]
n -= 1
yield (x,y)
points_gen = rand_gen(10, (3,4), 4)
for point in points_gen:
print(point)
points = [*rand_gen(10, (3,4), 4)]
print(points)
If you want to generate n random points, you can extend your function with a parameter and for-loop:
import math
import random
def rand_gen(R, c, n):
out = []
for i in range(n):
# random angle
alpha = 2 * math.pi * random.random()
# random radius
r = R * math.sqrt(random.random())
# calculating coordinates
x = r * math.cos(alpha) + c[0]
y = r * math.sin(alpha) + c[1]
out.append((x,y))
return out
print(rand_gen(10, (3, 4), 3))
Prints (for example):
[(-4.700562169626218, 5.62666979720004), (6.481518730246707, -1.849892172014873), (0.41713910134636345, -1.9065302305716285)]
But better approach in my opinion would be leave the function as is and generate n points using list comprehension:
lst = [rand_gen(10, (3, 4)) for _ in range(3)]
print(lst)
Prints:
[(-1.891474340814674, -2.922205399732765), (12.557063558442614, 1.9323688240857821), (-5.450160078420653, 7.974550456763403)]
I have two functions that compute the same metric. One ends up using a list comprehension to cycle through a calculation, the other uses only numpy tensor operations. The functions take in a (N, 3) array, where N is the number of points in 3D space. When N <~ 3000 the tensor function is faster, when N >~ 3000 the list comprehension is faster. Both seem to have linear time complexity in terms of N i.e two time-N lines cross at N=~3000.
def approximate_area_loop(section, num_area_divisions):
n_a_d = num_area_divisions
interp_vectors = get_section_interp_(section)
a1 = section[:-1]
b1 = section[1:]
a2 = interp_vectors[:-1]
b2 = interp_vectors[1:]
c = lambda u: (1 - u) * a1 + u * a2
d = lambda u: (1 - u) * b1 + u * b2
x = lambda u, v: (1 - v) * c(u) + v * d(u)
area = np.sum([np.linalg.norm(np.cross((x((i + 1)/n_a_d, j/n_a_d) - x(i/n_a_d, j/n_a_d)),\
(x(i/n_a_d, (j +1)/n_a_d) - x(i/n_a_d, j/n_a_d))), axis = 1)\
for i in range(n_a_d) for j in range(n_a_d)])
Dt = section[-1, 0] - section[0, 0]
return area, Dt
def approximate_area_tensor(section, num_area_divisions):
divisors = np.linspace(0, 1, num_area_divisions + 1)
interp_vectors = get_section_interp_(section)
a1 = section[:-1]
b1 = section[1:]
a2 = interp_vectors[:-1]
b2 = interp_vectors[1:]
c = np.multiply.outer(a1, (1 - divisors)) + np.multiply.outer(a2, divisors) # c_areas_vecs_divs
d = np.multiply.outer(b1, (1 - divisors)) + np.multiply.outer(b2, divisors) # d_areas_vecs_divs
x = np.multiply.outer(c, (1 - divisors)) + np.multiply.outer(d, divisors) # x_areas_vecs_Divs_divs
u = x[:, :, 1:, :-1] - x[:, :, :-1, :-1] # u_areas_vecs_Divs_divs
v = x[:, :, :-1, 1:] - x[:, :, :-1, :-1] # v_areas_vecs_Divs_divs
sub_area_norm_vecs = np.cross(u, v, axis = 1) # areas_crosses_Divs_divs
sub_areas = np.linalg.norm(sub_area_norm_vecs, axis = 1) # areas_Divs_divs (values are now sub areas)
area = np.sum(sub_areas)
Dt = section[-1, 0] - section[0, 0]
return area, Dt
Why does the list comprehension version work faster at large N? Surely the tensor version should be faster? I'm wondering if it's something to do with the size of the calculations meaning it's too big to be done in cache? Please ask if I haven't included enough information, I'd really like to get to the bottom of this.
The bottleneck in the fully vectorized function was indeed in the np.linalg.norm as #hpauljs comment suggested.
Norm was used only to get the magnitude of all the vectors contained in axis 1. A much simpler and faster method was to just:
sub_areas = np.sqrt((sub_area_norm_vecs*sub_area_norm_vecs).sum(axis = 1))
This gives exactly the same results and sped up the code by up to 25 times faster than the loop implementation (even when the loop doesn't use linalg.norm either).