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Runge-Kutta 4 for solving systems of ODEs Python

I wrote code for Runge-Kutta 4 for solving system of ODEs.
It works fine for 1-D ODE but when I try to solve x'' + kx = 0 I have a problem trying to define a vectorial function:
Let u1 = x and u2 = x' = u1', then the system looks like:
u1' = u2
u2' = -k*u1
If u = (u1,u2) and f(u, t) = (u2, -k*u1), then we need to solve:
u' = f(u, t)
def f(u,t, omega=2):
u, v = u
return np.asarray([v, -omega**2*u])
My entire code is:
import numpy as np
def ode_RK4(f, X_0, dt, T):
N_t = int(round(T/dt))
# Create an array for the functions ui
u = np.zeros((len(X_0),N_t+1)) # Array u[j,:] corresponds to the j-solution
t = np.linspace(0, N_t*dt, N_t + 1)
# Initial conditions
for j in range(len(X_0)):
u[j,0] = X_0[j]
# RK4
for j in range(len(X_0)):
for n in range(N_t):
u1 = f(u[j,n] + 0.5*dt* f(u[j,n], t[n])[j], t[n] + 0.5*dt)[j]
u2 = f(u[j,n] + 0.5*dt*u1, t[n] + 0.5*dt)[j]
u3 = f(u[j,n] + dt*u2, t[n] + dt)[j]
u[j, n+1] = u[j,n] + (1/6)*dt*( f(u[j,n], t[n])[j] + 2*u1 + 2*u2 + u3)
return u, t
def demo_exp():
import matplotlib.pyplot as plt
def f(u,t):
return np.asarray([u])
u, t = ode_RK4(f, [1] , 0.1, 1.5)
plt.plot(t, u[0,:],"b*", t, np.exp(t), "r-")
plt.show()
def demo_osci():
import matplotlib.pyplot as plt
def f(u,t, omega=2):
# u, v = u Here I've got a problem
return np.asarray([v, -omega**2*u])
u, t = ode_RK4(f, [2,0], 0.1, 2)
for i in [1]:
plt.plot(t, u[i,:], "b*")
plt.show()
In advance, thank you.

You are on the right path, but when applying time-integration methods such as RK to vector valued ODEs, one essentially does the exact same thing as in the scalar case, just with vectors.
Thus, you skip the for j in range(len(X_0)) loop and associated indexation and you make sure that you pass initial values as vectors (numpy arrays).
Also cleaned up the indexation for t a little and stored the solution in a list.
import numpy as np
def ode_RK4(f, X_0, dt, T):
N_t = int(round(T/dt))
# Initial conditions
usol = [X_0]
u = np.copy(X_0)
tt = np.linspace(0, N_t*dt, N_t + 1)
# RK4
for t in tt[:-1]:
u1 = f(u + 0.5*dt* f(u, t), t + 0.5*dt)
u2 = f(u + 0.5*dt*u1, t + 0.5*dt)
u3 = f(u + dt*u2, t + dt)
u = u + (1/6)*dt*( f(u, t) + 2*u1 + 2*u2 + u3)
usol.append(u)
return usol, tt
def demo_exp():
import matplotlib.pyplot as plt
def f(u,t):
return np.asarray([u])
u, t = ode_RK4(f, np.array([1]) , 0.1, 1.5)
plt.plot(t, u, "b*", t, np.exp(t), "r-")
plt.show()
def demo_osci():
import matplotlib.pyplot as plt
def f(u,t, omega=2):
u, v = u
return np.asarray([v, -omega**2*u])
u, t = ode_RK4(f, np.array([2,0]), 0.1, 2)
u1 = [a[0] for a in u]
for i in [1]:
plt.plot(t, u1, "b*")
plt.show()

The model is this:
enter image description here
From the Langtangen’s book Programming for Computations - Python.

Why is tempered mcmc fit not convering well?

I am trying to fit a simple straight line y=mx+c type to some synthetic data using parallel-tempered mcmc. My goal is to just be able to understand how to use it, so that I can apply to some more complex models later. The example I am trying is the replica of what has already been done in a simple emcee code :
http://dfm.io/emcee/current/user/line/
but instead of using mcmc, I want to use parallel-tempered mcmc:
http://dfm.io/emcee/current/user/pt/
Here is a working code:
import numpy as np
from emcee import PTSampler
import emcee
# Choose the "true" parameters.
m_true = -0.9594
b_true = 4.294
f_true = 0.534
# Generate some synthetic data from the model.
N = 50
x = np.sort(10*np.random.rand(N))
yerr = 0.1+0.5*np.random.rand(N)
y = m_true*x+b_true
y += np.abs(f_true*y) * np.random.randn(N)
y += yerr * np.random.randn(N)
def lnlike(theta, x, y, yerr):
m, b, lnf = theta
model = m * x + b
inv_sigma2 = 1.0/(yerr**2 + model**2*np.exp(2*lnf))
return -0.5*(np.sum((y-model)**2*inv_sigma2 - np.log(inv_sigma2)))
def lnprior(theta):
m, b, lnf = theta
if -5.0 < m < 0.5 and 0.0 < b < 10.0 and -10.0 < lnf < 1.0:
return 0.0
return -np.inf
def lnprob(theta, x, y, yerr):
lp = lnprior(theta)
if not np.isfinite(lp):
return -np.inf
return lp + lnlike(theta, x, y, yerr)
import scipy.optimize as op
nll = lambda *args: -lnlike(*args)
result = op.minimize(nll, [m_true, b_true, np.log(f_true)], args=(x, y, yerr))
m_ml, b_ml, lnf_ml = result["x"]
init = [0.5, m_ml, b_ml, lnf_ml]
ntemps = 10
nwalkers = 100
ndim = 3
from multiprocessing import Pool
pos = np.random.uniform(low=-1, high=1, size=(ntemps, nwalkers, ndim))
for i in range(ntemps):
#initialize parameters near scipy optima
pos[i:,] = np.array([result["x"] + 1e-4*np.random.randn(ndim) for i in range(nwalkers)])
pool = Pool(processes=4)
sampler=PTSampler(ntemps,nwalkers, ndim, lnlike, lnprior, loglargs=(x, y, yerr), pool=pool)# args=(x, y, yerr))
#burn-in
sampler.run_mcmc(pos, 1000)
sampler.reset()
sampler.run_mcmc(pos, 10000, thin=10)
samples = sampler.chain.reshape((-1, ndim))
print('Number of posterior samples is {}'.format(samples.shape[0]))
#print best fit value together with errors
print(map(lambda v: (v[1], v[2]-v[1], v[1]-v[0]),
zip(*np.percentile(samples, [16, 50, 84],
axis=0))))
import corner
fig = corner.corner(samples, labels=["$m$", "$b$", "$\ln\,f$"],
truths=[m_true, b_true, np.log(f_true)])
fig.savefig("triangle.png")
The only problem when running this code is I get optimal parameters value which are way off the true values. And increasing the number of walkers or samples is not helping in any sense. Can anyone please advice why tempered-mcmc is not working here?
Update:
I found out a useful package called ptemcee (https://pypi.org/project/ptemcee/#description), although the documentation of this package is non-existent. It seems that this package might be useful, any help on how to implement the same linear fitting with this package would also be highly appreciated.

I have modified some lines
import time
import numpy as np
from emcee import PTSampler
import corner
import matplotlib.pyplot as plt
import scipy.optimize as op
t1 = time.time()
np.random.seed(6) # To reproduce results
# Choose the "true" parameters.
m_true = -0.9594
b_true = 4.294
f_true = 0.534
# Generate some synthetic data from the model.
N = 50
x = np.sort(10 * np.random.rand(N))
yerr = 0.1 + 0.5 * np.random.rand(N)
y_1 = m_true * x + b_true
y = np.abs(f_true * y_1) * np.random.randn(N) + y_1
y += yerr * np.random.randn(N)
plt.plot(x, y, 'o')
# With emcee
def lnlike(theta, x, y, yerr):
m, b, lnf = theta
model = m * x + b
inv_sigma2 = 1.0/(yerr**2 + model**2*np.exp(2*lnf))
return -0.5*(np.sum((y-model)**2*inv_sigma2 - np.log(inv_sigma2)))
def lnprior(theta):
m, b, lnf = theta
if -5.0 < m < 0.5 and 0.0 < b < 10.0 and -10.0 < lnf < 1.0:
return 0.0
return -np.inf
def lnprob(theta, x, y, yerr):
lp = lnprior(theta)
if not np.isfinite(lp):
return -np.inf
return lp + lnlike(theta, x, y, yerr)
nll = lambda *args: -lnlike(*args)
result = op.minimize(nll, [m_true, b_true, np.log(f_true)], args=(x, y, yerr))
m_ml, b_ml, lnf_ml = result["x"]
init = [0.5, m_ml, b_ml, lnf_ml]
ntemps = 10
nwalkers = 100
ndim = 3
pos = np.random.uniform(low=-1, high=1, size=(ntemps, nwalkers, ndim))
for i in range(ntemps):
pos[i:, :] = np.array([result["x"] + 1e-4*np.random.randn(ndim) for i in range(nwalkers)])
sampler = PTSampler(ntemps, nwalkers, ndim, lnlike, lnprior, loglargs=(x, y, yerr), threads=4) # args=(x, y, yerr))
#burn-in
print(pos.shape)
sampler.run_mcmc(pos, 100)
sampler.reset()
sampler.run_mcmc(pos, 5000, thin=10)
samples = sampler.chain.reshape((-1, ndim))
print('Number of posterior samples is {}'.format(samples.shape[0]))
#print best fit value together with errors
p1, p2, p3 = map(lambda v: (v[1], v[2]-v[1], v[1]-v[0]),
zip(*np.percentile(samples, [16, 50, 84],
axis=0)))
print(p1, '\n', p2, '\n', p3)
fig = corner.corner(samples, labels=["$m$", "$b$", "$\ln\,f$"],
truths=[m_true, b_true, np.log(f_true)])
t2 = time.time()
print('It took {:.3f} s'.format(t2 - t1))
plt.show()
The figure I get with corner is:
The important line is
sampler = PTSampler(ntemps, nwalkers, ndim, lnlike, lnprior, loglargs=(x, y, yerr), threads=4)
I have used threads=4 instead of Pool.
Look closely at this line print(p1, '\n', p2, '\n', p3), it prints the values of m_true, b_true and f_true you get:
(-1.277782877669762, 0.5745273177144817, 2.0813620981463297)
(4.800481378230051, 3.1747356851201163, 2.245189235990341)
(-0.9391847529845194, 1.1196053087321716, 3.6017609114364273)
For f, you need np.exp(-0.93918), which is 0.3909, which is close to 0.534. The values you get are close (-1.277 compared to -0.9594 and 4.8 compared to 4.294), although the errors are not bad (except for f). I mean, are you expecting to get the exact numbers? With this method, in my computer, it takes 111 s to complete, is that normal?
Let's try something different. Let's be clear: the problem is not easy when f_true is added. I will use pymc3 (you don't need to know how to use pymc3, I want to check the results found by emcee).
import time
import numpy as np
import corner
import matplotlib.pyplot as plt
import pymc3 as pm
t1 = time.time()
np.random.seed(6)
# Choose the "true" parameters.
m_true = -0.9594
b_true = 4.294
f_true = 0.534
# Generate some synthetic data from the model.
N = 50
x = np.sort(10 * np.random.rand(N))
yerr = 0.1 + 0.5 * np.random.rand(N)
y_1 = m_true * x + b_true
y = np.abs(f_true * y_1) * np.random.randn(N) + y_1
y += yerr * np.random.randn(N)
plt.plot(x, y, 'o')
with pm.Model() as model: # model specifications in PyMC3 are wrapped in a with-statement
# Define priors
f = pm.HalfCauchy('f', beta=5)
m = pm.Normal('m', 0, sd=20)
b = pm.Normal('b', 0, sd=20)
mu2 = b + m * x
sigma2 = yerr**2 + f**2 * (y_1)**2
post = pm.Normal('y', mu=mu2, sd=pm.math.sqrt(sigma2), observed=y)
with model:
trace = pm.sample(2000, tune=2000)
print(pm.summary(trace))
pm.traceplot(trace)
all_values = np.stack([trace.get_values('b'), trace.get_values('m'), trace.get_values('f')], axis=1)
fig2 = corner.corner(all_values, labels=["$b$", "$m$", "$f$"],
truths=[b_true, m_true, f_true])
t2 = time.time()
print('It took {:.3f} s'.format(t2 - t1))
plt.show()
The summary is
mean sd mc_error hpd_2.5 hpd_97.5 n_eff Rhat
m -0.995545 0.067818 0.001174 -1.123187 -0.857653 2685.610018 1.000121
b 4.398158 0.332526 0.005585 3.767336 5.057909 2746.736563 1.000201
f 0.425442 0.063884 0.000904 0.311037 0.554446 4195.591204 1.000309
The important part is the column mean, you see that the values found by pymc3 are close to the true values. The columns hpd_2.5 and hpd_97.5 are the errors for f, b and m. And it took 14 s.
The figure I get with corner is
You will say that the results of emcee are not quite good, but if you really want more accuracy, you have to modify this function:
def lnprior(theta):
m, b, lnf = theta
if -5.0 < m < 0.5 and 0.0 < b < 10.0 and -10.0 < lnf < 1.0:
return 0.0
return -np.inf
The famous prior. In this case, it is flat, and since there are a lot of priors...

Using solve_ivp instead of odeint to solve initial problem value

Currently, I solve the following ODE system of equations using odeint
dx/dt = (-x + u)/2.0
dy/dt = (-y + x)/5.0
initial conditions: x = 0, y = 0
However, I would like to use solve_ivp which seems to be the recommended option for this type of problems, but honestly I don't know how to adapt the code...
Here is the code I'm using with odeint:
import numpy as np
from scipy.integrate import odeint, solve_ivp
import matplotlib.pyplot as plt
def model(z, t, u):
x = z[0]
y = z[1]
dxdt = (-x + u)/2.0
dydt = (-y + x)/5.0
dzdt = [dxdt, dydt]
return dzdt
def main():
# initial condition
z0 = [0, 0]
# number of time points
n = 401
# time points
t = np.linspace(0, 40, n)
# step input
u = np.zeros(n)
# change to 2.0 at time = 5.0
u[51:] = 2.0
# store solution
x = np.empty_like(t)
y = np.empty_like(t)
# record initial conditions
x[0] = z0[0]
y[0] = z0[1]
# solve ODE
for i in range(1, n):
# span for next time step
tspan = [t[i-1], t[i]]
# solve for next step
z = odeint(model, z0, tspan, args=(u[i],))
# store solution for plotting
x[i] = z[1][0]
y[i] = z[1][1]
# next initial condition
z0 = z[1]
# plot results
plt.plot(t,u,'g:',label='u(t)')
plt.plot(t,x,'b-',label='x(t)')
plt.plot(t,y,'r--',label='y(t)')
plt.ylabel('values')
plt.xlabel('time')
plt.legend(loc='best')
plt.show()
main()

It's important that solve_ivp expects f(t, z) as right-hand side of the ODE. If you don't want to change your ode function and also want to pass your parameter u, I recommend to define a wrapper function:
def model(z, t, u):
x = z[0]
y = z[1]
dxdt = (-x + u)/2.0
dydt = (-y + x)/5.0
dzdt = [dxdt, dydt]
return dzdt
def odefun(t, z):
if t < 5:
return model(z, t, 0)
else:
return model(z, t, 2)
Now it's easy to call solve_ivp:
def main():
# initial condition
z0 = [0, 0]
# number of time points
n = 401
# time points
t = np.linspace(0, 40, n)
# step input
u = np.zeros(n)
# change to 2.0 at time = 5.0
u[51:] = 2.0
res = solve_ivp(fun=odefun, t_span=[0, 40], y0=z0, t_eval=t)
x = res.y[0, :]
y = res.y[1, :]
# plot results
plt.plot(t,u,'g:',label='u(t)')
plt.plot(t,x,'b-',label='x(t)')
plt.plot(t,y,'r--',label='y(t)')
plt.ylabel('values')
plt.xlabel('time')
plt.legend(loc='best')
plt.show()
main()
Note that without passing t_eval=t, the solver will automatically choose the time points inside tspan at which the solution will be stored.

radial basis network derivatives are pushing means together

I am trying to make my own implementation of a simple neural network to classify points. I heard about a specific type of activation function that I am interested in testing, the Gaussian. I do not just want to use relus or sigmoids, I am trying to build a network that takes as input about 300 x and y values, then in the first layer computes the Gaussian function on these values with about 50 neurons which each have a separate x and y value as their means (I will keep the sigma constant). Mathematically I anticipate this to look like
exp(- [(x-Mx)^2 + (y-My)^2] / (2 * sigma^2) ) / (sqrt(2*pi*sigma))
then I will perform a weighted sum of these terms over all the neurons in the first layer, add a bias, and pass it through a sigmoid to get my prediction. I will perform this step for each training example and get a list of predictions. I think that I do the forward propagation but I will include the code for that in case someone can spot an obvious error in my implementation. Then I perform the back-propogation. I have tested my updating of the weights and bias, and I believe that they are not the problem. I think that there is something wrong with my implementation of the gradient for the means however because they always cluster to a single point which clearly does not maximize the cost function. I have already tried using a couple of different data sets, and varying some hyper parameters, all to no avail. Can anyone figure out what the problem is?
Here is my code.
# libraries
import matplotlib.patches as patches
import seaborn as sns; sns.set()
import matplotlib.pyplot as plt
import pandas as pd
import numpy as np
import pdb
# functions
def gaussian(sq_error, sigma):
return ((1/np.sqrt(2*np.pi*sigma**2))) * np.exp(-(sq_error)/(2*sigma**2))
def calc_X1(X0, Mx, My, m, sigma):
X1 = [] # shape will be (10, m)
for ex in range(0, m):
sq_error = (X0[0][ex] - Mx) **2 + (X0[1][ex] - My) **2
X1.append(gaussian(sq_error, sigma))
X1 = np.array(X1)
return X1.T
def sigmoid(Z):
return 1 / (1 + np.exp(-Z))
def calc_X2(W2, X1, b2):
return sigmoid(np.dot(W2, X1) + b2)
def cost(X2, Y, m):
return -1/m * ( np.dot(Y, np.log(X2.T)) + np.dot(1-Y, np.log(1-X2.T))) [0]
def calc_dZ2(X2, Y):
return X2 - Y
def calc_dM(dZ2, W2, X1, sigma, M, m, xOrY, X0):
cur_dM = np.zeros(M.shape)
for i in range(0, m):
# pdb.set_trace()
cur_dM += dZ2[0][i] * float(np.dot(W2, X1.T[i])) * 1/sigma**2 * (X0[xOrY][i] - M)
return cur_dM / m
def train_correct(X2, Y, m):
ct = 0
for i in range(0, m):
if np.round(X2[0][i]) == Y[i]:
ct += 1
return ct / m
# graphing functions
def plot_train_data(X, Y, m, ax):
for ex in range(0, m):
xCur = X[0][ex]
yCur = X[1][ex]
if Y[ex] == 1:
color=(1, 0, 0)
else:
color=(0,0,1)
ax.scatter(xCur, yCur, c=color)
def probability_hash(pr):
return (float(pr), float(np.round(pr)), float(1-pr))
def probability_hash_1d(pr):
return float(pr)
def plot_boundary(Mx, My, sigma, W2, b2, ax):
boundsx = [-5, 5]
boundsy = [-5, 5]
samples = [10, 10]
width = (boundsx[1] - boundsx[0]) / samples[0]
height = (boundsy[1] - boundsy[0]) / samples[1]
pt = np.zeros((2,1))
for x in np.linspace(boundsx[0], boundsx[1], samples[0]):
for y in np.linspace(boundsy[0], boundsy[1], samples[1]):
pt[0][0] = x
pt[1][0] = y
X1_cur = calc_X1(pt, Mx, My, 1, sigma)
X2_cur = calc_X2(W2, X1_cur, b2)
# ax.add_patch(patches.Rectangle((x, y), width, height, facecolor=probability_hash(X2_cur)))
ax.scatter(x, y, c=probability_hash(X2_cur))
def cool_plot_boundary(Mx, My, sigma, W2, b2, ax):
boundsx = [-2, 2]
boundsy = [-2, 2]
samples = [50, 50]
width = (boundsx[1] - boundsx[0]) / samples[0]
height = (boundsy[1] - boundsy[0]) / samples[1]
pt = np.zeros((2,1))
heats = []
xs = np.linspace(boundsx[0], boundsx[1], samples[0])
ys = np.linspace(boundsy[0], boundsy[1], samples[1])
for x in xs:
heats.append([])
for y in ys:
pt[0][0] = x
pt[1][0] = y
X1_cur = calc_X1(pt, Mx, My, 1, sigma)
X2_cur = calc_X2(W2, X1_cur, b2)
heats[-1].append(probability_hash_1d(X2_cur))
# xticks = []
# yticks = []
# for i in range(0, len(xs)):
# if i % 3 == 0:
# xticks.append(round(xs[i], 2))
# for i in range(0, len(ys)):
# if i % 3 == 0:
# yticks.append(round(ys[i], 2))
xticks = []
yticks = []
sns.heatmap(heats, ax=ax, cbar=True, xticklabels=xticks, yticklabels=yticks)
def plot_m(Mx, My, n1, ax):
for i in range(0, n1):
ax.scatter(Mx[i], My[i], c="k")
# initialize parameters
file = "data/disk2.csv"
df = pd.read_csv(file)
sigma = 2
itterations = 10000
learning_rate = 0.9
n0 = 2 # DO NOT CHANGE, formality
X0 = np.row_stack((df["0"], df["1"])) # shape is (2, m)
Y = np.array(df["2"])
m = len(Y)
n1 = 50
Mx = np.random.randn(n1)
My = np.random.randn(n1)
X1 = calc_X1(X0, Mx, My, m, sigma)
n2 = 1 # DO NOT CHANGE, formality
small_number = 0.01
W2 = np.random.randn(1, n1) * small_number
b2 = 0
X2 = calc_X2(W2, X1, b2)
J = cost(X2, Y, m)
Js = []
itters = []
fig = plt.figure()
plotGap = 200
for i in range(0, itterations):
# forward propogation
X1 = calc_X1(X0, Mx, My, m, sigma)
X2 = calc_X2(W2, X1, b2)
J = cost(X2, Y, m)
if i % plotGap == 0:
fig.clear()
costAx = fig.add_subplot(311)
plotAx = fig.add_subplot(312)
pointsAx = fig.add_subplot(313)
cool_plot_boundary(Mx, My, sigma, W2, b2, plotAx)
# plot_boundary(Mx, My, sigma, W2, b2, plotAx)
plot_train_data(X0, Y, m, pointsAx)
Js.append(J)
itters.append(i)
costAx.plot(itters, Js, c="k")
print("cost = " + str(J) + "\ttraining correct = " + str(train_correct(X2, Y, m)))
plot_m(Mx, My, n1, pointsAx)
plt.pause(0.1)
# back propogation
dZ2 = calc_dZ2(X2, Y)
dW2 = np.dot(dZ2, X1.T) / m
db2 = np.sum(dZ2) / m
dMx = calc_dM(dZ2, W2, X1, sigma, Mx, m, 0, X0)
dMy = calc_dM(dZ2, W2, X1, sigma, My, m, 1, X0)
b2 -= learning_rate * db2
W2 -= learning_rate * dW2
Mx -= learning_rate * dMx
My -= learning_rate * dMy
For data I have a csv with a bunch of point locations and labels. You can use this code to generate a similar csv. (Make sure you have a folder called data in the folder you run this from).
# makes data in R2 to learn
import matplotlib.pyplot as plt
import pandas as pd
import numpy as np
n = 2
# number of exaples
m = 300
X = []
Y = []
# hyperparamers for data
rApprox = 1
error = 0.4
noise = 0.1
name = "data/disk2"
plt.cla()
for ex in range(0, m):
xCur = np.random.randn(2)
X.append(xCur)
if abs(np.linalg.norm(xCur) + np.random.randn()*noise - rApprox) < error:
Y.append(1)
color="r"
else:
Y.append(0)
color="b"
plt.scatter(xCur[0], xCur[1], c=color)
if abs(np.random.randn()) < 0.01:
plt.pause(0.1)
plt.pause(1)
plt.savefig(name + ".png")
X = np.array(X)
Y = np.array(Y)
df = pd.DataFrame(X)
df[2] = Y
df.to_csv(name + ".csv", index=False)
Thanks for your help.

Substitute this function for the calculate dm function. You must be careful when multiplying, it is not just enough that the dimensions work out.
def calculuate_dMs(X0, X1, X2, Mx, My, W2, dZ2, sigma, m, n1):
# pdb.set_trace()
X0x_big = np.dot(np.ones((n1, 1)), X0[0].reshape(1, m))
X0y_big = np.dot(np.ones((n1, 1)), X0[1].reshape(1, m))
Mx_big = np.dot(Mx.reshape(n1, 1), np.ones((1, m)))
My_big = np.dot(My.reshape(n1, 1), np.ones((1, m)))
W2_big = np.dot(W2.reshape(n1, 1), np.ones((1, m)))
dZ2_big = np.dot(np.ones((n1, 1)), dZ2.reshape(1, m))
dxTemp = np.multiply(np.multiply(np.multiply((X0x_big - Mx_big), X1), W2_big), dZ2_big)
dyTemp = np.multiply(np.multiply(np.multiply((X0y_big - My_big), X1), W2_big), dZ2_big)
return (np.sum(dxTemp, axis=1)/m, np.sum(dyTemp, axis=1)/m)

Producing 2D perlin noise with numpy

I'm trying to produce 2D perlin noise using numpy, but instead of something smooth I get this :
my broken perlin noise, with ugly squares everywhere
For sure, I'm mixing up my dimensions somewhere, probably when I combine the four gradients ... But I can't find it and my brain is melting right now. Anyone can help me pinpoint the problem ?
Anyway, here is the code:
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
def perlin(x,y,seed=0):
# permutation table
np.random.seed(seed)
p = np.arange(256,dtype=int)
np.random.shuffle(p)
p = np.stack([p,p]).flatten()
# coordinates of the first corner
xi = x.astype(int)
yi = y.astype(int)
# internal coordinates
xf = x - xi
yf = y - yi
# fade factors
u = fade(xf)
v = fade(yf)
# noise components
n00 = gradient(p[p[xi]+yi],xf,yf)
n01 = gradient(p[p[xi]+yi+1],xf,yf-1)
n11 = gradient(p[p[xi+1]+yi+1],xf-1,yf-1)
n10 = gradient(p[p[xi+1]+yi],xf-1,yf)
# combine noises
x1 = lerp(n00,n10,u)
x2 = lerp(n10,n11,u)
return lerp(x2,x1,v)
def lerp(a,b,x):
"linear interpolation"
return a + x * (b-a)
def fade(t):
"6t^5 - 15t^4 + 10t^3"
return 6 * t**5 - 15 * t**4 + 10 * t**3
def gradient(h,x,y):
"grad converts h to the right gradient vector and return the dot product with (x,y)"
vectors = np.array([[0,1],[0,-1],[1,0],[-1,0]])
g = vectors[h%4]
return g[:,:,0] * x + g[:,:,1] * y
lin = np.linspace(0,5,100,endpoint=False)
y,x = np.meshgrid(lin,lin)
plt.imshow(perlin(x,y,seed=0))

Thanks to Paul Panzer and a good night of sleep it works now ...
import numpy as np
import matplotlib.pyplot as plt
def perlin(x, y, seed=0):
# permutation table
np.random.seed(seed)
p = np.arange(256, dtype=int)
np.random.shuffle(p)
p = np.stack([p, p]).flatten()
# coordinates of the top-left
xi, yi = x.astype(int), y.astype(int)
# internal coordinates
xf, yf = x - xi, y - yi
# fade factors
u, v = fade(xf), fade(yf)
# noise components
n00 = gradient(p[p[xi] + yi], xf, yf)
n01 = gradient(p[p[xi] + yi + 1], xf, yf - 1)
n11 = gradient(p[p[xi + 1] + yi + 1], xf - 1, yf - 1)
n10 = gradient(p[p[xi + 1] + yi], xf - 1, yf)
# combine noises
x1 = lerp(n00, n10, u)
x2 = lerp(n01, n11, u) # FIX1: I was using n10 instead of n01
return lerp(x1, x2, v) # FIX2: I also had to reverse x1 and x2 here
def lerp(a, b, x):
"linear interpolation"
return a + x * (b - a)
def fade(t):
"6t^5 - 15t^4 + 10t^3"
return 6 * t**5 - 15 * t**4 + 10 * t**3
def gradient(h, x, y):
"grad converts h to the right gradient vector and return the dot product with (x,y)"
vectors = np.array([[0, 1], [0, -1], [1, 0], [-1, 0]])
g = vectors[h % 4]
return g[:, :, 0] * x + g[:, :, 1] * y
lin = np.linspace(0, 5, 100, endpoint=False)
x, y = np.meshgrid(lin, lin) # FIX3: I thought I had to invert x and y here but it was a mistake
plt.imshow(perlin(x, y, seed=2), origin='upper')