Develop Reference

How to make a piecewise linear fit in Python with some constant pieces?

I'm trying to make a piecewise linear fit consisting of 3 pieces whereof the first and last pieces are constant. As you can see in this figure
don't get the expected fit, since the fit doesn't capture the 3 linear pieces clearly visual from the original data points.
I've tried following this question and expanded it to the case of 3 pieces with the two constant pieces, but I must have done something wrong.
Here is my code:
from scipy import optimize
import matplotlib.pyplot as plt
import numpy as np
%matplotlib inline
plt.rcParams['figure.figsize'] = [16, 6]
x = np.arange(0, 50, dtype=float)
y = np.array([50 for i in range(10)]
+ [50 - (50-5)/31 * i for i in range(1, 31)]
+ [5 for i in range(10)],
dtype=float)
def piecewise_linear(x, x0, y0, x1, y1):
return np.piecewise(x,
[x < x0, (x >= x0) & (x < x1), x >= x1],
[lambda x:y0, lambda x:(y1-y0)/(x1-x0)*(x-x0)+y0, lambda x:y1])
p , e = optimize.curve_fit(piecewise_linear, x, y)
xd = np.linspace(0, 50, 101)
plt.plot(x, y, "o", label='original data')
plt.plot(xd, piecewise_linear(xd, *p), label='piecewise linear fit')
plt.legend()
The accepted answer to the previous mentioned question suggest looking at segments_fit.ipynb for the case of N parts, but following that it doesn't seem that I can specify, that the first and last pieces should be constant.
Furthermore I do get the following warning:
OptimizeWarning: Covariance of the parameters could not be estimated
What do I do wrong?

You could directly copy the segments_fit implementation
from scipy import optimize
def segments_fit(X, Y, count):
xmin = X.min()
xmax = X.max()
seg = np.full(count - 1, (xmax - xmin) / count)
px_init = np.r_[np.r_[xmin, seg].cumsum(), xmax]
py_init = np.array([Y[np.abs(X - x) < (xmax - xmin) * 0.01].mean() for x in px_init])
def func(p):
seg = p[:count - 1]
py = p[count - 1:]
px = np.r_[np.r_[xmin, seg].cumsum(), xmax]
return px, py
def err(p):
px, py = func(p)
Y2 = np.interp(X, px, py)
return np.mean((Y - Y2)**2)
r = optimize.minimize(err, x0=np.r_[seg, py_init], method='Nelder-Mead')
return func(r.x)
Then you apply it as follows
import numpy as np;
# mimic your data
x = np.linspace(0, 50)
y = 50 - np.clip(x, 10, 40)
# apply the segment fit
fx, fy = segments_fit(x, y, 3)
This will give you (fx,fy) the corners your piecewise fit, let's plot it
import matplotlib.pyplot as plt
# show the results
plt.figure(figsize=(8, 3))
plt.plot(fx, fy, 'o-')
plt.plot(x, y, '.')
plt.legend(['fitted line', 'given points'])
EDIT: Introducing constant segments
As mentioned in the comments the above example doesn't guarantee that the output will be constant in the end segments.
Based on this implementation the easier way I can think is to restrict func(p) to do that, a simple way to ensure a segment is constant, is to set y[i+1]==y[i]. Thus I added xanchor and yanchor. If you give an array with repeated numbers you can bind multiple points to the same value.
from scipy import optimize
def segments_fit(X, Y, count, xanchors=slice(None), yanchors=slice(None)):
xmin = X.min()
xmax = X.max()
seg = np.full(count - 1, (xmax - xmin) / count)
px_init = np.r_[np.r_[xmin, seg].cumsum(), xmax]
py_init = np.array([Y[np.abs(X - x) < (xmax - xmin) * 0.01].mean() for x in px_init])
def func(p):
seg = p[:count - 1]
py = p[count - 1:]
px = np.r_[np.r_[xmin, seg].cumsum(), xmax]
py = py[yanchors]
px = px[xanchors]
return px, py
def err(p):
px, py = func(p)
Y2 = np.interp(X, px, py)
return np.mean((Y - Y2)**2)
r = optimize.minimize(err, x0=np.r_[seg, py_init], method='Nelder-Mead')
return func(r.x)
I modified a little the data generation to make it more clear the effect of the change
import matplotlib.pyplot as plt
import numpy as np;
# mimic your data
x = np.linspace(0, 50)
y = 50 - np.clip(x, 10, 40) + np.random.randn(len(x)) + 0.25 * x
# apply the segment fit
fx, fy = segments_fit(x, y, 3)
plt.plot(fx, fy, 'o-')
plt.plot(x, y, '.k')
# apply the segment fit with some consecutive points having the
# same anchor
fx, fy = segments_fit(x, y, 3, yanchors=[1,1,2,2])
plt.plot(fx, fy, 'o--r')
plt.legend(['fitted line', 'given points', 'with const segments'])

You can get a one line solution (not counting the import) using univariate splines of degree one. Like this
from scipy.interpolate import UnivariateSpline
f = UnivariateSpline(x,y,k=1,s=0)
Here k=1 means we interpolate using polynomials of degree one aka lines. s is the smoothing parameter. It decides how much you want to compromise on the fit to avoid using too many segments. Setting it to zero means no compromises i.e. the line HAS to go threw all points. See the documentation.
Then
plt.plot(x, y, "o", label='original data')
plt.plot(x, f(x), label='linear interpolation')
plt.legend()
plt.savefig("out.png", dpi=300)
gives

This i consider a funny non-linear approach that works quite well.
Note that even though this is highly non-linear it approximates the linear behavior very well. Moreover, the fit parameters provide the linear results. Only for the offset b a little transformation and according error propagation is required. (Also, I don't care about the value of p as long as it is somewhat larger than 5)
import matplotlib.pyplot as plt
import numpy as np
from scipy.optimize import curve_fit
np.set_printoptions( linewidth=250, precision=4)
np.set_printoptions( linewidth=250, precision=4)
### piecewise linear function for data generation
def pwl( x, m, b, a1, a2 ):
if x < a1:
out = pwl( a1, m, b, a1, a2 )
elif x > a2:
out = pwl( a2, m, b, a1, a2 )
else:
out = m * x + b
return out
### non-linear approximation
def func( x, m, b, a1, a2, p ):
out = b + np.log(
1 / ( 1 + np.exp( -m *( x - a1 ) )**p )
) / p - np.log(
1 / ( 1 + np.exp( -m * ( x - a2 ) )**p )
) / p
return out
### some data
nn = 36
xdata = np.linspace( -5, 19, nn )
ydata = np.fromiter( (pwl( x, -2.1, 11.6, -1.1, 12.7 ) for x in xdata ), float)
ydata += np.random.normal( size=nn, scale=0.2)
### dense grid for printing
xth = np.linspace( -5, 19, 150 )
###fitting
popt, cov = curve_fit( func, xdata, ydata, p0=[-2, 11, -1, 10, 1])
mF, betaF, a1F, a2F, pF = popt
bF = betaF - mF * a1F
sol=( mF, bF, a1F, a2F, pF )
### transforming the covariance due to the b' -> b mapping
J1 = np.identity(5)
J1[1,0] = -popt[2]
J1[1,2] = -popt[0]
cov2 = np.dot( J1, np.dot( cov, np.transpose( J1 ) ) )
### results
print( cov2 )
for i, v in enumerate( ("m", "b", "a1", "a2", "p" ) ):
print( "{:>2} = {:+2.4e} ± {:0.4e}".format( v, sol[i], np.sqrt( cov2[i,i] ) ) )
### plotting
fig = plt.figure()
ax = fig.add_subplot( 1, 1, 1 )
ax.plot( xdata, ydata, ls='', marker='+' )
ax.plot( xth, func( xth, -2, 11, -1, 10, 1 ) )
ax.plot( xth, func( xth, *popt ) )
plt.show()
Providing
[[ 1.3553e-04 -7.6291e-04 -4.3488e-04 4.5624e-04 1.2619e-01]
[-7.6291e-04 6.4126e-03 3.4560e-03 -1.5573e-03 -7.4983e-01]
[-4.3488e-04 3.4560e-03 3.4741e-03 -9.8284e-04 -4.2344e-01]
[ 4.5624e-04 -1.5573e-03 -9.8284e-04 3.0842e-03 -5.2739e+00]
[ 1.2619e-01 -7.4983e-01 -4.2344e-01 -5.2739e+00 3.1583e+05]]
m = -2.0810e+00 ± 9.7718e-03
b = +1.1463e+01 ± 6.7217e-02
a1 = -1.2545e+00 ± 5.0384e-02
a2 = +1.2739e+01 ± 4.7176e-02
p = +1.6840e+01 ± 2.9872e+02
and

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.

Nullcline Plot for Nonlinear System of ODEs

I am attempting to plot the nullcline (steady state) curves of the Oregonator model to assert the existence of a limit cycle by applying the Poincare-Bendixson Theorem. I am close, but for some reason the plot that is produced shows two straight lines. I think it has something to do with the plotting stage. Any ideas?
Also any hints for how to construct a quadrilateral to apply the theorem with would be most appreciated.
Code:
import numpy as np
import matplotlib.pyplot as plt
# Dimensionless parameters
eps = 0.04
q = 0.0008
f = 1
# Oregonator model as numpy array
def Sys(Y, t = 0):
return np.array((Y[0] * (1 - Y[0] - ((Y[0] - q) * f * Y[1]) / (Y[0] + q)) / eps, Y[0] - Y[1] ))
# Oregonator model steady states
def g(x,z):
return (x * (1 - x) + ((q - x) * f * z) / (q + x)) / eps
def h(x,z):
return x - z
# Initial lists containing values
x = []
z = []
def sys(iv1, iv2, dt, time):
# initial values:
x.append(iv1)
z.append(iv2)
# Compute and fill lists
for i in range(time):
x.append(x[i] + (g(x[i],z[i])) * dt)
z.append(z[i] + (h(x[i],z[i])) * dt)
return x, z
sys(1, 0.5, 0.01, 30)
# Locate and find equilibrium points
eqp = []
def find_fixed_points(r):
for x in range(r):
for z in range(r):
if ((g(x, z) == 0) and (h(x, z) == 0)):
eqp.append((x,z))
return eqp
# Plot nullclines
plt.plot([0,2],[2,0], 'r-', lw=2, label='x-nullcline')
plt.plot([1,1],[0,2], 'b-', lw=2, label='z-nullcline')
# Plot equilibrium points
for point in eqp:
plt.plot(point[0],point[1],"red", marker = "o", markersize = 10.0)
plt.legend(loc='best')
x = np.linspace(0, 2, 20)
z = np.linspace(0, 2, 20)
X1 , Z1 = np.meshgrid(x, z) # Create a grid
DX1, DZ1 = Sys([X1, Z1]) # Compute reaction rate on the grid
M = (np.hypot(DX1, DZ1)) # Norm reaction rate
M[ M == 0] = 1. # Avoid zero division errors
DX1 /= M # Normalise each arrows
DZ1 /= M
plt.quiver(X1, Z1, DX1, DZ1, M, pivot='mid')
plt.xlabel("x(\u03C4)")
plt.ylabel("z(\u03C4)")
plt.legend()
plt.grid()
plt.show()

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)

Lotka-Volterra equations(predator prey) using Runge-Kutta in Python

I am trying to write a program using the Lotka-Volterra equations for predator-prey interactions. Solve Using ODE's:
dx/dt = a*x - B*x*y
dy/dt = g*x*y - s*y
Using 4th order Runge-Kutta method
I need to plot a graph showing both x and y as a function of time from t = 0 to t=30.
a = alpha = 1
b = beta = 0.5
g = gamma = 0.5
s = sigma = 2
initial conditions x = y = 2
Here is my code so far but not display anything on the graph. Some help would be nice.
#!/usr/bin/env python
from __future__ import division, print_function
import matplotlib.pyplot as plt
import numpy as np
def rk4(f, r, t, h):
""" Runge-Kutta 4 method """
k1 = h*f(r, t)
k2 = h*f(r+0.5*k1, t+0.5*h)
k3 = h*f(r+0.5*k2, t+0.5*h)
k4 = h*f(r+k3, t+h)
return (k1 + 2*k2 + 2*k3 + k4)/6
def f(r, t):
alpha = 1.0
beta = 0.5
gamma = 0.5
sigma = 2.0
x, y = r[2], r[2]
fxd = x*(alpha - beta*y)
fyd = -y*(gamma - sigma*x)
return np.array([fxd, fyd], float)
tpoints = np.linspace(0, 30, 0.1)
xpoints = []
ypoints = []
r = np.array([2, 2], float)
for t in tpoints:
xpoints += [r[2]]
ypoints += [r[2]]
r += rk4(f, r, t, h)
plt.plot(tpoints, xpoints)
plt.plot(tpoints, ypoints)
plt.xlabel("Time")
plt.ylabel("Population")
plt.title("Lotka-Volterra Model")
plt.savefig("Lotka_Volterra.png")
plt.show()

A simple check of your variable tpoints after running your script shows it's empty:
In [7]: run test.py
In [8]: tpoints
Out[8]: array([], dtype=float64)
This is because you're using np.linspace incorrectly. The third argument is the number of elements desired in the output. You've requested an array of length 0.1.
Take a look at np.linspace's docstring. You won't have a problem figuring out how to adjust your code.

1) define 'h' variable.
2) use
tpoints = np.arange(30) #array([0, 1, 2, ..., 30])
not
np.linspace()
and don't forget to set time step size equal to h:
h=0.1
tpoints = np.arange(0, 30, h)
3) be careful with indexes:
def f(r,t):
...
x, y=r[0], r[1]
...
for t in tpoints:
xpoints += [r[0]]
ypoints += [r[1]]
...
and better use .append(x):
for t in tpoints:
xpoints.append(r[0])
ypoints.append(r[1])
...
Here's tested code for python 3.7 (I've set h=0.001 for more presize)
import matplotlib.pyplot as plt
import numpy as np
def rk4(r, t, h): #edited; no need for input f
""" Runge-Kutta 4 method """
k1 = h*f(r, t)
k2 = h*f(r+0.5*k1, t+0.5*h)
k3 = h*f(r+0.5*k2, t+0.5*h)
k4 = h*f(r+k3, t+h)
return (k1 + 2*k2 + 2*k3 + k4)/6
def f(r, t):
alpha = 1.0
beta = 0.5
gamma = 0.5
sigma = 2.0
x, y = r[0], r[1]
fxd = x*(alpha - beta*y)
fyd = -y*(gamma - sigma*x)
return np.array([fxd, fyd], float)
h=0.001 #edited
tpoints = np.arange(0, 30, h) #edited
xpoints, ypoints = [], []
r = np.array([2, 2], float)
for t in tpoints:
xpoints.append(r[0]) #edited
ypoints.append(r[1]) #edited
r += rk4(r, t, h) #edited; no need for input f
plt.plot(tpoints, xpoints)
plt.plot(tpoints, ypoints)
plt.xlabel("Time")
plt.ylabel("Population")
plt.title("Lotka-Volterra Model")
plt.savefig("Lotka_Volterra.png")
plt.show()
You can also try to plot "cycles":
plt.xlabel("Prey")
plt.ylabel("Predator")
plt.plot(xpoints, ypoints)
plt.show()
https://i.stack.imgur.com/NB9lc.png