I don't understand why this code (reference):
from numpy import zeros, linspace
import matplotlib.pyplot as plt
# Time unit: 1 h
beta = 10./(40*8*24)
gamma = 3./(15*24)
dt = 0.1 # 6 min
D = 30 # Simulate for D days
N_t = int(D*24/dt) # Corresponding no of hours
t = linspace(0, N_t*dt, N_t+1)
S = zeros(N_t+1)
I = zeros(N_t+1)
R = zeros(N_t+1)
# Initial condition
S[0] = 50
I[0] = 1
R[0] = 0
# Step equations forward in time
for n in range(N_t):
S[n+1] = S[n] - dt*beta*S[n]*I[n]
I[n+1] = I[n] + dt*beta*S[n]*I[n] - dt*gamma*I[n]
R[n+1] = R[n] + dt*gamma*I[n]
fig = plt.figure()
l1, l2, l3 = plt.plot(t, S, t, I, t, R)
fig.legend((l1, l2, l3), ('S', 'I', 'R'), 'upper left')
plt.xlabel('hours')
plt.show()
doesn't produce the same results as this code I made:
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
beta = 10. / (40 * 8 * 24)
gamma = 3. / (15 * 24)
def flu(y, t):
S = y[0]
P = y[1]
R = y[2]
S = - beta * S * P
P = beta * S * P - gamma * P
R = gamma * P
return [S, P, R]
C_I = [50, 1, 0]
t = np.linspace(0, 1000, 1000)
y = odeint(flu, C_I, t)
S = y[:, 0]
P = y[:, 1]
R = y[:, 2]
fig, ax = plt.subplots()
ax.plot(t, S, 'b--', label='S')
ax.plot(t, P, 'r--', label='I')
ax.plot(t, R, 'g--', label='R')
legend = ax.legend(loc='upper right', shadow=True, fontsize='x-large')
legend.get_frame().set_facecolor('#FFFCCC')
plt.show()
I used P instead of I to avoid confusion.
The equations solved with odeint should be the same as the ones provided in the reference link above. And if the equations I use are correct, which I am convinced they are, I don't understand where the mistake(s) lie(s).
Thank you for your help
You set S=y[0] then you set S=- beta * S * P. This overwrites y[0]!!! Similar problems for P and R
Try this:
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
beta = 10. / (40 * 8 * 24)
gamma = 3. / (15 * 24)
def flu(y, t):
S = y[0]
P = y[1]
R = y[2]
dS = - beta * S * P
dP = beta * S * P - gamma * P
dR = gamma * P
return [dS, dP, dR]
C_I = [50, 1, 0]
t = np.linspace(0, 1000, 1000)
y = odeint(flu, C_I, t)
S = y[:, 0]
P = y[:, 1]
R = y[:, 2]
fig, ax = plt.subplots()
ax.plot(t, S, 'b--', label='S')
ax.plot(t, P, 'r--', label='I')
ax.plot(t, R, 'g--', label='R')
legend = ax.legend(loc='upper right', shadow=True, fontsize='x-large')
legend.get_frame().set_facecolor('#FFFCCC')
plt.show()
Related
My following code can't add a common title, common legend also a common x and y label. When I add title, legend in between the for loop it works but its not working for the whole figure?
I want to add a common title, common legend, and shared x and y label in this plot. What is the best way to do that?
My code
#position vs Time
h = 0.05
nrow, ncol, count = 2,4,1
plt.figure(figsize=(20,10))
plt.title("Common Title") #does not working
with plt.style.context('ggplot'):
for inv in range(8):
plt.subplot(nrow, ncol, count)
x = [1]
v = [0]
rk4_x = [1] #position
rk4_v = [0] #velocity
t = np.arange(0, 30, 0.1)
for i in range(len(t)-1):
k1 = h*v[i]
l1 = h*-(x[i])
k2 = h*(v[i]+l1)
l2 = h*-(x[i]+k1)
x_value = x[i]+0.5*(k1+k2)
x.append(x_value)
v_value = v[i]+0.5*(l1+l2)
v.append(v_value)
k1 = h*rk4_v[i]
l1 = h*-(rk4_x[i])
k2 = h*(rk4_v[i]+0.5*l1)
l2 = h*-(rk4_x[i]+0.5*k1)
k3 = h*(rk4_v[i]+0.5*l2)
l3 = h*-(rk4_x[i]+0.5*k2)
k4 = h*(rk4_v[i]+l3)
l4 = h*-(rk4_x[i]+k3)
x_value = rk4_x[i]+(1/6)*(k1+2*k2+2*k3+k4)
rk4_x.append(x_value)
v_value = rk4_v[i]+(1/6)*(l1+2*l2+2*l3+l4)
rk4_v.append(v_value)
# with plt.style.context('ggplot'):
# plt.figure()
plt.plot(t, x,'r', label= "Position vs Time RK2")
plt.plot(t, rk4_x, 'k--', label = "Position vs Time RK4")
plt.ylim(-1.5, 1.5)
h = round(h,3)
plt.title("h="+str(h), fontsize= '11')
plt.grid('on')
plt.tick_params(labelcolor='none', which='both', top=False, bottom=False, left=False, right=False)
plt.xlabel('Time')
plt.ylabel('Position')
plt.legend()
h = h + 0.05
count = count+1
Use subplots with fig, axs. See the document on matplotlib.pyplot.subplots
And you should use Numpy array not the python list to speed up the numerical calculation. (Not this example, but generally).
Also, If the amount of calculation increases, it seems necessary to make the formula simple. Like x[i+1] = (1 - 0.5 * h**2) * x[i]+ h * v[i]
import numpy as np
import matplotlib.pyplot as plt
import matplotlib
matplotlib.use('Agg')
plt.style.use('ggplot')
def get_x_v(h, t):
x = np.ones_like(t)
v = np.zeros_like(t)
rk4_x = np.ones_like(t)
rk4_v = np.zeros_like(t)
for i, _ in enumerate(t[:-1]):
k1 = h * (v[i])
l1 = h * (x[i]) * -1
k2 = h * (v[i] + l1)
l2 = h * (x[i] + k1) * -1
x[i+1] = x[i] + 0.5 * (k1 + k2)
v[i+1] = v[i] + 0.5 * (l1 + l2)
rk1 = h * (rk4_v[i])
rl1 = h * (rk4_x[i]) * -1
rk2 = h * (rk4_v[i] + 0.5 * rl1)
rl2 = h * (rk4_x[i] + 0.5 * rk1) * -1
rk3 = h * (rk4_v[i] + 0.5 * rl2)
rl3 = h * (rk4_x[i] + 0.5 * rk2) * -1
rk4 = h * (rk4_v[i] + rl3)
rl4 = h * (rk4_x[i] + rk3) * -1
rk4_x[i+1] = rk4_x[i]+(1/6)*(rk1+2*rk2+2*rk3+rk4)
rk4_v[i+1] = rk4_v[i]+(1/6)*(rl1+2*rl2+2*rl3+rl4)
return x, v, rk4_x, rk4_v
h = 0.05
ncol, nrow = 4, 2
fig, axs = plt.subplots(nrows=nrow, ncols=ncol, figsize=(20, 10))
fig.suptitle('Common Title', y=0.94, fontsize=20)
fig.supxlabel('Common X Label', y=0.05, fontsize=14)
t = np.arange(0, 30, 0.1)
for ax in axs.flat:
x, v, rk4_x, rk_v = get_x_v(h, t)
ax.plot(t, x, 'r', label="Position vs Time RK2")
ax.plot(t, rk4_x, 'k--', label="Position vs Time RK4")
ax.set_ylim(-1.5, 1.5)
ax.set_title('h={0}'.format(str(round(h,3))), fontsize='11')
ax.grid('on')
ax.tick_params(labelcolor='none', which='both',
top=False, bottom=False, left=False, right=False)
ax.set_xlabel('Time')
ax.set_ylabel('Position')
ax.legend(loc=1)
ax.set_facecolor('gray')
h += 0.05
plt.savefig('test.png')
I made an animation of an orbit that plots an orbit using an initial altitude and velocity.
Here is the code:
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import ode
import matplotlib.animation as animation
plt.style.use('dark_background')
mu_earth = 398600
# Planet radius, km
R = 6371
fig = plt.figure()
ax1 = fig.add_subplot(111)
orbit_alt = 500
v_mag = 7.7 # km/s
# time step (delta-time)
dt = 10
satellite_size = 100
r_mag = R + orbit_alt
v_esc = np.sqrt(2*mu_earth*1e9/(R*1000 + orbit_alt*1000))
earth = plt.Circle((0, 0), 6371, color='blue')
ax1.add_artist(earth)
def diff_eq(t, y):
rx, ry, vx, vy = y # state vectors
r = np.array([rx, ry])
radial_vector = np.linalg.norm(r)
ax, ay = -r * mu_earth / radial_vector ** 3
return [vx, vy, ax, ay]
# Initial states
r0 = [r_mag, 0]
v0 = [0, v_mag]
specific_energy = (v_mag * 1000 * v_mag * 1000) / 2 - ((mu_earth * 1e9) / ((R + orbit_alt) * 1000))
semi_major = -mu_earth * 1e9 / (2 * specific_energy)
orbit_period = 2 * np.pi * np.sqrt(semi_major ** 3 / (mu_earth * 1e9))
if v_mag * 1000 < v_esc:
tspan = 5 * orbit_period
else:
tspan = 1500000
eccen = ((v_mag * 1000) ** 2 * (r_mag)*1000)/(mu_earth*1e9) - 1
perigee = (semi_major * (1 - np.abs(eccen)))/1000
apogee = (semi_major * (1 + np.abs(eccen)))/1000
n_steps = int(np.ceil(tspan/dt))
ts = np.zeros((n_steps, 1))
ys = np.zeros((n_steps, 4))
ys0 = r0 + v0
ts[0] = 0
ys[0] = ys0
solver = ode(diff_eq)
solver.set_integrator('lsoda')
solver.set_initial_value(ys0, 0)
rs = ys[:, :2]
def animate(i):
solver.integrate(solver.t + dt)
ts[i] = solver.t
ys[i] = solver.y
point = plt.Circle((ys[i][0], ys[i][1]), satellite_size, facecolor=(1, 1, 1))
ax1.add_artist(point)
satellite, = ax1.plot(rs[:, 0], rs[:, 1], color='r', alpha = 1)
return satellite, point
ax1.set_xlim([-apogee, apogee])
ax1.set_ylim([-apogee, apogee])
ax1.set_xlabel('X (km)')
ax1.set_ylabel('Y (km)')
ax1.set_title("ORBIT SIMULATION")
plt.gca().set_aspect('equal', adjustable='box')
ani = animation.FuncAnimation(fig, animate, interval=0.01, blit=True)
plt.show()
And here is a screen shot of the output:
I want to omit the line that goes to the origin (0, 0). That line is a part of the satellite line object. What I believe is happening is that the origin is considered the first point, and the satellite is the second point, and matplotlib is connecting them. So how could that line leading to the middle be omitted?
The main problem is that your data is a full vector of zeros and your code plots the full dataset. Change from rs[:, 0] to rs[:, 0][:i] to only plot the points that have been simulated so far.
Second consideration is that FuncAnimation is usually used to update data already plotted. See small change below in animate(i).
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import ode
import matplotlib.animation as animation
plt.style.use('dark_background')
mu_earth = 398600
# Planet radius, km
R = 6371
fig = plt.figure()
ax1 = fig.add_subplot(111)
orbit_alt = 500
v_mag = 7.7 # km/s
# time step (delta-time)
dt = 10
satellite_size = 100
r_mag = R + orbit_alt
v_esc = np.sqrt(2*mu_earth*1e9/(R*1000 + orbit_alt*1000))
earth = plt.Circle((0, 0), 6371, color='blue')
ax1.add_artist(earth)
def diff_eq(t, y):
rx, ry, vx, vy = y # state vectors
r = np.array([rx, ry])
radial_vector = np.linalg.norm(r)
ax, ay = -r * mu_earth / radial_vector ** 3
return [vx, vy, ax, ay]
# Initial states
r0 = [r_mag, 0]
v0 = [0, v_mag]
specific_energy = (v_mag * 1000 * v_mag * 1000) / 2 - ((mu_earth * 1e9) / ((R + orbit_alt) * 1000))
semi_major = -mu_earth * 1e9 / (2 * specific_energy)
orbit_period = 2 * np.pi * np.sqrt(semi_major ** 3 / (mu_earth * 1e9))
if v_mag * 1000 < v_esc:
tspan = 5 * orbit_period
else:
tspan = 1500000
eccen = ((v_mag * 1000) ** 2 * (r_mag)*1000)/(mu_earth*1e9) - 1
perigee = (semi_major * (1 - np.abs(eccen)))/1000
apogee = (semi_major * (1 + np.abs(eccen)))/1000
n_steps = int(np.ceil(tspan/dt))
ts = np.zeros((n_steps, 1))
ys = np.zeros((n_steps, 4))
ys0 = r0 + v0
ts[0] = 0
ys[0] = ys0
solver = ode(diff_eq)
solver.set_integrator('lsoda')
solver.set_initial_value(ys0, 0)
rs = ys[:, :2]
orbit, = ax1.plot(rs[:, 0], rs[:, 1], color='r', alpha=1)
point = plt.Circle((ys[0][0], ys[0][1]), satellite_size, facecolor=(1, 1, 1))
ax1.add_artist(point)
def animate(i):
solver.integrate(solver.t + dt)
ts[i] = solver.t
ys[i] = solver.y
orbit.set_data(rs[:, 0][:i], rs[:, 1][:i])
point.set_center((ys[i][0], ys[i][1]))
return orbit, point
ax1.set_xlim([-apogee, apogee])
ax1.set_ylim([-apogee, apogee])
ax1.set_xlabel('X (km)')
ax1.set_ylabel('Y (km)')
ax1.set_title("ORBIT SIMULATION")
plt.gca().set_aspect('equal', adjustable='box')
ani = animation.FuncAnimation(fig, animate, interval=10, blit=True)
plt.show()
I'm trying to plot a diagram of Coulomb damping mass-spring model, see image bellow. mi*N is the damping friction force.
Becuase the damping force switches direction depending on the direction of speed vector, we have 2 different equations of motion x(t), I named them x_a(t) and x_b(t). If you look closely on the diagram above, the use of x(t) depends on the period, for the first period which is from 0 to pi/omega_n x_a(t) is used, for the second period which is from pi/omega_n to 2*pi/omega_n we use x_b(t) and so on.
I was thinking to use a piece of code which would go something like
t = i * (pi/omega_n)
for i = 1, 3, 5, 7,...
x_a(t)
for i = 2, 4, 6, 8,...
x_b(t)
But I have no clue how to implement this.
I managed to define the x_a(t) part of the code and plot it with the undampled model, see bellow.
import numpy as np
import matplotlib.pyplot as plt
#constants
k = 2 #(N/m), spring coef
m = 0.04 #(kg), mass
x0 = -0.1 #(m), preload
mi = 0.3 #(), dry dynamic friction coef. ABS-ABS
N = 0.3 #(N), normal contact force
f_tr = mi * N / k #friction force/pring coef - equivalent distance
omega_0 = np.sqrt(k/m)
#time
t = np.linspace(0,5,100)
#undamped model
x_undamp = x0*np.cos(omega_0*t)
dx_undamp = -omega_0*x0*np.sin(omega_0*t)
#damped model
x_damp = (x0+f_tr)*np.cos(omega_0*t)-f_tr
dx_damp = -omega_0*(x0+f_tr)*np.sin(omega_0*t)
#time to x=0
t0 = np.arccos(f_tr/(x0+f_tr))/omega_0
print t0
#plotting
fig, (ax1, ax2) = plt.subplots(1, 2)
fig.suptitle('position on left, velocity on right')
ax1.plot(t, x_undamp,'r', label='x_undamp')
ax1.plot(t, x_damp, 'b', label = 'x_damp')
ax2.plot(t, dx_undamp, 'r', label = 'dx_undamp')
ax2.plot(t, dx_damp, 'b', label = 'dx_damp')
#grids, titles, legends, axis labels
ax1.grid()
ax2.grid()
ax1.set_title('Position vs time')
ax2.set_title('Velocity vs time')
ax1.legend()
ax2.legend()
ax1.set_xlabel('t(s)')
ax1.set_ylabel('x(m)')
ax2.set_xlabel('t(s)')
ax2.set_ylabel('dx(m/s)')
plt.show()
I'm gonna put plot a diagram of Coulomb damping mass-spring model solution here till someone comes with a better one:
from scipy.integrate import odeint
import numpy as np
import matplotlib.pyplot as plt
m = 0.2
k = 2.0
c = 0.1
mus = 0.3
muk = 0.2
g = 9.8
vf = 0.01
v0 = 0.0
t1 = 10
sign = lambda x: np.tanh(100*x)
def Xi(t):
if t < 1 :
return 0
else:
return 1
vXi = np.vectorize(Xi)
def eq(X, t, Xi):
Ff = k * (Xi(t) - X[0])
if np.abs(X[1]) < vf and np.abs(Ff) < mus * m * g :
Ff = k * (Xi(t) - X[0])
else:
Ff = sign(X[1]) * muk * m * g
d2x = (k * (Xi(t) - X[0]) - Ff) / m
return [X[1], d2x]
t = np.linspace(0, t1, 1000)
X0 = [v0, 0]
sol = odeint(func = eq, y0 = X0, t = t, args = (Xi, ), mxstep = 50000, atol = 1e-5)
plt.plot(t, sol[:, 0], 'r-', label = 'Output (x(t))')
plt.plot(t, vXi(t), 'b-', label = 'Input (xi(t))')
plt.ylabel('values')
plt.xlabel('time')
plt.legend(loc='best')
plt.show()
How can I plot in 3D in python?
I am trying to plot orbital trajectories. Plotting Orbital Trajectories
From the link above, I was able to get help with setting up the function. However I don't know how to plot in 3D.
When this is run, it doesn't generate the correct trajectory.
Switching np.linspace to np.arnage cause a memory error and I am running this on a 64bit system running Xubuntu with 16 GB of Ram.
So I tried converting Distance Units and Time Units but something isn't correct. Maybe my math or something else.
I let 149.6 * 10 ** 6 = 1 DU. A TU is defined as mu = DU ** 3 / TU ** 2 so 1TU = 2241.15 and DU/TU = 66751.4 Using these conversion, I have: I also tried using x2,y2,z2 to see if that would work.
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
from numpy import linspace
from mpl_toolkits.mplot3d import Axes3D
mu = 1
# r0 = [-149.6 * 10 ** 6, 0.0, 0.0] # Initial position
# v0 = [29.9652, -5.04769, 0.0] # Initial velocity
u0 = [-1, 0.0, 0.0, 0.000448907, -0.0000756192, 0.0]
def deriv(u, dt):
n = -mu / np.sqrt(u[0] ** 2 + u[1] ** 2 + u[2] ** 2)
return [u[3], # dotu[0] = u[3]'
u[4], # dotu[1] = u[4]'
u[5], # dotu[2] = u[5]'
u[0] * n, # dotu[3] = u[0] * n
u[1] * n, # dotu[4] = u[1] * n
u[2] * n] # dotu[5] = u[2] * n
dt = np.arange(0.0, 20, .0001) # Time to run code in seconds'
u = odeint(deriv, u0, dt)
x, y, z, x2, y2, z2 = u.T
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot(x2, y2, z2)
plt.show()
but this plot isn't correct either. It should be an ellipse that stays on the same trajectory.
#!/usr/bin/env python
# This program solves the 3 Body Problem numerically and plots the
# trajectories
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
from numpy import linspace
from mpl_toolkits.mplot3d import Axes3D
mu = 132712000000.0
# r0 = [-149.6 * 10 ** 6, 0.0, 0.0] # Initial position
# v0 = [29.9652, -5.04769, 0.0] # Initial velocity
u0 = [-149.6 * 10 ** 6, 0.0, 0.0, 29.9652, -5.04769, 0.0]
def deriv(u, dt):
n = -mu / np.sqrt(u[0] ** 2 + u[1] ** 2 + u[2] ** 2)
return [u[3], # dotu[0] = u[3]'
u[4], # dotu[1] = u[4]'
u[5], # dotu[2] = u[5]'
u[0] * n, # dotu[3] = u[0] * n
u[1] * n, # dotu[4] = u[1] * n
u[2] * n] # dotu[5] = u[2] * n
dt = np.linspace(0.0, 86400 * 700, 5000) # Time to run code in seconds'
u = odeint(deriv, u0, dt)
x, y, z, x2, y2, z2 = u.T
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot(x, y, z)
plt.show()
You can literally take the first several lines from that page that #sashkello, and plug in the x,y, and z that you got from the ode solver.
Copied from http://matplotlib.org/mpl_toolkits/mplot3d/tutorial.html :
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
#<<solve for x, y, z here>>#
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot(x, y, z)
plt.show()
I'd like to plot pulse propagation in such a way at each step, it plots the pulse shape. In other words, I want a serie of x-z plots, for each values of y. Something like this (without color):
How can I do this using matplotlib (or Mayavi)? Here is what I did so far:
def drawPropagation(beta2, C, z):
""" beta2 in ps / km
C is chirp
z is an array of z positions """
T = numpy.linspace(-10, 10, 100)
sx = T.size
sy = z.size
T = numpy.tile(T, (sy, 1))
z = numpy.tile(z, (sx, 1)).T
U = 1 / numpy.sqrt(1 - 1j*beta2*z * (1 + 1j * C)) * numpy.exp(- 0.5 * (1 + 1j * C) * T * T / (1 - 1j*beta2*z*(1 + 1j*C)))
fig = pyplot.figure()
ax = fig.add_subplot(1,1,1, projection='3d')
surf = ax.plot_wireframe(T, z, abs(U))
Change to:
ax.plot_wireframe(T, z, abs(U), cstride=1000)
and call:
drawPropagation(1.0, 1.0, numpy.linspace(-2, 2, 10))
will create the following graph:
If you need the curve been filled with white color:
import numpy
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import pyplot
from matplotlib.collections import PolyCollection
def drawPropagation(beta2, C, z):
""" beta2 in ps / km
C is chirp
z is an array of z positions """
T = numpy.linspace(-10, 10, 100)
sx = T.size
sy = z.size
T = numpy.tile(T, (sy, 1))
z = numpy.tile(z, (sx, 1)).T
U = 1 / numpy.sqrt(1 - 1j*beta2*z * (1 + 1j * C)) * numpy.exp(- 0.5 * (1 + 1j * C) * T * T / (1 - 1j*beta2*z*(1 + 1j*C)))
fig = pyplot.figure()
ax = fig.add_subplot(1,1,1, projection='3d')
U = numpy.abs(U)
verts = []
for i in xrange(T.shape[0]):
verts.append(zip(T[i, :], U[i, :]))
poly = PolyCollection(verts, facecolors=(1,1,1,1), edgecolors=(0,0,1,1))
ax.add_collection3d(poly, zs=z[:, 0], zdir='y')
ax.set_xlim3d(numpy.min(T), numpy.max(T))
ax.set_ylim3d(numpy.min(z), numpy.max(z))
ax.set_zlim3d(numpy.min(U), numpy.max(U))
drawPropagation(1.0, 1.0, numpy.linspace(-2, 2, 10))
pyplot.show()