I am trying to solve for the following equation with a simple algorithm. I am not sure if the algorithm that I'm using is the best one or not but it is the only way that I could think of.
In this equation, everything other than P is known, and I am trying to solve for that. N is an array of counts, an i is the channel number. S(n) is the probability of having a certain n and C is binomial coefficient of (n, r). Pi is the probability in i channel and Pj is the probability in the previous channels with D distance to i. The code itself is not working but I believe that the main problem is in the way that I am trying to solve for it.
import numpy as np
import matplotlib.pyplot as plt
import math as ms
from scipy.misc import derivative
import scipy as sp
def next_guess(f, x):
slop = derivative(f, x, dx = 0.01)
return x - float(f(x))/slop
def my_newton(f, guess):
for i in range(30):
#print(guess)
guess = next_guess(f, guess)
return guess
def binomial(n, r):
dif = ms.factorial(n - r)
n = ms.factorial(n)
r = ms.factorial(r)
return (n/(r*dif))
def wrap_func(x, S = np.array([0.1, 0.5, 0.2, 0.1, 0.1]), D = 1, N = np.array([10, 15, 20, 1, 13])):
if type(x) == float:
z = np.zeros(1)
else:
z = np.zeros(x.shape[0])
N_tot = N.sum()
n_max = S.shape[0]
for i in range(z.shape[0]):
z[i] += my_newton(func(x, S = S, D = 1, N = N[i], n_max = n_max, N_tot = N_tot, i = i), i/100)
return z
def func(x, S = np.array([0.1, 0.5, 0.2, 0.1, 0.1]), D = 1, N = 0, n_max = 5, N_tot = 10, i = 0):
S_sum = 0
binom_sum = 0
y = 0
for n in range(n_max):
S_sum += S[n]
for r in range(n):
binom_sum += binomial(n, r)
y += S_sum * binom_sum * (x**r) * (1 - x - summ_x(x, D, i, S, N, n_max, N_tot))**(n-r)
return N_tot * y - N
def summ_x(x, D, i, S, N, n_max, N_tot):
j_min = max(i - D - 1, 0)
j_max = i - 1
x_values = 0
if i == 0:
return x_values
else:
for j in range(j_min, j_max):
x_values += func(x, S, D, N, n_max, N_tot, i)
return x_values
x = np.linspace(0, 1, 1000)
S = np.array([0.1, 0.5, 0.2, 0.1, 0.1])
N = np.random.choice(50, size = 1000)
print(my_newton(wrap_func, 0.1))
plt.plot(x, wrap_func(x, S = S, D = 1, N = N ))
plt.axhline(0, lw = 0.5, color = 'grey')
#plt.plot(my_newton(wrap_func, 1), wrap_func(my_newton(wrap_func, 1), S = S, D = 1, N = N), 'd')
plt.show()
Related
I am trying to figure to find out an alternative way to rewrite
Initial and boundary condition for the 1 dimensional heat eqaution
I have the following examples for 1D
#Methodd used during classes
import math
import numpy as np
import matplotlib.pyplot as plt
#D, Nx, Nt, L, T = 1.0, 14, 100, 1.0, 0.1
D, Nx, Nt, L, T = 1.0, 20, 250, 1.0, 0.1
#D, Nx, Nt, L, T = 1.0, 40, 500, 1.0, 0.1
t = np.linspace(0, T, num=Nt+1, dtype = float)
x = np.linspace(0, L, num=Nx+1, dtype = float)
dx = x[1] - x[0]
dt = t[1] - t[0]
r = D*dt/ (dx*dx)
print('r = {}'.format(r))
assert r < 0.5
#u = np.zeros((Nx+1, Nt+1), dtype = float)
u = np.empty((Nx+1, Nt+1), dtype = float)
#initial condition
#u[:,0] = 4.0*x*(1-x)
#u[:,0] = 0.5(np.sign(x-0.4) + np.sign(0.6 - x))
#u[:,0] = np.where(x < 0.5, 2*x, 2*(x -1))
u[:,0] = np.where(x < 0.5, 0, 1)
#boundary conditon
u[0,:] = 0.0
u[Nx,:] = 0.0
#u[Nx,:] = 1.0
#iteration solution
for j in range (Nt):
#for i in range(1, Nx):
#u[i, j+1] = r*u[i - 1, j] + (1 -2r)*u[i, j] + r*u[i + 1, j]
#vectorization
u[1:-1,j+1] = r*u[:-2,j] + (1-2*r)*u[1:-1,j] + r*u[2:,j]
#visualization
print(u)
plt.title("1D heat equation")
plt.xlabel("time")
plt.ylabel("X")
plt.imshow(u[:,::3], cmap = "hot", interpolation = "nearest")
#plt.imshow(u[:,::3], cmap = "hot", interpolation = "bilinear")
#plt.imshow(u[:,::3], cmap = "hot", interpolation = "hamming")
#plt.imshow(u[:,::5], cmap = "hot", interpolation = "nearest")
As you could see I have different choices to pick up. But I would like to add some alternative initial and boundary conditions. The graphics I should get are like the following one:
Could anyone possibly suggest any alternative patterns for them? Thank you so much.
This program creates a cube of size Gridsize**3 with user choice of starting point and space between point (even if they are not function parameters there isn't difficult to implement).
import numpy as np
def CreateMap(Gridsize):
X = Y = Z = Gridsize
M = np.zeros(shape=(X*Y*Z, 3))
d_x = 5 / Gridsize # increment of the cube x dimension
d_y = 5 / Gridsize
d_z = 5 / Gridsize
x0 = -1.0
y0 = 1.0
z0 = 0
x = np.arange(x0, X * d_x, d_x, dtype=float)
y = np.arange(y0, Y * d_y, d_y, dtype=float)
z = np.arange(z0, Z * d_z, d_z, dtype=float)
g = 0
for i in range(X):
for j in range(Y):
for k in range(Z):
M[g, 0] = x[i]
M[g, 1] = y[j]
M[g, 2] = z[k]
g = g + 1
print(M)
return 0
I was wondering what was the best method to create an hyper cube of size Gridsize**n were n will also be user defined?
Check out np.meshgrid. Instead of your for loops, you can just do
M = np.stack(np.meshgrid(x, y, z))
If you guys have optimization advice...
import numpy as np
def CreateMap(Gridsize, x0, xf):
k = np.shape(x0)[0]
M = np.zeros(shape=(Gridsize**k, k))
d_x = np.zeros(k)
for i in range(k):
d = 0
j = 0
d_x[i] = (xf[i] - x0[i]) / (Gridsize - 1) # increment of the cube x dimension
x = np.arange(x0[i], xf[i]+d_x[i], d_x[i], dtype=float)
for v in range(Gridsize ** (k - i - 1)):
for j in range(Gridsize):
temp = x[j]
for z in range(Gridsize ** i):
M[d, i] = temp
d = d + 1
print(M)
return 0
x0 = np.array([-1, 0, 1])
xf = np.array([10, 2, 5])
CreateMap(4, x0, xf)
I try to minimize the vectors x,y, but they just satisfied constraints with no work for minimizing.
e.g:
input init: x=[0.2,0.3,0.5] (sum of elements is 1) ,feedback: res.x=[0.,0.3,0.5],it hasn't changed at all!
# -*- coding:utf8-*-
import random
import numpy as np
from scipy import optimize
import networkx as nx
import matplotlib.pyplot as plt
def Ud(x, X, Aj, G, p):
lost = G.node[Aj[-1]]["weight"]
Aj = Aj[:-1]
P = 1
sum = 0
for xi in x:
for Xi in X:
N = set(Aj).intersection(set(Xi))
# N=[random.randint(90,120) for _ in range(0,1)]
for n in N:
P *= (1 - p[n - 1])
sum += xi * P
P = 1
return -lost * sum
### objective function for defenders ###
def min_Ud(x, X, A, G, p):
min = float("inf")
for Aj in A:
temp_min = Ud(x, X, Aj, G, p)
if temp_min < min:
min = temp_min
return -min
### objective function for attackers ###
def Ua(a, X, A, G, p):
sum = 0
P = 1
for aj, Aj in a, A:
for Xi in X:
N = set(Aj[:-1]).intersection(set(Xi))
for n in N:
P *= (1 - p[n - 1])
sum += G.node[Aj[-1]]["weight"] * aj * P
P = 1
return sum
### fun for LP ###
def coreLP(X, A, G, p):
x0 = np.array([0.7, 0.2, 0.1])
a0 = np.array([0.5, 0.2, 0.3])
x_res = float("inf")
a_res = float("inf")
def c1(x):
return x.sum() - 1
cons = ({'type': 'eq', 'fun': c1})
Ud_star = optimize.minimize(min_Ud, x0, args=(X, A, G, p), constraints=cons, bounds=((0, 1), (0, 1), (0, 1)))
Ua_star = optimize.minimize(min_Ud, a0, args=(X, A, G, p), constraints=cons, bounds=((0, 1), (0, 1), (0, 1)))
print Ud_star
print Ua_star
return Ud_star.x, Ua_star.x
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
At first, thank you everybody for the amazing work on stackoverflow... you guys are amazing and have helped me out quite some times already. Regarding my problem: I have a series of vectors in the format (VectorX, VectorY, StartingpointX, StartingpointY)
data = [(-0.15304757819399128, -0.034405679205349315, -5.42877197265625, 53.412933349609375), (-0.30532995491023485, -0.21523935094046465, -63.36669921875, 91.832427978515625), (-0.15872430479453215, -0.077999419482978283, -67.805389404296875, 81.001983642578125), (-0.36415549211687903, -0.33757147194808113, -59.015228271484375, 82.976226806640625), (0.0, 0.0, 0.0, 0.0), (-0.052973530805275004, 0.098212384392411423, 19.02667236328125, -13.72125244140625), (-0.34318724086483599, 0.17123742336019632, 80.0394287109375, 108.58499145507812), (0.19410169197834648, -0.17635303976555861, -55.603790283203125, -76.298828125), (-0.38774018337716143, -0.0824692384322816, -44.59942626953125, 68.402496337890625), (0.062202543524108478, -0.37219011831012949, -79.828826904296875, -10.764404296875), (-0.56582988168383963, 0.14872365390732512, 39.67657470703125, 97.303192138671875), (0.12496832467900276, -0.12216653754859408, 24.65948486328125, -30.92584228515625)]
When I plot the vectorfield it looks like this:
import numpy as np
import matplotlib.pyplot as plt
def main():
# Format Data...
numdata = len(data)
x = np.zeros(numdata)
y = np.zeros(numdata)
u = np.zeros(numdata)
v = np.zeros(numdata)
for i,el in enumerate(data):
x[i] = el[2]
y[i] = el[3]
# length of vector
z[i] = math.sqrt(el[0]**2+el[1]**2)
u[i] = el[0]
v[i] = el[1]
# Plot
plt.quiver(x,y,u,v )
# showing the length with color
plt.scatter(x, y, c=z)
plt.show()
main()
I want to create a polynomial function to fit a continous vector field for the whole area. After some research I found the following functions for fitting polynoms in two dimensions. The problem is, that it only accepts one value for the value that is fitted.
def polyfit2d(x, y, z, order=3):
ncols = (order + 1)**2
G = np.zeros((x.size, ncols))
ij = itertools.product(range(order+1), range(order+1))
for k, (i,j) in enumerate(ij):
G[:,k] = x**i * y**j
m, _, _, _ = np.linalg.lstsq(G, z)
return m
def polyval2d(x, y, m):
order = int(np.sqrt(len(m))) - 1
ij = itertools.product(range(order+1), range(order+1))
z = np.zeros_like(x)
for a, (i,j) in zip(m, ij):
z += a * x**i * y**j
return z
Also when I tried to fit the one dimensional length of the vectors, the values returned from the polyval2d were completely off. Does anybody know a method to get a fitted function that will return a vector (x,y) for any point in the grid?
Thank you!
A polynomial to fit a 2-d vector field will be two bivariate polynomials - one for the x-component and one for the y-component. In other words, your final polynomial fitting will look something like:
P(x,y) = ( x + x*y, 1 + x + y )
So you will have to call polyfit2d twice. Here is an example:
import numpy as np
import itertools
def polyfit2d(x, y, z, order=3):
ncols = (order + 1)**2
G = np.zeros((x.size, ncols))
ij = itertools.product(range(order+1), range(order+1))
for k, (i,j) in enumerate(ij):
G[:,k] = x**i * y**j
m, _, _, _ = np.linalg.lstsq(G, z)
return m
def fmt1(x,i):
if i == 0:
return ""
elif i == 1:
return x
else:
return x + '^' + str(i)
def fmt2(i,j):
if i == 0:
return fmt1('y',j)
elif j == 0:
return fmt1('x',i)
else:
return fmt1('x',i) + fmt1('y',j)
def fmtpoly2(m, order):
for (i,j), c in zip(itertools.product(range(order+1), range(order+1)), m):
yield ("%f %s" % (c, fmt2(i,j)))
xs = np.array([ 0, 1, 2, 3] )
ys = np.array([ 0, 1, 2, 3] )
zx = np.array([ 0, 2, 6, 12])
zy = np.array([ 1, 3, 5, 7])
mx = polyfit2d(xs, ys, zx, 2)
print "x-component(x,y) = ", ' + '.join(fmtpoly2(mx,2))
my = polyfit2d(xs, ys, zy, 2)
print "y-component(x,y) = ", ' + '.join(fmtpoly2(my,2))
In this example our vector field is:
at (0,0): (0,1)
at (1,1): (2,3)
at (2,2): (6,5)
at (3,3): (12,7)
Also, I think I found a bug in polyval2d - this version gives more accurate results:
def polyval2d(x, y, m):
order = int(np.sqrt(len(m))) - 1
ij = itertools.product(range(order+1), range(order+1))
z = np.zeros_like(x)
for a, (i,j) in zip(m, ij):
z = z + a * x**i * y**j
return z