I am getting a horrible fit when I am trying to fit a parabola to this data.
I am initially making a histogram of the data which is the position of an object and then plotting the negative log values of the histogram bin counts to the position using a parabola fit.
the code I am using is this:
time,pos=postime()
plt.plot(time, pos)
poslen=len(pos)
plt.xlabel('Time')
plt.ylabel('Positions')
plt.show()
n,bins,patches = plt.hist(pos,bins=100)
n=n.tolist()
plt.show()
l=len(bins)
s=len(n)
posx=[]
i=0
j=0
pbin=[]
sig=[]
while j < (l-1):
pbin.append((bins[j]+bins[j+1])/2)
j=j+1
while i < s:
if n[i]==0:
pbin[i]=0
else:
sig.append(np.power(1/n[i],2))
n[i]=n[i]/poslen
n[i]=np.log(n[i])
n[i]=n[i]*(-1)
i=i+1
n[:]=[y for y in n if y != 0]
pbin[:]=[y for y in pbin if y != 0]
from scipy.optimize import curve_fit
def parabola(x, a , b):
return a * (np.power(x,2)) + b
popt, pcov = curve_fit(parabola, pbin, n)
print popt
plt.plot(pbin,n)
plt.plot(pbin, parabola(pbin, *popt), 'r-')
I am not sure why you are computing the histogram... But here is a working example which does not require histogram computation.
import numpy as np
from scipy.optimize import curve_fit
from matplotlib import pyplot
time_ = np.arange(-5, 5, 0.1)
pos = time_**2 + np.random.rand(len(time_))*5
def parabola(x, a, b):
return a * (np.power(x, 2)) + b
popt, pcov = curve_fit(parabola, time_, pos)
yfit = parabola(time_, *popt)
pyplot.plot(time_, pos, 'o')
pyplot.plot(time_, yfit)
Also, if your time_ vector is not uniformly sampled, and you want it to be uniformly sampled for the fit, you can do: fittime_ = np.linsapce(np.min(time_), np.max(time_)) and then yfit = parabola(fittime_, *popt).
You can also use matrix inversion.
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-5,5,100)
Y = (np.power(x,2) + np.random.normal(0,1,x.shape)).reshape(-1,1)
X = np.c_[np.ones(x.shape), x, np.power(x,2)]
A = np.linalg.inv(X.transpose().dot(X)).dot(X.transpose().dot(Y))
Yp = X.dot(A)
fig = plt.figure()
ax = fig.add_subplot()
plt.plot(x,Y,'o',alpha=0.5)
plt.plot(x,Yp)
plt.show()
The matrix form is
X*A=Y
A=(Xt*X)-1*Xt*Y
You can have a better idea here if needed. It does not always work out and you may want to apply some form of regularization.
Related
I use the following code to plot a scatter plot. I have been trying to plot the lower boundary for it. I tried following the other question but I was unable to replicate it for my objective function and data. The code is as follows :
from numpy import arange
import pandas as pd
from pandas import read_csv
from scipy.optimize import curve_fit
from matplotlib import pyplot
def objective(x, a, b):
return a * x + b
events = pd.read_excel('values.xlsx')
x = events.loc[:, 'Value']
y = events.loc[:, 'Frame']
popt, _ = curve_fit(objective, x, y)
a, b = popt
pyplot.scatter(x, y)
x_line = arange(min(x), max(x), 1)
y_line = objective(x_line, a, b)
pyplot.plot(x_line, y_line, '--', color='purple')
xmin, xmax = pyplot.xlim() # the limits of the x-axis for drawing the line
ymin, ymax = pyplot.ylim()
pos_min = np.argmin(x)
pos_max = np.argmax(x)
alpha_min = x[pos_min]
alpha_max = x[pos_max]
pyplot.show()
I want to plot the lower boundary of the points like .
As you have a linear function, your upper and lower bound will have the same slope a but different b-values. So, we calculate them for all points and choose the lowest and highest:
import numpy as np
from scipy.optimize import curve_fit
from matplotlib import pyplot
def objective(x, a, b):
return a * x + b
#sample data
rng = np.random.default_rng(123)
x = np.linspace(2, 10, 150)
y = objective(x, -2, 3)
y += 5 * rng.random(len(x))
popt, _ = curve_fit(objective, x, y)
a, b = popt
pyplot.scatter(x, y, label="raw data")
x_line = np.asarray([np.min(x), np.max(x)])
y_line = objective(x_line, a, b)
pyplot.plot(x_line, y_line, '--', color='purple', label=f"y={a:.2f}x+{b:.2f}")
b_values = y - a * x
pyplot.plot(x_line, objective(x_line, a, np.min(b_values)), '--', color='red', label="lower bound")
pyplot.plot(x_line, objective(x_line, a, np.max(b_values)), '--', color='orange', label="upper bound")
pyplot.legend()
pyplot.show()
Sample output:
Given:
and
We have formula:
I make 3D model, but I can't give the condition like when x = 0 u(0,t) = 0
import math
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
def u(x,t,n):
for i in range(1,n):
alpha=((6*(-1)**i-30)/(i**2*np.pi**2))
e=np.exp((-(np.pi**2)*(i**2)*t))
sin=np.sin((i*np.pi*x)/3)
u=alpha*e*sin
return u
N=20
L = 4 # length
att = 20 # iteration
x = np.linspace(0, L ,N) #x-array
t = np.linspace(0, L, N) #t-array
X, Y = np.meshgrid(x, t)
Z = u(X, Y, att)
fig = plt.figure(figsize = (10,10))
ax = fig.add_subplot(111, projection='3d')
ax.plot_wireframe(X, Y, Z, rstride=10, cstride=1000)
plt.show()
My 3D model:
It would help if you actually computed a sum in the partial Fourier sum calculation, at the moment you just return the last term of that sum.
def u(x,t,n):
u = 0*x
for i in range(1,n):
alpha=((6*(-1)**i-30)/(i**2*np.pi**2))
e=np.exp((-(np.pi**2)*(i**2)*t))
sin=np.sin((i*np.pi*x)/3)
u+=alpha*e*sin
return u
Are you sure about the Fourier coefficients? The number 30 in it is for me somewhat suspicious. Also the frequency seems strange, the continuation of u(x,0) should be an odd rectangular wave of period 8. Notice, it is a=3 but L=4.
I have written a code that reads in my data file and plots it and then fits it and finds the peaks however I have 6 peaks and the code is only currently fitting 2 of the peaks and isn't returning any data on them by code is as follows:
from scipy.optimize import curve_fit
import numpy as np
import matplotlib.pyplot as plt
from scipy.signal import find_peaks
data = np.genfromtxt("C:\\Users\\lenovo laptop\\practice_data_ll16ame1.dat", skip_header = 15)
x = data[: , 0]
y = data[: , 1]
plt.plot(x,y)
plt.show()
def func(x, *params):
y = np.zeros_like(x)
for i in range(0, len(params), 3):
ctr = params[i]
amp = params[i+1]
wid = params[i+2]
y = y + amp * np.exp( -((x - ctr)/wid)**2)
return y
guess = [0, 60000, 80, 1000, 60000, 80]
for i in range(12):
guess += [60+80*i, 46000, 25]
popt, pcov = curve_fit(func, x, y, p0=guess)
fit = func(x, *popt)
plt.plot(x, y)
plt.plot(x, fit , 'r-')
plt.show()
When I looked at the plot of your custom function, it was clear that the majority of points were in a more-or-less horizontal line, so the function wouldn't fit well to your peaks. Because there is no noise and the peaks are so prominent, you just need to pass the y values and a threshold to the find_peaks function.
By implementing find_peaks instead of your custom function, you get the following code:
from scipy.optimize import curve_fit
import numpy as np
import matplotlib.pyplot as plt
from scipy.signal import find_peaks
data = np.genfromtxt("C:\\Users\\lenovo laptop\\practice_data_ll16ame1.dat", skip_header = 15)
x = data[: , 0]
y = data[: , 1]
points = find_peaks(y, height = 100)
plt.plot(x, y)
for i in points[0]:
plt.scatter(x[i], y[i])
plt.show()
Find_peaks returns a tuple consisting of two things:
1. The index of the peaks ( points[0] in the code above)
2. The height of each peak (points[1])
The code yields the following plot, which I believe is what you want:
I am trying to plot a 3D surface but I am having some trouble because the documentation for matplotlib does not appear to be very thorough and is lacking in examples. Anyways the program I have written is to solve the Heat Equation Numerically via Method of Finite Differences. Here is my code:
## This program is to implement a Finite Difference method approximation
## to solve the Heat Equation, u_t = k * u_xx,
## in 1D w/out sources & on a finite interval 0 < x < L. The PDE
## is subject to B.C: u(0,t) = u(L,t) = 0,
## and the I.C: u(x,0) = f(x).
import numpy as np
import matplotlib.pyplot as plt
from matplotlib import cm
from mpl_toolkits.mplot3d import Axes3D
# Parameters
L = 1 # length of the rod
T = 10 # terminal time
N = 40 # spatial values
M = 1600 # time values/hops; (M ~ N^2)
s = 0.25 # s := k * ( (dt) / (dx)^2 )
# uniform mesh
x_init = 0
x_end = L
dx = float(x_end - x_init) / N
x = np.arange(x_init, x_end, dx)
x[0] = x_init
# time discretization
t_init = 0
t_end = T
dt = float(t_end - t_init) / M
t = np.arange(t_init, t_end, dt)
t[0] = t_init
# time-vector
for m in xrange(0, M):
t[m] = m * dt
# spatial-vector
for j in xrange(0, N):
x[j] = j * dx
# definition of the solution u(x,t) to u_t = k * u_xx
u = np.zeros((N, M+1)) # array to store values of the solution
# Finite Difference Scheme:
u[:,0] = x * (x - 1) #initial condition
for m in xrange(0, M):
for j in xrange(1, N-1):
if j == 1:
u[j-1,m] = 0 # Boundary condition
elif j == N-1:
u[j+1,m] = 0 # Boundary Condition
else:
u[j,m+1] = u[j,m] + s * ( u[j+1,m] -
2 * u[j,m] + u[j-1,m] )
This is what I have written to try and plot a 3D surface graph:
# for 3D graph
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
surf = ax.plot_surface(x, t, u, rstride=1, cstride=1, cmap=cm.coolwarm, linewidth=0, antialiased=False)
fig.colorbar(surf, shrink=0.5, aspect=5)
plt.show()
I am getting this error when I run the code to plot the graph: "ValueError: shape mismatch: two or more arrays have incompatible dimensions on axis 1."
Please, any and all help is very greatly appreicated. I think the error comes up because I defined u to be a Nx(M+1) matrix but it is necessary to make the original program run. I am unsure of how to correct this so the graph plots properly. Thanks!
Use this code (look at the comments):
# plot 3d surface
# create a meshgrid of (x,t) points
# T and X are 2-d arrays
T, X = np.meshgrid(t,x)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
# Use X and T arrays to plot u
# shape of X, T and u must to be the same
# but shape of u is [40,1601] and I will skip last row while plotting
surf = ax.plot_surface(X, T, u[:,:1600], rstride=1, cstride=1, cmap=cm.coolwarm, linewidth=0, antialiased=False)
fig.colorbar(surf, shrink=0.5, aspect=5)
plt.show()
Result:
because the documentation for matplotlib does not appear to be very thorough and is lacking in examples
http://matplotlib.org/examples/mplot3d/index.html
It's helpful to print out the shapes of the variables x, t, and u:
x.shape == (40,)
t.shape == (1600,)
u.shape == (40, 1601)
So there are two problems here.
The first one is that x and t are 1-dimensional, even though they need to be 2-dimensional.
And the second one is that u has one more element than t in the second dimension.
You can fix both by running
t, x = np.meshgrid(t, x)
u = u[:,:-1]
before creating the 3d plot.
Would anything have to be changed in the answer for Gaussian fit for Python to fit data in log-log space? Specifically, for both x and y data covering several orders of magnitude and this code snippet:
from scipy.optimize import curve_fit
from scipy import asarray as ar,exp
def gaus(x,a,x0,sigma):
return a*exp(-(x-x0)**2/(2*sigma**2))
b=np.genfromtxt('Stuff.dat', delimiter=None, filling_values=0)
x = b[:,0]
y = b[:,1]
n = len(x) #the number of data
mean = sum(x*y)/n #note this correction
sigma = sum(y*(x-mean)**2)/n #note this correction
popt,pcov = curve_fit(gaus,x,y,p0=[max(y),mean,sigma])
ax = pl.gca()
ax.plot(x, y, 'r.-')
ax.plot(x,gaus(x,*popt),'ro:')
ax.set_xscale('log')
ax.set_yscale('log')
The "fits" are horizontal lines and I am not sure whether I am missing something in my code, or if my data simply isn't fittable by a Gaussian. Any help will be appreciated!
This is what I was missing: the data needs to be transformed before doing the fitting, then transformed back to plot on log axes:
from scipy.optimize import curve_fit
from scipy import asarray as ar,exp
import numpy as np
def gaus(x,a,x0,sigma):
return a*exp(-(x-x0)**2/(2*sigma**2))
b=np.genfromtxt('Stuff.dat', delimiter=None, filling_values=0)
x = np.log(b[:,0])
y = np.log(b[:,1])
n = len(x) #the number of data
mean = sum(x*y)/n #note this correction
sigma = sum(y*(x-mean)**2)/n #note this correction
popt,pcov = curve_fit(gaus,x,y,p0=[max(y),mean,sigma])
ax = pl.gca()
ax.plot(x, y, 'r.-')
ax.plot(10**x,10**(gaus(x,*popt)),'ro:')
ax.set_xscale('log')
ax.set_yscale('log')