Exponential Fit Between 2 Lists - python

I have two lists that I am trying to do an exponential fit of form y=a*e^(bx) between. I am using an approach similar to the second answer from here but the results are not matching what I know to be true from testing with excel. Here is my code:
import numpy as np
from scipy.optimize import curve_fit
exp_constants = [62.5, 87.5, 112.5, 137.5, 162.5, 187.5, 212.5, 237.5, 262.5, 287.5]
means = [211.94, 139.30, 80.09, 48.29, 26.94, 12.12, 3.99, 1.02, 0.09, 0.02]
def func(x1, a, b):
return a * np.exp(b * x1)
popt, pcov = curve_fit(func, exp_constants, means)
When returning popt[0] and popt[1] I get 3.222e-127 and 1.0 respectively. However, when checking with excel the correct exponential equation should be y=7231.3e^(-0.04x). I am not very familiar with the curve_fit approach, is there something that I am missing in my code or a better approach to getting the correct exponential fit?
Edit: Here is the plot that is made with the following code:
plt.figure()
plt.plot(exp_constants, means, 'ko', label="Data")
plt.plot(exp_constants, func(exp_constants, *popt), 'r-', label="Fitted Curve")
plt.legend()
plt.show


I guess the problem is that you do not provide an initial guess for the parameters, so as per the manual, curve_fit uses [1, 1] as a guess. The optimization might then get stuck at some local minimum. One other thing you should do is to change your xdata and ydata lists to numpy arrays, as shown by this answer:
import numpy as np
from scipy.optimize import curve_fit
exp_constants = np.array([62.5, 87.5, 112.5, 137.5, 162.5, 187.5, 212.5,
237.5, 262.5, 287.5])
means = np.array([211.94, 139.30, 80.09, 48.29, 26.94, 12.12, 3.99,
1.02, 0.09, 0.02])
def func(x1, a, b):
return a * np.exp(b * x1)
guess = [100, -0.1]
popt, pcov = curve_fit(func, exp_constants, means, p0 = guess)
The exact value of the guess is not important, but you should probably have at least the order of magnitude and the signs right, so that the optimization can converge to the optimal value. I just used some random numbers close to the 'correct answer' you mentioned. When you don't know what to guess, you can do a polyfit(xdata, log(ydata), 1) and some basic math to get an initial value, as shown by this answer to the question you linked.
Quick plot:
x = np.linspace(exp_constants[0], exp_constants[-1], 1000)
plt.plot(exp_constants, means, 'ko', x, popt[0]*np.exp(popt[1]*x), 'r')
plt.show()
Result:

Related

How to get results from odeint function using an array of time values

I am trying to solve the following ODE to get displacement values wrt. time using scipy.integrate.ode. Here is the code I am working with.
import numpy as np
import scipy as sp
from scipy.integrate import odeint
import matplotlib.pyplot as plt
%matplotlib inline
rho_water = 0.9970470
h=7.5
def Lor(Z, t,args):
h = args[0]
g = args[1]
Omeg = args[2]
if Z[1]>0:
return [Z[1], 1/Z[0] - 1 -Omeg*Z[1] - (Z[1])**2/Z[0]]
else:
return [Z[1], 1/Z[0] - 1 -Omeg*Z[1] ]
def osc(h=7.5, g=9.80665e2,R=0.05,eta=0.0000889):
n= (16)*(eta)*((h)**(1/2))
d=(rho_water)*(R)**2*(g)**(1/2)
Omeg = (n/d)
params = (h,g,Omeg)
t=np.arange(0,500,0.001)
Z=sp.integrate.odeint(Lor, [0.17, 0.00], t, args=(params,))
z=Z[:,0]*h
n= (16)*(eta)*((h)**(1/2))
d=(rho_water)*(R)**2*(g)**(1/2)
Omeg = (n/d)
params = (h,g,Omeg)
return z
z_0_10=osc(h=7.5, g=9.80665e2,R=0.05,eta=0.0000889)
t=np.arange(0,500,0.001)
This code works when I use np.arange and create an array of time values that way but as I have data that I want to fit to the model described by the ODE, I need to solve the ODE over an array of very specific time values, so that I can get displacement values at those times. I want to use those solutions to calculate residuals for my data wrt the model. The array of time values I want to use is:
t_0_10=np.array([0.000, 0.067, 0.100, 0.133, 0.167, 0.200, 0.233, 0.267, 0.300, 0.333, 0.367, 0.400, 0.433, 0.467, 0.500, 0.533, 0.567, 0.600, 0.633, 0.667, 0.700, 0.767, 0.800, 0.833, 0.867, 0.900, 0.933, 0.967, 1.000, 1.033, 1.067, 1.133, 1.167, 1.200, 1.233, 1.333, 1.367, 1.400, 1.433, 1.467, 1.500, 1.533, 1.567, 1.600, 1.633, 1.733, 1.767, 1.867, 1.967, 2.000, 2.033, 2.067, 2.100])
When I use the array of time values as my argument for time in the ODE instead of using np.arange, I do not get the correct results for displacement anymore. The shape of the graph I plot, is not correct. Here is the code for graph I get when I use np.arange, which is the correct shape. When I use my time values, I get a different graph.
graph= plt.subplots(nrows=1, ncols=1, figsize=(14,8))
fig, ax1 =graph
ax1.plot(t*(h*1e-2/9.8)**0.5, z_0_10, c='orange', label='Model Solution', marker="o")
ax1.set_ylabel('Fluid level (Z) / cm',fontsize=12)
ax1.set_xlabel('Time Elapsed (t) / s',fontsize=12)
ax1.tick_params(axis="x", labelsize=12)
ax1.set_xlim([0,2.10])
How can I fix this issue and get the correct values for displacement using my array of time values? Or, if this is not possible, how can I obtain an array of the correct displacement values at those specific times, while still using np.arange to solve the ODE?

How to fit data logarithmic in python?

I've been trying to fit some data I have gained from some simulations. From the curve, I guess a logarithmic fit would be ideal. However, the curve comes looking out quite funky. I've also tried higher order polynomials and np.polyfit, but I couldn't get either to work.
Any help would be appreciated!
from scipy.optimize import curve_fit
import numpy as np
import matplotlib.pyplot as plt
xdata=[9.24104360013e-06, 4.72619458107e-06, 4.03957328857e-06, 9.78301182748e-06, 1.36994566431e-05, 1.16294573409e-05, 7.70899546232e-06, 2.72587766232e-06, 2.19089955631e-06, 5.34851640035e-06, 7.84434545123e-06, 7.6524185787e-06, 1.00592536363e-05, 6.08711035578e-07, 4.08259572135e-07, 5.74424798328e-07, 6.20036326494e-07, 4.34755225756e-06, 4.72832211908e-06, 1.25156011417e-06, 1.44996714816e-05, 3.79992166335e-06, 4.45935911838e-06, 6.6307841155e-06, 2.38540191336e-06, 9.4649801666e-07, 9.11518608157e-06, 3.1944675219e-06, 5.32674287313e-06, 1.48463901861e-05, 3.41127723277e-06, 3.40027150288e-06, 3.33064781566e-06, 2.12828505238e-06, 7.22565690506e-06, 7.86527964811e-06, 2.25791582571e-06, 1.94875869207e-05, 1.54712884424e-05, 5.82300791075e-06, 9.5783833758e-06, 1.89519143607e-05, 1.03731970283e-05, 2.53090894753e-05, 9.26047056658e-06, 1.05428610146e-05, 2.89162870493e-05, 4.78624726782e-05, 1.00005855557e-05, 6.88617910928e-05]
ydata=[0.00281616449359, 0.00257023004939, 0.00250030932407, 0.00284317789756, 0.00300158447316, 0.00291690879783, 0.00274898865728, 0.0023625485679, 0.0023018015629, 0.00259860025555, 0.00269155777824, 0.00265941197135, 0.0028073724168, 0.00192920496041, 0.00182900945464, 0.00191452746379, 0.00193227563253, 0.00253266811688, 0.00255961306471, 0.00212426145702, 0.00285906942634, 0.00247877245272, 0.0025348504727, 0.00269881922057, 0.00232270371493, 0.00204672286703, 0.00281306442303, 0.00241938445736, 0.00261083321385, 0.00287440363274, 0.00244324770882, 0.00244364989768, 0.00244593671433, 0.00228714406931, 0.00263301289418, 0.00269385915315, 0.0022968948347, 0.00313898537645, 0.00305650121575, 0.00265291893623, 0.00278748794063, 0.00312801724905, 0.00289450806538, 0.00313176225397, 0.00284010926578, 0.0028957865422, 0.00335438183977, 0.00360421739757, 0.00270734995952, 0.00377301191882]
plt.plot(xdata,ydata,'o')
x = np.array(xdata, dtype=float) #transform your data in a numpy array of floats
y = np.array(ydata, dtype=float) #so the curve_fit can work
#def func(x,a,b,c):
# return a*x**2+ b*x +c
def func(x,a,b):
return a*np.log(x)+ b
popt, pcov = curve_fit(func, x, y)
plt.plot(x, func(x, *popt), label="Fitted Curve")
plt.show()

Sort x before plotting
x_sorted = np.sort(x)
plt.plot(x_sorted, func(x_sorted, *popt), label="Fitted Curve")
plt.show()

How to uniformly resample a non-uniform signal using SciPy?

I have an (x, y) signal with non-uniform sample rate in x. (The sample rate is roughly proportional to 1/x). I attempted to uniformly re-sample it using scipy.signal's resample function. From what I understand from the documentation, I could pass it the following arguments:
scipy.resample(array_of_y_values, number_of_sample_points, array_of_x_values)
and it would return the array of
[[resampled_y_values],[new_sample_points]]
I'd expect it to return an uniformly sampled data with a roughly identical form of the original, with the same minimal and maximalx value. But it doesn't:
# nu_data = [[x1, x2, ..., xn], [y1, y2, ..., yn]]
# with x values in ascending order
length = len(nu_data[0])
resampled = sg.resample(nu_data[1], length, nu_data[0])
uniform_data = np.array([resampled[1], resampled[0]])
plt.plot(nu_data[0], nu_data[1], uniform_data[0], uniform_data[1])
plt.show()
blue: nu_data, orange: uniform_data
It doesn't look unaltered, and the x scale have been resized too. If I try to fix the range: construct the desired uniform x values myself and use them instead, the distortion remains:
length = len(nu_data[0])
resampled = sg.resample(nu_data[1], length, nu_data[0])
delta = (nu_data[0,-1] - nu_data[0,0]) / length
new_samplepoints = np.arange(nu_data[0,0], nu_data[0,-1], delta)
uniform_data = np.array([new_samplepoints, resampled[0]])
plt.plot(nu_data[0], nu_data[1], uniform_data[0], uniform_data[1])
plt.show()
What is the proper way to re-sample my data uniformly, if not this?

Please look at this rough solution:
import matplotlib.pyplot as plt
from scipy import interpolate
import numpy as np
x = np.array([0.001, 0.002, 0.005, 0.01, 0.02, 0.05, 0.1, 0.2, 0.5, 1, 2, 5, 10, 20])
y = np.exp(-x/3.0)
flinear = interpolate.interp1d(x, y)
fcubic = interpolate.interp1d(x, y, kind='cubic')
xnew = np.arange(0.001, 20, 1)
ylinear = flinear(xnew)
ycubic = fcubic(xnew)
plt.plot(x, y, 'X', xnew, ylinear, 'x', xnew, ycubic, 'o')
plt.show()
That is a bit updated example from scipy page. If you execute it, you should see something like this:
Blue crosses are initial function, your signal with non uniform sampling distribution. And there are two results - orange x - representing linear interpolation, and green dots - cubic interpolation. Question is which option you prefer? Personally I don't like both of them, that is why I usually took 4 points and interpolate between them, then another points... to have cubic interpolation without that strange ups. That is much more work, and also I can't see doing it with scipy, so it will be slow. That is why I've asked about size of the data.

morse potential fit using python and curve fit from scipy

I am trying to fit a morse potential using a python and scipy.
The morse potential is defined as:
V = D*(exp(-2*m*(x-u)) - 2*exp(-m*(x-u)))
where D, m and u are the parameters I need to extract.
Unfortunately the fit is not satisfactory as you can see below (sorry I do not have 10 reputation so the image has to be clicked). Could anyone help me please? I must say I am not the best programmer with python.
Here is my code:
from scipy.optimize import curve_fit
import numpy as np
import matplotlib.pyplot as plt
xdata2=np.array([1.0 ,1.1 ,1.2 ,1.3 ,1.4 ,1.5 ,1.6 ,1.7 ,1.8 ,1.9 ,2.0 ,2.1 ,2.2 ,2.3 ,2.4 ,2.5 ,2.6 ,2.7 ,2.8 ,2.9 ,3.0 ,3.1 ,3.2 ,3.3 ,3.4 ,3.5 ,3.6 ,3.7 ,3.8 ,3.9 ,4.0 ,4.1 ,4.2 ,4.3 ,4.4 ,4.5 ,4.6 ,4.7 ,4.8 ,4.9 ,5.0 ,5.1 ,5.2 ,5.3 ,5.4 ,5.5 ,5.6 ,5.7 ,5.8 ,5.9])
ydata2=[-1360.121815,-1368.532641,-1374.215047,-1378.090480,-1380.648178,-1382.223113,-1383.091562,-1383.479384,-1383.558087,-1383.445803,-1383.220380,-1382.931531,-1382.609269,-1382.273574,-1381.940879,-1381.621299,-1381.319042,-1381.036231,-1380.772039,-1380.527051,-1380.301961,-1380.096257,-1379.907700,-1379.734621,-1379.575837,-1379.430693,-1379.299282,-1379.181303,-1379.077272,-1378.985220,-1378.903626,-1378.831588,-1378.768880,-1378.715015,-1378.668910,-1378.629996,-1378.597943,-1378.572742,-1378.554547,-1378.543296,-1378.539843,-1378.543593,-1378.554519,-1378.572747,-1378.597945,-1378.630024,-1378.668911,-1378.715015,-1378.768915,-1378.831593]
t=np.linspace(0.1,7)
def morse(q, m, u, x ):
return (q * (np.exp(-2*m*(x-u))-2*np.exp(-m*(x-u))))
popt, pcov = curve_fit(morse, xdata2, ydata2, maxfev=40000000)
yfit = morse(t,popt[0], popt[1], popt[2])
print popt
plt.plot(xdata2, ydata2,"ro")
plt.plot(t, yfit)
plt.show()
Old fit before gboffi's comment

I am guessing the exact depth of the morse potential does not interest you overly much. So I added an additional parameter to shift the morse potential up and down (v), includes #gboffis comment. Furthermore, the first argument of your function must be the arguments, not the parameters you want to fit (see http://docs.scipy.org/doc/scipy-0.16.1/reference/generated/scipy.optimize.curve_fit.html)
In addition, such fits are dependent on your starting position. The following should give you what you want.
from scipy.optimize import curve_fit
import numpy as np
import matplotlib.pyplot as plt
xdata2=np.array([1.0 ,1.1 ,1.2 ,1.3 ,1.4 ,1.5 ,1.6 ,1.7 ,1.8 ,1.9 ,2.0 ,2.1 ,2.2 ,2.3 ,2.4 ,2.5 ,2.6 ,2.7 ,2.8 ,2.9 ,3.0 ,3.1 ,3.2 ,3.3 ,3.4 ,3.5 ,3.6 ,3.7 ,3.8 ,3.9 ,4.0 ,4.1 ,4.2 ,4.3 ,4.4 ,4.5 ,4.6 ,4.7 ,4.8 ,4.9 ,5.0 ,5.1 ,5.2 ,5.3 ,5.4 ,5.5 ,5.6 ,5.7 ,5.8 ,5.9])
ydata2=[-1360.121815,-1368.532641,-1374.215047,-1378.090480,-1380.648178,-1382.223113,-1383.091562,-1383.479384,-1383.558087,-1383.445803,-1383.220380,-1382.931531,-1382.609269,-1382.273574,-1381.940879,-1381.621299,-1381.319042,-1381.036231,-1380.772039,-1380.527051,-1380.301961,-1380.096257,-1379.907700,-1379.734621,-1379.575837,-1379.430693,-1379.299282,-1379.181303,-1379.077272,-1378.985220,-1378.903626,-1378.831588,-1378.768880,-1378.715015,-1378.668910,-1378.629996,-1378.597943,-1378.572742,-1378.554547,-1378.543296,-1378.539843,-1378.543593,-1378.554519,-1378.572747,-1378.597945,-1378.630024,-1378.668911,-1378.715015,-1378.768915,-1378.831593]
t=np.linspace(0.1,7)
tstart = [1.e+3, 1, 3, 0]
def morse(x, q, m, u , v):
return (q * (np.exp(-2*m*(x-u))-2*np.exp(-m*(x-u))) + v)
popt, pcov = curve_fit(morse, xdata2, ydata2, p0 = tstart, maxfev=40000000)
print popt # [ 5.10155662 1.43329962 1.7991549 -1378.53461345]
yfit = morse(t,popt[0], popt[1], popt[2], popt[3])
#print popt
#
#
#
plt.plot(xdata2, ydata2,"ro")
plt.plot(t, yfit)
plt.show()

curve fitting with a known function numpy

I have a x and y one-dimension numpy array and I would like to reproduce y with a known function to obtain "beta". Here is the code I am using:
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
y = array([ 0.04022493, 0.04287536, 0.03983657, 0.0393201 , 0.03810298,
0.0363814 , 0.0331144 , 0.03074823, 0.02795767, 0.02413816,
0.02180802, 0.01861309, 0.01632699, 0.01368056, 0.01124232,
0.01005323, 0.00867196, 0.00940864, 0.00961282, 0.00892419,
0.01048963, 0.01199101, 0.01533408, 0.01855704, 0.02163586,
0.02630014, 0.02971127, 0.03511223, 0.03941218, 0.04280329,
0.04689105, 0.04960554, 0.05232003, 0.05487037, 0.05843364,
0.05120701])
x= array([ 0., 0.08975979, 0.17951958, 0.26927937, 0.35903916,
0.44879895, 0.53855874, 0.62831853, 0.71807832, 0.80783811,
0.8975979 , 0.98735769, 1.07711748, 1.16687727, 1.25663706,
1.34639685, 1.43615664, 1.52591643, 1.61567622, 1.70543601,
1.7951958 , 1.88495559, 1.97471538, 2.06447517, 2.15423496,
2.24399475, 2.33375454, 2.42351433, 2.51327412, 2.60303391,
2.6927937 , 2.78255349, 2.87231328, 2.96207307, 3.05183286,
3.14159265])
def func(x,beta):
return 1.0/(4.0*np.pi)*(1+beta*(3.0/2*np.cos(x)**2-1.0/2))
guesses = [20]
popt,pcov = curve_fit(func,x,y,p0=guesses)
y_fit = 1/(4.0*np.pi)*(1+popt[0]*(3.0/2*np.cos(x)**2-1.0/2))
plt.figure(1)
plt.plot(x,y,'ro',x,y_fit,'k-')
plt.show()
The code works but the fitting is completely off (see picture). Any idea why?
It looks like the formula to use contains an additional parameter, i.e. p
def func(x,beta,p):
return p/(4.0*np.pi)*(1+beta*(3.0/2*np.cos(x)**2-1.0/2))
guesses = [20,5]
popt,pcov = curve_fit(func,x,y,p0=guesses)
y_fit = func(angle_plot,*popt)
plt.figure(2)
plt.plot(x,y,'ro',x,y_fit,'k-')
plt.show()
print popt # [ 1.23341604 0.27362069]
In the popt which one is beta and which one is p?

This is perhaps not what you want but, if you are just trying to get a good fit to the data, you could use np.polyfit:
fit = np.polyfit(x,y,4)
fit_fn = np.poly1d(fit)
plt.scatter(x,y,label='data',color='r')
plt.plot(x,fit_fn(x),color='b',label='fit')
plt.legend(loc='upper left')
Note that fit gives the coefficient values of, in this case, a 4th order polynomial:
>>> fit
array([-0.00877534, 0.05561778, -0.09494909, 0.02634183, 0.03936857])

This is going to be as good as you can get (assuming you get the equation right as #mdurant suggested), an additional intercept term is required to further improve the fit:
def func(x,beta, icpt):
return 1.0/(4.0*np.pi)*(1+beta*(3.0/2*np.cos(x)**2-1.0/2))+icpt
guesses = [20, 0]
popt,pcov = curve_fit(func,x,y,p0=guesses)
y_fit = func(x, *popt)
plt.figure(1)
plt.plot(x,y,'ro', x,y_fit,'k-')
print popt #[ 0.33748816 -0.05780343]

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