Curve_fit not fitting to SIR model - python

My question was to fit an SIR model to data. The two sets of data are xdata, and ydata1 in the code. For some reason the curve fit is coming to a point rather than being rounded at the top. I have tried playing around with the def of fit_odeint, 'popt,pcov', and 'fitted'. With no luck to getting the curve to change, i only have gotten errors when playing around with the code.
N = 763
I0 = 1/N
S0 = N - I0
R0 = 0.0
ydata1 = [3,8, 28,75, 221, 291, 255, 235, 190, 125, 70, 28, 12, 5]
ydata = [y/N for y in ydata1]
xdata = ['1','2','3','4','5','6','7','8','9','10','11','12','13','14']
xdata = [float(t) for t in xdata]
ydata = np.array(ydata, dtype=float)
xdata = np.array(xdata, dtype=float)
def sir_model(y, x, beta, gamma):
S = -beta * y[0] * y[1] / N
R = gamma * y[1]
I = -(S + R)
return S, I, R
def fit_odeint(x, beta, gamma):
return odeint(sir_model, (S0, I0, R0), x, args=(beta, gamma))[:,1]
popt, pcov = curve_fit(fit_odeint, xdata, ydata)
fitted = fit_odeint(xdata, *popt)
plt.plot(xdata, ydata, 'o')
plt.plot(xdata, fitted)
plt.show()
The photo below is the graph [1] that I am receiving with my current code.
[1]: https://i.stack.imgur.com/kQ7Nn.png
Where am I going wrong for the fit curve?

Related

Fit a curve with a SIR model and predict

I have searched and referenced the code for solving the SIR model from others on this website, but the fitting effect is very poor. Is there something wrong with my data? Still what? How should I predict given new data for this SIR model?
import numpy as np
import matplotlib.pyplot as plt
from scipy import integrate, optimize
import pandas as pd
y_total = [0.0, 0.0010131712259371835, 0.0035460992907801418, 0.00911854103343465,
0.008611955420466059, 0.021783181357649443, 0.00911854103343465, 0.07852077001013172, 0.4397163120567376,
0.21681864235055726, 0.232016210739615, 0.5278622087132725, 0.13576494427558258, 0.2988855116514691, 0.37436676798378926,
0.4209726443768997, 0.544579533941236, 0.7254305977710233, 1.0, 0.7740628166160081, 0.43617021276595747, 0.48226950354609927]
x_total = range(0,22)
ydata = np.array(y_total, dtype=float)
xdata = np.array(x_total, dtype=float)
# IO + SO + R0 is always 1 regardless of "value"
I0 = 0.3
S0 = 1 - I0
R0 = 0
def sir_model(y, x, beta, gamma):
S = -beta * y[0] * y[1] / N
R = gamma * y[1]
I = -(S + R)
return S, I, R
def fit_odeint(x, beta, gamma):
return integrate.odeint(sir_model, (S0, I0, R0), x, args=(beta, gamma))[:,1]
N = 1.0
I0 = ydata[0]
S0 = N - I0
R0 = 0.0
popt, pcov = optimize.curve_fit(fit_odeint, xdata, ydata)
fitted = fit_odeint(xdata, *popt)
plt.plot(xdata, ydata, 'o')
plt.plot(xdata, fitted)
plt.show()
Make the initial infected number also a variable, which is easily done by shift the computation of the initial state into the target function
def sir_model(y, x, beta, gamma):
N = sum(y)
S = -beta * y[0] * y[1] / N
R = gamma * y[1]
I = -(S + R)
return S, I, R
def fit_odeint(x, beta, gamma, I0):
# IO + SO + R0 is always 1 regardless of "value"
S0 = 1 - I0
R0 = 0
return integrate.odeint(sir_model, (S0, I0, R0), x, args=(beta, gamma))[:,1]
popt, pcov = optimize.curve_fit(fit_odeint, xdata, ydata,(1/5,1/8,0.1))
There were some other changes, especially adding an initial guess to curve_fit. Still this gets a warning on some difficulty in odeint. But a result is reached anyway, with popt = [0.36714402, 0.04176973, 0.01311579], 1/popt = [2.72372678, 23.94078424, 76.2439491]. Changing the initial guess to values close to this, (1/3, 1/24, 0.05), eliminates the warning.

Fitting sinusoidal data in Python

I am trying to fit experimental data
with a function of the form:
A * np.sin(w*t + p) * np.exp(-g*t) + c
However, the fitted curve (the line in the following image) is not accurate:
If I leave out the exponential decay part, it works and I get a sinus function that is not decaying:
The function that I use is from this thread:
def fit_sin(tt, yy):
'''Fit sin to the input time sequence, and return fitting parameters "amp", "omega", "phase", "offset", "freq", "period" and "fitfunc"'''
tt = np.array(tt)
yy = np.array(yy)
ff = np.fft.fftfreq(len(tt), (tt[1]-tt[0])) # assume uniform spacing
Fyy = abs(np.fft.fft(yy))
guess_freq = abs(ff[np.argmax(Fyy[1:])+1]) # excluding the zero frequency "peak", which is related to offset
guess_amp = np.std(yy) * 2.**0.5
guess_offset = np.mean(yy)
guess_phase = 0.
guess_damping = 0.5
guess = np.array([guess_amp, 2.*np.pi*guess_freq, guess_phase, guess_offset, guess_damping])
def sinfunc(t, A, w, p, g, c): return A * np.sin(w*t + p) * np.exp(-g*t) + c
popt, pcov = scipy.optimize.curve_fit(sinfunc, tt, yy, p0=guess)
A, w, p, g, c = popt
f = w/(2.*np.pi)
fitfunc = lambda t: A * np.sin(w*t + p) * np.exp(-g*t) + c
return {"amp": A, "omega": w, "phase": p, "offset": c, "damping": g, "freq": f, "period": 1./f, "fitfunc": fitfunc, "maxcov": np.max(pcov), "rawres": (guess,popt,pcov)}
res = fit_sin(x, y)
x_fit = np.linspace(np.min(x), np.max(x), len(x))
plt.plot(x, y, label='Data', linewidth=line_width)
plt.plot(x_fit, res["fitfunc"](x_fit), label='Fit Curve', linewidth=line_width)
plt.show()
I am not sure if I implemented the code incorrectly or if the function is not able to describe my data correctly. I appreciate your help!
You can load the txt file from here:
GitHub
and manipulate the data like this to compare it with the post:
file = 'A2320_Data.txt'
column = 17
data = np.loadtxt(file, float)
start = 270
end = 36000
time_scale = 3600
x = []
y = []
for i in range(len(data)):
if start < data[i][0] < end:
x.append(data[i][0]/time_scale)
y.append(data[i][column])
x = np.array(x)
y = np.array(y)
plt.plot(x, y, label='Pyro Oscillations', linewidth=line_width)
Your fitted curve will look like this
import numpy as np
import matplotlib.pyplot as plt
import scipy.optimize
def sinfunc(t, A, w, p, g, c): return A * np.sin(w*t + p) * np.exp(-g*t) + c
tt = np.linspace(0, 10, 1000)
yy = sinfunc(tt, -1, 10, 2, 0.3, 2)
plt.plot(tt, yy)
g stretches the envelope horizontally, c moves the center vertically, w determines the oscillation frequency, A stretches the envelope vertically.
So it can't accurately model the data you have.
Also, you will not be able to reliably fit w, to determine the oscilation frequency it is better to try an FFT.
Of course you could adjust the function to look like your data by adding a few more parameters, e.g.
import numpy as np
import matplotlib.pyplot as plt
import scipy.optimize
def sinfunc(t, A, w, p, g, c1, c2, c3): return A * np.sin(w*t + p) * (np.exp(-g*t) - c1) + c2 * np.exp(-g*t) + c3
tt = np.linspace(0, 10, 1000)
yy = sinfunc(tt, -1, 20, 2, 0.5, 1, 1.5, 1)
plt.plot(tt, yy)
But you will still have to give a good guess for the frequency.

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

Why is tempered mcmc fit not convering well?

I am trying to fit a simple straight line y=mx+c type to some synthetic data using parallel-tempered mcmc. My goal is to just be able to understand how to use it, so that I can apply to some more complex models later. The example I am trying is the replica of what has already been done in a simple emcee code :
http://dfm.io/emcee/current/user/line/
but instead of using mcmc, I want to use parallel-tempered mcmc:
http://dfm.io/emcee/current/user/pt/
Here is a working code:
import numpy as np
from emcee import PTSampler
import emcee
# Choose the "true" parameters.
m_true = -0.9594
b_true = 4.294
f_true = 0.534
# Generate some synthetic data from the model.
N = 50
x = np.sort(10*np.random.rand(N))
yerr = 0.1+0.5*np.random.rand(N)
y = m_true*x+b_true
y += np.abs(f_true*y) * np.random.randn(N)
y += yerr * np.random.randn(N)
def lnlike(theta, x, y, yerr):
m, b, lnf = theta
model = m * x + b
inv_sigma2 = 1.0/(yerr**2 + model**2*np.exp(2*lnf))
return -0.5*(np.sum((y-model)**2*inv_sigma2 - np.log(inv_sigma2)))
def lnprior(theta):
m, b, lnf = theta
if -5.0 < m < 0.5 and 0.0 < b < 10.0 and -10.0 < lnf < 1.0:
return 0.0
return -np.inf
def lnprob(theta, x, y, yerr):
lp = lnprior(theta)
if not np.isfinite(lp):
return -np.inf
return lp + lnlike(theta, x, y, yerr)
import scipy.optimize as op
nll = lambda *args: -lnlike(*args)
result = op.minimize(nll, [m_true, b_true, np.log(f_true)], args=(x, y, yerr))
m_ml, b_ml, lnf_ml = result["x"]
init = [0.5, m_ml, b_ml, lnf_ml]
ntemps = 10
nwalkers = 100
ndim = 3
from multiprocessing import Pool
pos = np.random.uniform(low=-1, high=1, size=(ntemps, nwalkers, ndim))
for i in range(ntemps):
#initialize parameters near scipy optima
pos[i:,] = np.array([result["x"] + 1e-4*np.random.randn(ndim) for i in range(nwalkers)])
pool = Pool(processes=4)
sampler=PTSampler(ntemps,nwalkers, ndim, lnlike, lnprior, loglargs=(x, y, yerr), pool=pool)# args=(x, y, yerr))
#burn-in
sampler.run_mcmc(pos, 1000)
sampler.reset()
sampler.run_mcmc(pos, 10000, thin=10)
samples = sampler.chain.reshape((-1, ndim))
print('Number of posterior samples is {}'.format(samples.shape[0]))
#print best fit value together with errors
print(map(lambda v: (v[1], v[2]-v[1], v[1]-v[0]),
zip(*np.percentile(samples, [16, 50, 84],
axis=0))))
import corner
fig = corner.corner(samples, labels=["$m$", "$b$", "$\ln\,f$"],
truths=[m_true, b_true, np.log(f_true)])
fig.savefig("triangle.png")
The only problem when running this code is I get optimal parameters value which are way off the true values. And increasing the number of walkers or samples is not helping in any sense. Can anyone please advice why tempered-mcmc is not working here?
Update:
I found out a useful package called ptemcee (https://pypi.org/project/ptemcee/#description), although the documentation of this package is non-existent. It seems that this package might be useful, any help on how to implement the same linear fitting with this package would also be highly appreciated.
I have modified some lines
import time
import numpy as np
from emcee import PTSampler
import corner
import matplotlib.pyplot as plt
import scipy.optimize as op
t1 = time.time()
np.random.seed(6) # To reproduce results
# Choose the "true" parameters.
m_true = -0.9594
b_true = 4.294
f_true = 0.534
# Generate some synthetic data from the model.
N = 50
x = np.sort(10 * np.random.rand(N))
yerr = 0.1 + 0.5 * np.random.rand(N)
y_1 = m_true * x + b_true
y = np.abs(f_true * y_1) * np.random.randn(N) + y_1
y += yerr * np.random.randn(N)
plt.plot(x, y, 'o')
# With emcee
def lnlike(theta, x, y, yerr):
m, b, lnf = theta
model = m * x + b
inv_sigma2 = 1.0/(yerr**2 + model**2*np.exp(2*lnf))
return -0.5*(np.sum((y-model)**2*inv_sigma2 - np.log(inv_sigma2)))
def lnprior(theta):
m, b, lnf = theta
if -5.0 < m < 0.5 and 0.0 < b < 10.0 and -10.0 < lnf < 1.0:
return 0.0
return -np.inf
def lnprob(theta, x, y, yerr):
lp = lnprior(theta)
if not np.isfinite(lp):
return -np.inf
return lp + lnlike(theta, x, y, yerr)
nll = lambda *args: -lnlike(*args)
result = op.minimize(nll, [m_true, b_true, np.log(f_true)], args=(x, y, yerr))
m_ml, b_ml, lnf_ml = result["x"]
init = [0.5, m_ml, b_ml, lnf_ml]
ntemps = 10
nwalkers = 100
ndim = 3
pos = np.random.uniform(low=-1, high=1, size=(ntemps, nwalkers, ndim))
for i in range(ntemps):
pos[i:, :] = np.array([result["x"] + 1e-4*np.random.randn(ndim) for i in range(nwalkers)])
sampler = PTSampler(ntemps, nwalkers, ndim, lnlike, lnprior, loglargs=(x, y, yerr), threads=4) # args=(x, y, yerr))
#burn-in
print(pos.shape)
sampler.run_mcmc(pos, 100)
sampler.reset()
sampler.run_mcmc(pos, 5000, thin=10)
samples = sampler.chain.reshape((-1, ndim))
print('Number of posterior samples is {}'.format(samples.shape[0]))
#print best fit value together with errors
p1, p2, p3 = map(lambda v: (v[1], v[2]-v[1], v[1]-v[0]),
zip(*np.percentile(samples, [16, 50, 84],
axis=0)))
print(p1, '\n', p2, '\n', p3)
fig = corner.corner(samples, labels=["$m$", "$b$", "$\ln\,f$"],
truths=[m_true, b_true, np.log(f_true)])
t2 = time.time()
print('It took {:.3f} s'.format(t2 - t1))
plt.show()
The figure I get with corner is:
The important line is
sampler = PTSampler(ntemps, nwalkers, ndim, lnlike, lnprior, loglargs=(x, y, yerr), threads=4)
I have used threads=4 instead of Pool.
Look closely at this line print(p1, '\n', p2, '\n', p3), it prints the values of m_true, b_true and f_true you get:
(-1.277782877669762, 0.5745273177144817, 2.0813620981463297)
(4.800481378230051, 3.1747356851201163, 2.245189235990341)
(-0.9391847529845194, 1.1196053087321716, 3.6017609114364273)
For f, you need np.exp(-0.93918), which is 0.3909, which is close to 0.534. The values you get are close (-1.277 compared to -0.9594 and 4.8 compared to 4.294), although the errors are not bad (except for f). I mean, are you expecting to get the exact numbers? With this method, in my computer, it takes 111 s to complete, is that normal?
Let's try something different. Let's be clear: the problem is not easy when f_true is added. I will use pymc3 (you don't need to know how to use pymc3, I want to check the results found by emcee).
import time
import numpy as np
import corner
import matplotlib.pyplot as plt
import pymc3 as pm
t1 = time.time()
np.random.seed(6)
# Choose the "true" parameters.
m_true = -0.9594
b_true = 4.294
f_true = 0.534
# Generate some synthetic data from the model.
N = 50
x = np.sort(10 * np.random.rand(N))
yerr = 0.1 + 0.5 * np.random.rand(N)
y_1 = m_true * x + b_true
y = np.abs(f_true * y_1) * np.random.randn(N) + y_1
y += yerr * np.random.randn(N)
plt.plot(x, y, 'o')
with pm.Model() as model: # model specifications in PyMC3 are wrapped in a with-statement
# Define priors
f = pm.HalfCauchy('f', beta=5)
m = pm.Normal('m', 0, sd=20)
b = pm.Normal('b', 0, sd=20)
mu2 = b + m * x
sigma2 = yerr**2 + f**2 * (y_1)**2
post = pm.Normal('y', mu=mu2, sd=pm.math.sqrt(sigma2), observed=y)
with model:
trace = pm.sample(2000, tune=2000)
print(pm.summary(trace))
pm.traceplot(trace)
all_values = np.stack([trace.get_values('b'), trace.get_values('m'), trace.get_values('f')], axis=1)
fig2 = corner.corner(all_values, labels=["$b$", "$m$", "$f$"],
truths=[b_true, m_true, f_true])
t2 = time.time()
print('It took {:.3f} s'.format(t2 - t1))
plt.show()
The summary is
mean sd mc_error hpd_2.5 hpd_97.5 n_eff Rhat
m -0.995545 0.067818 0.001174 -1.123187 -0.857653 2685.610018 1.000121
b 4.398158 0.332526 0.005585 3.767336 5.057909 2746.736563 1.000201
f 0.425442 0.063884 0.000904 0.311037 0.554446 4195.591204 1.000309
The important part is the column mean, you see that the values found by pymc3 are close to the true values. The columns hpd_2.5 and hpd_97.5 are the errors for f, b and m. And it took 14 s.
The figure I get with corner is
You will say that the results of emcee are not quite good, but if you really want more accuracy, you have to modify this function:
def lnprior(theta):
m, b, lnf = theta
if -5.0 < m < 0.5 and 0.0 < b < 10.0 and -10.0 < lnf < 1.0:
return 0.0
return -np.inf
The famous prior. In this case, it is flat, and since there are a lot of priors...

How to Fit to The Outer Shell of a Function

I am trying to make a gaussian fit on a function that is messy. I want to only fit the exterior outer shell (these are not just the max values at each x, because some of the max values will be too low too, because the sample size is low).
from scipy.optimize import curve_fit
def Gauss(x, a, x0, sigma, offset):
return a * np.exp(-np.power(x - x0,2) / (2 * np.power(sigma,2))) + offset
def fitNormal(x, y):
popt, pcov = curve_fit(Gauss, x, y, p0=[np.max(y), np.median(x), np.std(x), np.min(y)])
return popt
plt.plot(xPlot,yPlot, 'k.')
plt.xlabel('x')
plt.ylabel('y')
plt.title('Y(x)')
x,y = xPlot,yPlot
popt = fitNormal(x, y)
minx, maxx = np.min(x), np.max(x)
xFit = np.arange(start=minx, stop=maxx, step=(maxx-minx)/1000)
yFitTest = Gauss(xPlot, popt[0], popt[1], popt[2], popt[3])
print('max fit test: ',np.max(yFitTest))
print('max y: ',np.max(yPlot))
maxIndex = np.where(yPlot==np.max(yPlot))[0][0]
factor = yPlot[maxIndex]/yFitTest[maxIndex]
yFit = Gauss(xPlot, popt[0], popt[1], popt[2], popt[3]) * factor
plt.plot(xFit,yFit,'r')
This is an iterative approach similar to this post. It is different in the sense that the shape of the graph does not permit the use of convex hull. So the idea is to create a cost function that tries to minimize the area of the graph while paying high cost if a point is above the graph. Depending on the type of the graph in OP the cost function needs to be adapted. One also has to check if in the final result all points are really below the graph. Here one can fiddle with details of the cost function. One my, e.g., include an offset in the tanh like tanh( slope * ( x - offset) ) to push the solution farther away from the data.
import matplotlib.pyplot as plt
import numpy as np
from scipy.optimize import leastsq
def g( x, a, s ):
return a * np.exp(-x**2 / s**2 )
def cost_function( params, xData, yData, slope, val ):
a,s = params
area = 0.5 * np.sqrt( np.pi ) * a * s
diff = np.fromiter ( ( y - g( x, a, s) for x, y in zip( xData, yData ) ), np.float )
cDiff = np.fromiter( ( val * ( 1 + np.tanh( slope * d ) ) for d in diff ), np.float )
out = np.concatenate( [ [area] , cDiff ] )
return out
xData = np.linspace( -5, 5, 500 )
yData = np.fromiter( ( g( x, .77, 2 ) * np.sin( 257.7 * x )**2 for x in xData ), np.float )
sol=[ [ 1, 2.2 ] ]
for i in range( 1, 6 ):
solN, err = leastsq( cost_function, sol[-1] , args=( xData, yData, 10**i, 1 ) )
sol += [ solN ]
print sol
fig = plt.figure()
ax = fig.add_subplot( 1, 1, 1)
ax.scatter( xData, yData, s=1 )
for solN in sol:
solY = np.fromiter( ( g( x, *solN ) for x in xData ), np.float )
ax.plot( xData, solY )
plt.show()
giving
>> [0.8627445 3.55774814]
>> [0.77758636 2.52613376]
>> [0.76712184 2.1181137 ]
>> [0.76874125 2.01910211]
>> [0.7695663 2.00262339]
and
Here is a different approach using scipy's Differental Evolution module combined with a "brick wall", where if any predicted value during the fit is greater than the corresponding Y value, the fitting error is made extremely large. I have shamelessly poached code from the answer of #mikuszefski to generate the data used in this example.
import numpy as np
import matplotlib
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
import warnings
from scipy.optimize import differential_evolution
def g( x, a, s ):
return a * np.exp(-x**2 / s**2 )
xData = np.linspace( -5, 5, 500 )
yData = np.fromiter( ( g( x, .77, 2 )* np.sin( 257.7 * x )**2 for x in xData ), np.float )
def Gauss(x, a, x0, sigma, offset):
return a * np.exp(-np.power(x - x0,2) / (2 * np.power(sigma,2))) + offset
# function for genetic algorithm to minimize (sum of squared error)
def sumOfSquaredError(parameterTuple):
warnings.filterwarnings("ignore") # do not print warnings by genetic algorithm
val = Gauss(xData, *parameterTuple)
multiplier = 1.0
for i in range(len(val)):
if val[i] < yData[i]: # ****** brick wall ******
multiplier = 1.0E10
return np.sum((multiplier * (yData - val)) ** 2.0)
def generate_Initial_Parameters():
# min and max used for bounds
maxX = max(xData)
minX = min(xData)
maxY = max(yData)
minY = min(yData)
minData = min(minX, minY)
maxData = max(maxX, maxY)
parameterBounds = []
parameterBounds.append([minData, maxData]) # parameter bounds for a
parameterBounds.append([minData, maxData]) # parameter bounds for x0
parameterBounds.append([minData, maxData]) # parameter bounds for sigma
parameterBounds.append([minData, maxData]) # parameter bounds for offset
# "seed" the numpy random number generator for repeatable results
result = differential_evolution(sumOfSquaredError, parameterBounds, seed=3, polish=False)
return result.x
# generate initial parameter values
geneticParameters = generate_Initial_Parameters()
# create values for display of fitted function
y_fit = Gauss(xData, *geneticParameters)
plt.scatter(xData, yData, s=1 ) # plot the raw data
plt.plot(xData, y_fit) # plot the equation using the fitted parameters
plt.show()
print('parameters:', geneticParameters)

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