Curve failed - function not working

Question

Curve:

import numpy as np
import scipy.stats as sp
from scipy.optimize import curve_fit
from lmfit import minimize, Parameters, Parameter, report_fit#
import xlwings as xw
import os
import pandas as pd

I tried using just a curve suitable for scipy: This returns

Out[156]: 
(array([ 1.,  1.]), array([[ inf,  inf],
        [ inf,  inf]]))

Ok, so I thought that I need to start limiting my parameters, which is what the solver does. I used lmfit for this:

params = Parameters()

params.add

I tried pushing python in the right direction by changing the initial parameters:

or using a curve:

Answer 1

Curve fitting runs smoothly when we provide a good starting point. We can get one

• linear regression on sp.norm.ppf(x_data)andnp.log(y_data)
• or by installing a free (not cropped) model first

Alternatively, if you want the computer to find a solution without "help"

• , (, , 100% , )
• ( , )

, excel.

import numpy as np
import scipy.stats as sp
from scipy.optimize import curve_fit, basinhopping, brute

def func1(x, n1, n2):
    return np.clip(np.exp(n2 + n1*(sp.norm.ppf(x))),25e4,10e6)

def func_free(x, n1, n2):
    return np.exp(n2 + n1*(sp.norm.ppf(x)))

def sqerr(n12, x, y):
    return ((func1(x, *n12) - y)**2).sum()

x_data = np.array((1e-04,9e-01,9.5835e-01,9.8e-01,9.9e-01,9.9e-01))
y_data = np.array((250e3,1e6,2.5e6,5e6,7.5e6,10e6))

# get a good starting point

# either by linear regression
lin = sp.linregress(sp.norm.ppf(x_data), np.log(y_data))

# or by using the free (non-clipped) version of the formula
(n1f, n2f), infof = curve_fit(func_free, x_data, y_data, (1, 1))

# use those on the original problem 
(n1, n2), info = curve_fit(func1, x_data, y_data, (lin.slope, lin.intercept))
(n12, n22), info2 = curve_fit(func1, x_data, y_data, (n1f, n2f))

# OR

# use basin hopping
hop = basinhopping(sqerr, (1, 1), minimizer_kwargs=dict(args=(x_data, y_data)), stepsize=10)

# OR 

# brute force it
brt = brute(sqerr, ((-100, 100), (-100, 100)), (x_data, y_data), 201, full_output=True)

# all four solutions are essentially the same:
assert np.allclose((n1, n2), (n12, n22))
assert np.allclose((n1, n2), hop.x)
assert np.allclose((n1, n2), brt[0])

# we are actually a bit better than excel
n1excel, n2excel = 1.7925, 11.6771

print('solution', n1, n2)
print('error', ((func1(x_data, n1, n2) - y_data)**2).sum())
print('excel', ((func1(x_data, n1excel, n2excel) - y_data)**2).sum())

:

solution 2.08286042997 11.1397332743
error 3.12796761241e+12
excel 5.80088578059e+12

. , , , , - sp.norm.ppf . , . , , - sp.norm.ppf(x_data) - -, .

sp.norm.ppf(x_data) .

Answer 2

, , 5 , 2 x, . . , , n2 ~ 10, .

min max , , , . .

, (, , ), :

import numpy as np
import scipy.stats as sp
from lmfit import Parameters, minimize, fit_report

import matplotlib.pyplot as plt

x_data = np.array((1e-04,9e-01,9.5835e-01,9.8e-01,9.9e-01,9.9e-01))
y_data = np.array((250e3,1e6,2.5e6,5e6,7.5e6,10e6))

x_data = x_data[1:]
y_data = y_data[1:]

def func(params, x, data, m1=250e3,m2=10e6):
    n1 = params['n1'].value
    n2 = params['n2'].value

    model = n2 + n1*(sp.norm.ppf(x))
    return np.log(data) - model

params = Parameters()

params.add('n1', value= 1, min=0.01, max=20)
params.add('n2', value= 10, min=0, max=20)

result = minimize(func, params, args=(x_data[1:-1], y_data[1:-1]))

print(fit_report(result))
n1 = result.params['n1'].value
n2 = result.params['n2'].value 
ym = np.exp(n2 + n1*(sp.norm.ppf(x_data)))

plt.plot(x_data, y_data, 'o')
plt.plot(x_data, ym, '-')
plt.legend(['data', 'fit'])
plt.show()

[[Fit Statistics]]
    # fitting method   = leastsq
    # function evals   = 15
    # data points      = 3
    # variables        = 2
    chi-square         = 0.006
    reduced chi-square = 0.006
    Akaike info crit   = -14.438
    Bayesian info crit = -16.241
[[Variables]]
    n1:   1.85709072 +/- 0.190473 (10.26%) (init= 1)
    n2:   11.5455736 +/- 0.390805 (3.38%) (init= 10)
[[Correlations]] (unreported correlations are <  0.100)
    C(n1, n2)                    = -0.993

Answer 3

scipy.optimize.curve_fit:

import matplotlib.pyplot as plt
import scipy.optimize as opt
import scipy.stats as stats
import numpy as np

% matplotlib inline


# Objective
def model(x, n1, n2):
    return np.exp(n2 + n1*(stats.norm.ppf(x)))

# Data
x_samp = np.array((1e-04, 9e-01, 9.5835e-01, 9.8e-01, 9.9e-01, 9.9e-01))
y_samp = np.array((250e3, 1e6, 2.5e6, 5e6 ,7.5e6, 10e6))

x_lin = np.linspace(min(x_samp), max(x_samp), 50)  # for fitting

# Regression
p0 = [5, 5]                                        # guessed params
w, cov = opt.curve_fit(model, x_samp, y_samp, p0=p0)     
print("Estimated Parameters", w)  
y_fit = model(x_lin, *w)

# Visualization
plt.plot(x_samp, y_samp, "ko", label="Data")
plt.plot(x_lin, y_fit, "k--", label="Fit")
plt.title("Curve Fitting")
plt.legend(loc="upper left")

Estimated Parameters [  2.08285048  11.13975585]

---

Your data was built as is, without conversion. The easiest way to perform such a regression is if the model and initial parameters p0reasonably match your data. When these elements are delivered to scipy.optimize.curve_fit, the tuple of the weights or optimized estimation parameters w, as well as the covariance matrix , are returned . We can calculate one standard deviation from the diagonals of the matrix:

p_stdev = np.sqrt(np.diag(cov)) 
print("Std. Dev. of Params:", p_stdev)
# Std. Dev. of Params: [ 0.42281111  0.95945127]

We visually evaluate the quality of the fit, again supplying these evaluative parameters of the model and building the appropriate line.