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I asked this question in stack Overflow, but no one gave me an answer.I managed to optimize a line in order to get a line of best fit using curve_fit, but I can't seem to get the R squared value the way I can for linear regression, this is my code:

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
%matplotlib inline
from scipy.optimize import *
from scipy.integrate import *
from scipy.interpolate import * 
df=pd.read_csv('F:/Data32.csv')
df2=df['Temperature']
df3=df['CO2-Rh']
def f(x,a,b,c) :
   return a*np.exp(b*x)+c
params, extras = curve_fit(f, df2, df3)
print('a=%g,b=%g, c=%g' %(params[0],df2[1],df3[2]))
plt.plot(df2,df3,'o')
plt.plot(df2,f(df2,params[0],params[1],params[2]))
plt.legend(['data','fit'],loc='best')
plt.show()
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  • $\begingroup$ I do not see any code which tries and fails to get $r^2$ although this may be because I do not use Python. $\endgroup$
    – mdewey
    Commented Aug 5, 2016 at 7:33
  • $\begingroup$ This is what i use for linear regression @mdewey: result = sm.ols(formula="x ~ y", data=df2).fit() print (result.params) print (result.summary()) $\endgroup$
    – Sergio Espejo
    Commented Aug 5, 2016 at 7:35
  • $\begingroup$ you probably don't want to compute $R^2$ for a non-linear fit, see here: stackoverflow.com/a/14530791/6353406 $\endgroup$
    – bipartitebipartite
    Commented Aug 5, 2016 at 8:35

1 Answer 1

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You can obtain an estimate of $R^2$ for non linear regression by calculate the square of the correlation value between the fitted values and the real values of the response variable.

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