Gabriel - 2 years ago 496

Python Question

I'm trying to obtain a confidence interval on an exponential fit to some

`x,y`

`from pylab import *`

from scipy.optimize import curve_fit

# Read data.

x, y = np.loadtxt('exponential_data.dat', unpack=True)

def func(x, a, b, c):

'''Exponential 3-param function.'''

return a * np.exp(b * x) + c

# Find best fit.

popt, pcov = curve_fit(func, x, y)

print popt

# Plot data and best fit curve.

scatter(x, y)

x = linspace(11, 23, 100)

plot(x, func(x, *popt), c='r')

show()

which produces:

How can I obtain the 95% (or some other value) confidence interval on this fit preferably using either pure

`python`

`numpy`

`scipy`

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Answer Source

After some research (see here, here and 1.96) I came up with my own solution.

It accepts an arbitrary X% confidence interval and plots upper and lower curves.

Here's the MWE:

```
from pylab import *
from scipy.optimize import curve_fit
from scipy import stats
def func(x, a, b, c):
'''Exponential 3-param function.'''
return a * np.exp(b * x) + c
# Read data.
x, y = np.loadtxt('exponential_data.dat', unpack=True)
# Define confidence interval.
ci = 0.95
# Convert to percentile point of the normal distribution.
# See: https://en.wikipedia.org/wiki/Standard_score
pp = (1. + ci) / 2.
# Convert to number of standard deviations.
nstd = stats.norm.ppf(pp)
print nstd
# Find best fit.
popt, pcov = curve_fit(func, x, y)
# Standard deviation errors on the parameters.
perr = np.sqrt(np.diag(pcov))
# Add nstd standard deviations to parameters to obtain the upper confidence
# interval.
popt_up = popt + nstd * perr
popt_dw = popt - nstd * perr
# Plot data and best fit curve.
scatter(x, y)
x = linspace(11, 23, 100)
plot(x, func(x, *popt), c='g', lw=2.)
plot(x, func(x, *popt_up), c='r', lw=2.)
plot(x, func(x, *popt_dw), c='r', lw=2.)
text(12, 0.5, '{}% confidence interval'.format(ci * 100.))
show()
```

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