CODEWITHSUNDEEP
pythonoptimizationscipy
Jasper van den Bosch
Jasper van den Bosch

Reputation: 3218

What is the difference between scipy.optimize.fminbound and scipy.optimize.minimize_scalar(bounds=(0,1))

scipy.optimize.fminbound(func, 0, 1) http://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.fminbound.html#scipy.optimize.fminbound

scipy.optimize.minimize_scalar(func, bounds=(0,1)) http://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.minimize_scalar.html#scipy.optimize.minimize_scalar`

Are these equal?

Upvotes: 0

Views: 1161

Answers (1)

Keith Arora-Williams
Keith Arora-Williams

Reputation: 76

Sometimes it is, and sometimes it isn't. It depends on the function that you define and the method you choose. The main difference is that the defined func must return a scalar to use minimize_scalar. I assume that this is because it allows it to return results more rapidly.

See the following for more concrete explanation:

from scipy import optimize

def func(x):
    return (x - 2) * (x + 2)**2

def func2(x):
    return (x - 2) * x * (x + 2)**2

min = 0
max = 1

res1 = optimize.fminbound(func, min, max)
res2 = optimize.minimize_scalar(func, bounds=(min,max))
res3 = optimize.fminbound(func2, min, max)
res4 = optimize.minimize_scalar(func2, bounds=(min,max))

print res1, res2.x
print res3, res4.x

import matplotlib.pyplot as plt
import numpy as np

xaxis = np.arange(-15,15)

plt.plot(xaxis, func(xaxis))
plt.plot(xaxis, func2(xaxis))
plt.scatter(res2.x, res2.fun)
plt.scatter(res4.x, res4.fun)
plt.show()

You can take a look at the plots if you want to see the function traces and where the minimum locations. The printed output is:

0.666666844366 0.666666656734
0.999994039139 1.28077640403

Upvotes: 1

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