Python: minimize function in respect to i-th variable

Question

I have a function func(x) where the argument is a vector of length n. I would like to minimize it in respect to i-th component of x while keeping the other components fixed. So to express it as a function of a single component I would do something like

import numpy as np
from scipy.optimize import minimize_scalar

def func(x):
    #do some calculations
    return function_value

def func_i(x_i, x0, i):
    x = np.copy(x0)
    x[i] = x_i
    return func(x)

res = minimize_scalar(func_i, args=(x0, i))

Is there a more efficient way of doing this? This kind of calculations will be done repeatedly, cycling over variables, and I worry that x = np.copy(x0), x[i] = x_i slow down the calculation. (The problem emerges in the context of Gibbs sampling, so minimizing in respect to all the variables simultaneously is not what I want.)

You're varying one cell, so other cells don't change, right? Then you can only make a copy per cell, func_i will set a value in that cell to x_i as you do. — STerliakov
– STerliakov, Commented Jan 6 at 21:30
@nocomment a measurement makes sense only when compared with an alternative. — Roger V.
– Roger V., Commented Jan 6 at 21:47
No. If you measure and find out that those actions only take 1% of the total time, then you know they don't (significantly) slow down the calculation. — no comment
– no comment, Commented Jan 6 at 21:53
It is worth looking at alternatives, but in Python it is possible that there isn't a faster way. Changing language to one that is fully compiled and optimised for native hardware is your other alternative if a significant performance improvement is essential. — Martin Brown
– Martin Brown, Commented Jan 8 at 13:02

Martin Brown · Accepted Answer · 2025-01-06 21:29:52Z

1

One possible go faster option rather than a full copy of x0 is to require the function to put x0 back into the state it found it after making the evaluation. IOW func_i becomes:

 def func_i(x_i, x0, i)
     temp = x0[i]
     x0[i] = x_i
     result = func(x0)
     x0[i] = temp
     return result

This avoids the copy to x but requires 3 extra assignments instead. I don't know the length of your array x0 but I'd hazard a guess that it will be faster than a copy for lengths where n is large.

answered Jan 6 at 21:29

Martin Brown

3,6361 gold badge6 silver badges27 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Python: minimize function in respect to i-th variable

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related