Iteratively generate restriction of multivariate function using Python

Question

I'm trying to write a function that generates the restrictions of a function g at a given point p.

For example, let's say g(x, y, z) = 2x + 3y + z and p = (5, 10, 15). I'm trying to create a function that would return [lambda x : g(x, 10, 15), lambda y: g(5, y, 15), lambda z: g(5, 10, z)]. In other words, I want to take my multivariate function and return a list of univariate functions.

I wrote some Python to describe what I want, but I'm having trouble figuring out how to pass the right inputs from p into the lambda properly.

def restriction_generator(g, p):
    restrictions = []
    for i in range(len(p)):
        restriction = lambda x : g(p[0], p[1], ..., p[i-1], p[x], p[i+1], .... p[-1])
        restrictions.append(restriction)
    return restrictions

Purpose: I wrote a short function to estimate the derivative of a univariate function, and I'm trying to extend it to compute the gradient of a multivariate function by computing the derivative of each restriction function in the list returned by restriction_generator.

Apologies if this question has been asked before. I couldn't find anything after some searching, but I'm having trouble articulating my problem without all of this extra context. Another title for this question would probably be more appropriate.

Do you want to do this for a function of three variables or the general case of n variables? — Michael Ruth, Sep 03 '22 at 03:15
General case of n variables. ACA provided a great approach below, but I'd love to see more ways of tackling the problem. — bandicoot12, Sep 03 '22 at 03:16

score 2 · Accepted Answer · answered Sep 03 '22 at 03:34

2

Since @bandicoot12 requested some more solutions, I will try to fix up your proposed code. I'm not familiar with the ... notation, but I think this slight change should work:

def restriction_generator(g, p):
    restrictions = []
    for i in range(len(p)):
        restriction = lambda x : g(*p[: i], x, *p[i+1:])
        restrictions.append(restriction)
    return restrictions

Although I am not familiar with the ... notation, if I had to guess, your original code doesn't work because it probably always inputs p[0]. Maybe it can be fixed by changing it from p[0], p[1], ..., p[i-1] to p[0], ..., p[i-1].

answered Sep 03 '22 at 03:34

Adam Oppenheimer

397
1
7

This is great! I was looking for a short lambda expression. The ... notation was just to help express what I was going for, but it didn't work. There's a slight bug in the code, which is resolved by changing the lambda to `lambda x, i=i : g(*p[: i], x, *p[i+1:])` The bug is explained in the posts below, just pasting for reference: https://stackoverflow.com/questions/3431676/creating-functions-or-lambdas-in-a-loop-or-comprehension https://stackoverflow.com/questions/452610/how-do-i-create-a-list-of-lambdas-in-a-list-comprehension-for-loop?noredirect=1&lq=1 – bandicoot12 Sep 03 '22 at 04:24

score 1 · Answer 2 · answered Sep 03 '22 at 03:09

1

try something like this:

def foo(fun, p, i):
    def bar(x):
        p[i] = x
        return fun(*p)
    return bar

and

def restriction_generator(g, p):
    restrictions = []
    for i in range(len(p)):
        restrictions.append(foo(g, p, i))
    return restrictions

answered Sep 03 '22 at 03:09

ACA

151
6

Nice! This was less tricky than I thought; I shouldn't have focused on using a lambda for this. Thanks for the help :) – bandicoot12 Sep 03 '22 at 03:16

Iteratively generate restriction of multivariate function using Python

2 Answers2