我认为最好的方法是使用MemoizeJac装饰器。这正是scipy.optimize.minimize在jac=True的引擎盖下所做的：

import numpy as np
from scipy.optimize import least_squares
from scipy.optimize._optimize import MemoizeJac

def fun_and_jac(x):
    return x**2 - 5 * x + 3, 2 * x - 5

fun = MemoizeJac(fun_and_jac)
jac = fun.derivative

res = least_squares(fun, x0=0, jac=jac)
print(res)

你可以做一点黑客：

val_cache = {}
jac_cache = {}

def val_fun(*args):
    try:
        return val_cache.pop(args)
    except KeyError:
        (val, jac) = fun(*args)
        jac_cache[args] = jac
        return val

def jac_fun(*args):
    try:
        return jac_cache.pop(args)
    except KeyError:
        (val, jac) = fun(*args)
        val_cache[args] = val
        return jac

从scipy.optimize.minimize的文档中：
如果jac是布尔值且为True，则假定fun返回包含目标函数和梯度的元组（f，g）。
https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.minimize.html?highlight=minimize
因此，您可以简单地这样做：

from scipy.optimize import minimize

def function(x):
    '''Function that returns both fun and jac'''
    return x**2 - 5 * x + 3, 2 * x - 5

print(minimize(function, 0, jac=True))

编辑，重新阅读您的问题，似乎此选项也适用于least_squares，但未记录。
这也是可行的：

from scipy.optimize import least_squares

def function(x):
    '''Function that returns both fun and jac'''
    return x**2 - 5 * x + 3, 2 * x - 5

print(least_squares(function, 0, jac=True))