Python-Scipy fmin，为fmin提供参数

Question

I'm a bit of a newbie in Python. I'm writing a little piece of code in order to find the minimum of a function:

import os,sys,matplotlib,pylab
import numpy as np
from scipy.optimize import fmin

par = [2., 0.5, 0.008]
x1 = 0.4
f2_2 = lambda x, param: param[0] * x**2 + param[1] * x + param[2]
xmin = fmin(f2_2,x1,args = (par))

print xmin

it should be very simple, however I am getting this error:

"Traceback (most recent call last):

  File "prova.fmin.py", line 9, in <module>

 xmin = fmin(f2_2,x1,args = (par))

  File "/usr/lib/python2.7/dist-packages/scipy/optimize/optimize.py", line 257, in fmin

 fsim[0] = func(x0)

  File "/usr/lib/python2.7/dist-packages/scipy/optimize/optimize.py", line 176, in function_wrapper

 return function(x, *args)

TypeError: <lambda>() takes exactly 2 arguments (4 given)"

Could someone help me in understanding this please?

Answer 1

I just tried this out. Looks like you need to say (par,) and not just (par) . Note that (par,) is a tuple, with the variable par as a single element, whereas (par) just evaluates to par : no tuple. The " args " keyword of fmin expects to find a tuple, not par , which in this case is a list.

Edit: Well, actually, it would seem that args doesn't mind receiving a list either. But then, inside of fmin , when the function f2_2 is called, args is unpacked , meaning its contents are now passed as arguments to f2_2 . This means that f2_2 ends up getting four arguments, viz. x , 2 , 0.5 and 0.008 in this case, as opposed to getting just the two arguments x and [2, 0.5, 0.008] .

Answer 2

您需要定义lambda函数以接受更多参数，如下所示：

f2_2 = lambda x, *param: param[0] * x**2 + param[1] * x + param[2]