在scipy中传递args optimize.minimize目标函数(获取参数数量的错误)

Question

我正在尝试使用scipy的optimizer.minimize函数,但我无法弄清楚将args传递给目标函数的确切方法.我有以下代码,根据我应该工作正常但是给我错误的参数数量.

result = minimize(compute_cost, x0, args=(parameter), method='COBYLA',constraints=cons, options={'maxiter':10000,'rhobeg':20})

这是目标函数的函数签名: def compute_cost(x,parameter)

parameter 是一个有51键值对的字典.

这会出现以下错误:

capi_return is NULL Call-back cb_calcfc_in__cobyla__user__routines failed. Traceback (most recent call last): File "C:\..\resource_optimizer.py", line 138, in <module> result = minimize(compute_cost, x0, args=(parameter), method='COBYLA',constraints=cons, options={'maxiter':10000,'rhobeg':20}) File "C:\Python27\lib\site-packages\scipy\optimize\_minimize.py", line 432, in minimize return _minimize_cobyla(fun, x0, args, constraints, **options) File "C:\Python27\lib\site-packages\scipy\optimize\cobyla.py", line 246, in _minimize_cobyla dinfo=info) File "C:\Python27\lib\site-packages\scipy\optimize\cobyla.py", line 238, in calcfc f = fun(x, *args) TypeError: compute_cost() takes exactly 2 arguments (52 given)

有人可以帮我解决这个问题.

Answer 1

更改args=(parameter)为args=(parameter,),args包含单个元素的元组也是如此.

args=(parameter)相当于args=parameter.当你这样做时,每个元素parameter作为一个单独的参数传递给你的目标函数.