[英]Passing a dictonary to `scipy.optimize.least_squares`
我有幾個在外部庫中定義的函數。 我無法更改 arguments 或這些函數的內容。 以以下函數為例(雖然原始函數要復雜得多):
def func1(info: dict) -> float:
return 1 - (1.5 * info["b"] - info["a"])
def func2(info: dict) -> float:
return 1 - (np.exp(info["c"]) - info["a"])
我有一個初步的猜測,我正在嘗試應用scipy.optimize.least_squares
來找到最小化func1
或func2
(不是同時)的最佳值,即目標是這樣的
import scipy
def func1(info: dict) -> float:
return 1 - (1.5 * info["b"] - info["a"])
def func2(info: dict) -> float:
return 1 - (np.exp(info["c"]) - info["a"])
initial_dict = {"a" : 5, "b" : 7}
result = scipy.optimize.least_squares(func1, initial_dict)
initial_dict["c"] = 3
result2 = scipy.optimize.least_squares(func1, initial_dict)
問題是least_squares
只接受floats
,而不接受dicts
。 我認為可以將字典的值轉換為列表並編寫一個“包裝器” function 將列表轉換回字典,即
def func1_wrapped(lst: list[float]) -> float:
a, b, c = lst
tmp_dict = {"a" : a, "b": b, "c": c}
return func1(tmp_dict)
result1 = scipy.optimize.least_squares(func1_wrapped,[5, 7, 3])
這樣的事情合理嗎? 有沒有更好、更有效的方法來做到這一點?
一種可能的包裝方式是
import scipy.optimize
import numpy as np
def dict_least_squares(fn, dict0, *args, **kwargs):
keys = list(dict0.keys());
result = scipy.optimize.least_squares(
lambda x: fn({k:v for k,v in zip(keys, x)}), # wrap the argument in a dict
[dict0[k] for k in keys], # unwrap the initial dictionary
*args, # pass position arguments
**kwargs # pass named arguments
)
# wrap the solution in a dictionary
try:
result.x = {k:v for k,v in zip(keys, result.x)}
except:
pass;
return result;
This maintains the interface of original least squares function, by forwarding arbitrary position arguments *args
, or named arguments **kwargs
.
使用示例
def func1(info: dict) -> float:
return 1 - (1.5 * info["b"] - info["a"])
initial_dict = {"a" : 5, "b" : 7}
dict_least_squares(func1, initial_dict)
給
active_mask: array([0., 0.])
cost: 1.2378255801353088e-15
fun: array([4.97559158e-08])
grad: array([ 4.97559158e-08, -7.46338734e-08])
jac: array([[ 1. , -1.49999999]])
message: '`xtol` termination condition is satisfied.'
nfev: 37
njev: 16
optimality: 7.463387344664022e-08
status: 3
success: True
x: {'a': 6.384615399632931, 'b': 4.92307689991801}
然后
def func2(info: dict) -> float:
return 1 - (np.exp(info["c"]) - info["a"])
initial_dict["c"] = 3
dict_least_squares(func2, initial_dict)
給
active_mask: array([0., 0., 0.])
cost: 4.3463374554994224e-17
fun: array([-9.32345157e-09])
grad: array([-9.32345157e-09, 0.00000000e+00, 2.12348517e-11])
jac: array([[ 1. , 0. , -0.00227757]])
message: '`gtol` termination condition is satisfied.'
nfev: 41
njev: 20
optimality: 9.323451566345398e-09
status: 1
success: True
x: {'a': -0.9977224357504928, 'b': 7.0, 'c': -6.0846446250890684}
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