迭代器用于遍历字典键子集中的参数范围

Question

我正在尝试编写一个函数，该函数允许我灵活运行字典中参数子集的网格搜索。我想实现的具体行为如下：

def my_grid_searching_function(fiducial_dict, **param_iterators): 
    for params in desired_iterator: 
     fiducial_dict.update(params) 
     # compute chi^2 
     # write new fiducial_dict values and associated chi^2 value to disk 

我的具体目标是要弄清楚如何写desired_iterator。

函数my_grid_searching_function接受关键字参数的任意子集，每个关键字参数将被解释为参数fiducial_dict。

这似乎是itertools.product的任务，但我遇到了问题。在下面的实现中，我使用product我能够在值有效的输入迭代器到单个环路的改造嵌套循环：

from itertools import product 
def my_failed_grid_searching_function(fiducial_dict, **param_iterators): 
    desired_iterator = product(*list(param_iterators.values())) 
    for params in desired_iterator: 
     print(params) 
fiducial_dict = {'x': 0, 'y': 0, 'z': 9} 
my_failed_grid_searching_function(fiducial_dict, x=[4, 5, 6], y=[1, 2])  

(1, 4) 
(1, 5) 
(1, 6) 
(2, 4) 
(2, 5) 
(2, 6) 

当然，这个问题是输入param_iterators已摔成一个普通的字典，所以在my_failed_grid_searching_function的命名空间内，我不知道值的顺序是什么。

任何人都可以提供关于如何编写desired_iterator的任何提示，以便产生足够的信息来更新fiducial_dict，如上所示？

Answer 1

由于您使用的是任意关键字参数，因此您可以抓住param_iterators字典的键来同步param产品的位置。或者，我会推荐使用sklearn软件包来执行grid search。

不管怎么说，尝试此解决方案：

from itertools import product 

def my_grid_searching_function(fiducial_dict, **param_iterators): 
    keys = param_iterators.keys() 
    desired_iterator = list(product(*list(param_iterators.values()))) 
    for i in range(len(desired_iterator)): 
     print("Epoch: ", i) 
     for loc in range(len(desired_iterator[i])): 
      print(keys[loc], desired_iterator[i][loc]) 
     # update your fiducial_dict here 

my_grid_searching_function({'x': 0, 'y': 0}, x=[1,2,3,4], y=[6,7,8]) 

输出：

('Epoch: ', 0) 
('y', 6) 
('x', 1) 
('Epoch: ', 1) 
('y', 6) 
('x', 2) 
('Epoch: ', 2) 
('y', 6) 
('x', 3) 
('Epoch: ', 3) 
('y', 6) 
('x', 4) 
('Epoch: ', 4) 
('y', 7) 
('x', 1) 
('Epoch: ', 5) 
('y', 7) 
('x', 2) 
('Epoch: ', 6) 
('y', 7) 
('x', 3) 
('Epoch: ', 7) 
('y', 7) 
('x', 4) 
('Epoch: ', 8) 
('y', 8) 
('x', 1) 
('Epoch: ', 9) 
('y', 8) 
('x', 2) 
('Epoch: ', 10) 
('y', 8) 
('x', 3) 
('Epoch: ', 11) 
('y', 8) 
('x', 4) 

***Repl Closed*** 

Answer 2

由于Scratch'N'Purr用于指出序列顺序可以简单地从.keys（）确定方法。

from itertools import product 
def param_grid_search_generator(**param_iterators): 
    param_names = list(param_iterators.keys()) 
    param_combination_generator = product(*list(param_iterators.values())) 
    for param_combination in param_combination_generator: 
     yield {param_names[i]: param_combination[i] for i in range(len(param_names))}