循环遍历字典键子集中参数范围的迭代器

Question

我正在尝试编写一个函数，它允许我灵活地在字典中的参数子集上运行网格搜索。我试图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

我的具体目标是找出如何写desired_iterator。

函数my_grid_searching_function接受关键字参数的任意子集，每个参数都将被解释为fiducial_dict的参数。你知道吗

这似乎是itertools.product的任务，但我遇到了一个问题。在下面的实现中，我能够使用product将输入迭代器的值上的嵌套循环有效地转换为单个循环：

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，这样做的问题是输入param_iterators已经被放到了一个普通的字典中，因此在my_failed_grid_searching_function的名称空间中，我不知道值的顺序是什么。你知道吗

有人能给我提供一些关于如何编写desired_iterator的技巧吗？这样就可以产生足够的信息来更新fiducial_dict，如上图所示？你知道吗

Answer 1

因为您使用的是任意关键字参数，所以可以获取param_iterators字典的键来同步param产品的位置。或者，我建议使用sklearn包来执行grid search。你知道吗

无论如何，试试这个解决方案：

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

感谢Scratch'N'Purr指出序列顺序可以通过.keys（）方法简单地确定。你知道吗

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))}