Question

我想创建一个字典，将三组的所有组合映射为一个整数。是否可以在单行和不使用任何导入的情况下执行此操作？

使用itertools可以通过以下方式完成：

colours = ['red','green','blue']
shapes = ['circle','square','triangle']
sizes = ['small','medium','large']
import itertools as it
lookup = {key:val for val,key in enumerate(it.product(colours,shapes,sizes))}

但是，我无法弄清楚如何使用嵌套的for循环和列表理解进行枚举。例如，下面的语法是一种尝试，但是不会在for循环的每个级别上递增：

lookup = {(c,s,z):i for i,c in enumerate(colours) for s in shapes for z in sizes}

输出应如下所示：

{('red', 'circle', 'small'): 0,
 ('red', 'circle', 'medium'): 1,
 ('red', 'circle', 'large'): 2,
 ('red', 'square', 'small'): 3,
 ('red', 'square', 'medium'): 4,
...

Answer 1

您可以像这样枚举内部迭代器：

>>> {x: i for i, x in enumerate(((c, s, z) for c in colours for s in shapes for z in sizes))}
{('red', 'circle', 'small'): 0,
 ('red', 'circle', 'medium'): 1,
 ('red', 'circle', 'large'): 2,
 ('red', 'square', 'small'): 3,
 ('red', 'square', 'medium'): 4,
 ('red', 'square', 'large'): 5,
 ('red', 'triangle', 'small'): 6,
 ('red', 'triangle', 'medium'): 7,
 ('red', 'triangle', 'large'): 8,
 ('green', 'circle', 'small'): 9,
 ('green', 'circle', 'medium'): 10,
 ('green', 'circle', 'large'): 11,
 ('green', 'square', 'small'): 12,
 ('green', 'square', 'medium'): 13,
 ('green', 'square', 'large'): 14,
 ('green', 'triangle', 'small'): 15,
 ('green', 'triangle', 'medium'): 16,
 ('green', 'triangle', 'large'): 17,
 ('blue', 'circle', 'small'): 18,
 ('blue', 'circle', 'medium'): 19,
 ('blue', 'circle', 'large'): 20,
 ('blue', 'square', 'small'): 21,
 ('blue', 'square', 'medium'): 22,
 ('blue', 'square', 'large'): 23,
 ('blue', 'triangle', 'small'): 24,
 ('blue', 'triangle', 'medium'): 25,
 ('blue', 'triangle', 'large'): 26}

使代码更具可读性：

{
    obj: index for index, obj in enumerate(
        (
            (color, shape, size) for color in colours
                                 for shape in shapes
                                 for size in sizes
        )
    )
}

Answer 2

您可以尝试：

lookup = {(c,s,z): i+3*j+9*k for i,c in enumerate(colours) for j,s in enumerate(shapes) for k,z in enumerate(sizes)}

输出：

{
('red', 'circle', 'small'): 0, 
('red', 'circle', 'medium'): 9, 
('red', 'circle', 'large'): 18, 
('red', 'square', 'small'): 3, 
('red', 'square', 'medium'): 12, 
('red', 'square', 'large'): 21, 
('red', 'triangle', 'small'): 6, 
('red', 'triangle', 'medium'): 15, 
('red', 'triangle', 'large'): 24, 
('green', 'circle', 'small'): 1, 
('green', 'circle', 'medium'): 10, 
('green', 'circle', 'large'): 19, 
('green', 'square', 'small'): 4, 
('green', 'square', 'medium'): 13, 
('green', 'square', 'large'): 22, 
('green', 'triangle', 'small'): 7, 
('green', 'triangle', 'medium'): 16, 
('green', 'triangle', 'large'): 25, 
('blue', 'circle', 'small'): 2, 
('blue', 'circle', 'medium'): 11, 
('blue', 'circle', 'large'): 20, 
('blue', 'square', 'small'): 5, 
('blue', 'square', 'medium'): 14, 
('blue', 'square', 'large'): 23, 
('blue', 'triangle', 'small'): 8, 
('blue', 'triangle', 'medium'): 17, 
('blue', 'triangle', 'large'): 26
}

然后为了按值对字典进行排序：

lookup=dict(sorted(lookup.items(), key=lambda x: x[1] ))

用Python中的列表理解枚举嵌套的for循环

2 个答案: