枚举嵌套的 for 循环与 Python 中的列表推导

Question

我想创建一个字典，将三个组的所有组合映射到一个整数。是否可以在一行中执行此操作而无需使用任何导入？

使用迭代工具，可以使用以下内容完成：

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，我不知道如何使用嵌套的 for 循环和列表理解进行枚举。例如，以下语法是一种尝试，但对于 for 循环的每个级别都不会递增：

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}

要使代码更具可读性，请执行以下操作：

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

Answer 2

你可以试试：

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] ))