如何有效地提高列表理解能力

Question

下面是一个(工作的)尝试组成元组，使第一个元组的第一个成员从第二个元组中提取，其余的从第一个元组中提取。

just_no =  ('Gruyere', 'Danish Blue', 'Cheshire')
missing =  ('Caerphilly', 'Red Windsor', 'Camembert')
shop_window = tuple(tuple(no if i!=j else gone 
                          for i, no in enumerate(just_no)) 
                    for j, gone in enumerate(missing))
print(shop_window, "\nWhat a senseless waste of human life.") 
# (('Caerphilly', 'Danish Blue', 'Cheshire'), ('Gruyere', 'Red Windsor', 'Cheshire'), ('Gruyere', 'Danish Blue', 'Camembert'))
# What a senseless waste of human life.

它起了很大作用，我想知道是否有一个更优雅的解决方案，可能使用迭代工具？

Answer 1

基于Python索引/切片的一行

just_no =  ('Gruyere', 'Danish Blue', 'Cheshire')
missing =  ('Caerphilly', 'Red Windsor', 'Camembert')
shop_window = tuple(just_no[:i] + (gone,) + just_no[i+1:] for i, gone in enumerate(missing))
print(shop_window)

产出：

(('Caerphilly', 'Danish Blue', 'Cheshire'), ('Gruyere', 'Red Windsor', 'Cheshire'), ('Gruyere', 'Danish Blue', 'Camembert'))

Answer 2

您可以使用元组的解压缩功能并编写以下内容：

just_no =  ('Gruyere', 'Danish Blue', 'Cheshire')
missing =  ('Caerphilly', 'Red Windsor', 'Camembert')
shop_window = tuple((*just_no[0:i], missing[i], *just_no[i+1:]) for i in range(0, len(just_no)))

Answer 3

你也可以和zip一起去

just_no =  ('Gruyere', 'Danish Blue', 'Cheshire')
missing =  ('Caerphilly', 'Red Windsor', 'Camembert')
shop_window = tuple((tuple(map(lambda x : x[1][1] if x[0] == i else x[1][0], enumerate(zip(just_no, missing)))) for i in range(0, len(just_no))))