python - 如何以pythonicly的方式改进列表理解

Question

接下来是（工作）尝试组合元组，使得第 i 个元组的第 i 个成员取自第二个元组，其余成员取自第一个元组。

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.

它很有效，我想知道是否有更优雅的解决方案，可能使用 itertools？

Answer 1

基于 Python索引/切片的单线：

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)

输出：

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

Answer 2

您可以使用元组的解包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

您也可以使用zip：

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))))