如何在Python中基于元组内容拆分列表

你可以这样做:

my_list = [(1,2,x), (2,3,x), (1,2,z), (2,20,z)]

list_x = [item for item in my_list if item[2] == x]
list_y = [item for item in my_list if item[2] == y]
list_z = [item for item in my_list if item[2] == z]

编辑:如果x、y和z不是数字,那么这应该可以:

my_list = [(1,2,'x'), (2,3,'x'), (1,2,'z'), (2,20,'z')]

list_x = [item for item in my_list if item[2] == 'x']
list_y = [item for item in my_list if item[2] == 'y']
list_z = [item for item in my_list if item[2] == 'z']

如果你想让他们在同一个列表中,那么就这样做

final_list = list((list_x,list_z))

类似于@eshirvana,但使用defaultdict,这对于此类问题来说非常有效且常见。我会制作一个defaultdict,并根据最后一个元素确定的键保存元组列表。然后你们所有的小组都会在 values()

from collections import defaultdict

l = [(1,2,'x'), (2,3,'x'), (1,2,'z'), (2,20,'z'), (3,4,'x')]

groups = defaultdict(list)

for t in l:
    groups[t[-1]].append(t)
    
list(groups.values())
# [[(1, 2, 'x'), (2, 3, 'x'), (3, 4, 'x')], [(1, 2, 'z'), (2, 20, 'z')]]

嵌套列表理解 如果你不知道 third_indices 先验:

my_list = [(1,2,'x'), (2,3,'x'), (1,2,'z'), (2,20,'z')]
third_indices = set([ix[-1] for ix in my_list])
[[item for item in my_list if item[-1]==ix] for ix in third_indices]

这里有一种方法可以解释O(n)的复杂性:

lt = [(1,2,'x'), (2,3,'x'), (1,2,'z'), (2,20,'z')]

res = {}
for i in lt:
  res.setdefault(i[2], []).append(i)

print(list(res.values()))

输出:

>>>
[[(1, 2, 'x'), (2, 3, 'x')], [(1, 2, 'z'), (2, 20, 'z')]]