透视python列表

必须有一个较短的&更简单的方法,但我就是这么做的( demo ):

a = [('A', '1'),
 ('A', '2'),
 ('A', '3'),
 ('B', '4'),
 ('B', '5'),
 ('C', '6'),
 ('C', '7')]


unique_keys = tuple(sorted(set(k for k,v in a)))
print(unique_keys)

values = tuple(sorted(v for k,v in a))
print(values)

ls = [unique_keys, ]
default = ''

# the sorted part allows me to iterate without checking which key I'm adding 
while len(a):
    tmp = []
    for uk in unique_keys:
        uk_added = False
        for k,v in a:
            if k == uk:
                tmp.append(v)
                a.remove((k,v))
                uk_added = True
                break
        if not uk_added:
            tmp.append(default)

    ls.append(tmp)

print(ls)

对于僵尸帖子,我很抱歉,但根据python docs的建议,我自己解决了这个问题,使用以下列表:

>>> matrix = [
...     [1, 2, 3, 4],
...     [5, 6, 7, 8],
...     [9, 10, 11, 12],
... ]

>>> [[row[i] for row in matrix] for i in range(4)]

[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]

Data Structures: Nested List Comprehensions