使用 None 作为函数参数进行转置（仅限 python 2.x）

from itertools import imap
from future_builtins import map as fmap # Different name to highlight differences

image = [[1, 2, 3],
         [4, 5, 6],
         [7, 8, 9]]

list(map(None, *image))
# Out: [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
list(fmap(None, *image))
# Out: [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
list(imap(None, *image))
# Out: [(1, 4, 7), (2, 5, 8), (3, 6, 9)]

image2 = [[1, 2, 3],
          [4, 5],
          [7, 8, 9]]
list(map(None, *image2))
# Out: [(1, 4, 7), (2, 5, 8), (3, None, 9)]  # Fill missing values with None
list(fmap(None, *image2))
# Out: [(1, 4, 7), (2, 5, 8)]                # ignore columns with missing values
list(imap(None, *image2))
# Out: [(1, 4, 7), (2, 5, 8)]                # dito

Python 3.x >= 3.0.0

list(map(None, *image))

TypeError：‘NoneType’对象不可调用

但有一种解决方法可以得到类似的结果：

def conv_to_list(*args):
    return list(args)

list(map(conv_to_list, *image))
# Out: [[1, 4, 7], [2, 5, 8], [3, 6, 9]]