Question

例如，我有这个：

list_of_lists = [
            [[1,2,3],       [4,5,6],       [7,8,9]      ],
            [[11,22,33],    [44,55,66],    [77,88,99]   ],
            [[111,222,333], [444,555,666], [777,888,999]],         
        ]

如何以最好的方式实现这一目标：

expected_result =  [
            [1,2,3,4,5,6,7,8,9],
            [11,22,33,44,55,66,77,88,99],
            [111,222,333,444,555,666,777,888,999]         
        ]

Answer 1

这对于一对嵌套列表推导并不太难：

result = [[x for inner in middle for x in inner] for middle in list_of_lists]

Answer 2

最好的办法是将itertools.chain.from_iterable与列表理解结合使用：

>>> from itertools import chain
>>> [list(chain.from_iterable(x)) for x in list_of_lists]
[[1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 22, 33, 44, 55, 66, 77, 88, 99], [111, 222, 333, 444, 555, 666, 777, 888, 999]]

或者如果NumPy是一个选项：

In [47]: arr = np.array(list_of_lists)                                  

In [48]: a, b, c = arr.shape                                            

In [49]: arr.flatten().reshape(a, b*c)                                  
Out[49]: 
array([[  1,   2,   3,   4,   5,   6,   7,   8,   9],
       [ 11,  22,  33,  44,  55,  66,  77,  88,  99],                   
       [111, 222, 333, 444, 555, 666, 777, 888, 999]])

Answer 3

使用operator.concat:

list_of_lists = [
            [[1,2,3],       [4,5,6],       [7,8,9]      ],
            [[11,22,33],    [44,55,66],    [77,88,99]   ],
            [[111,222,333], [444,555,666], [777,888,999]],         
        ]

import operator
expected_result = [reduce(operator.concat, List) for List in list_of_lists)]

Answer 4

我会这样做

expected_result1 = list(map(lambda row: sum(row, []), list_of_lists))

print(expected_result)
print(expected_result1)

The output is:
#[[1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 22, 33, 44, 55, 66, 77, 88, 99], [111, 222, 333, 444, 555, 666, 777, 888, 999]]
#[[1, 2, 3, 4, 5, 6, 7, 8, 9], [11, 22, 33, 44, 55, 66, 77, 88, 99], [111, 222, 333, 444, 555, 666, 777, 888, 999]]

水平加入列表列表的最佳方式

4 个答案: