将元组列表转换为简单的平面列表

Question

可能重复:
如何使用列表推导将元组元组转换为一维列表？

假设我有以下元组列表:

[(1,2), (1,3), (1,4), (1,5), (1,6)]

我试图将它转换为一个简单的列表,如下所示:

[1,2,1,3,1,4,1,5,1,6]

如何将其转换为如上所示的简单列表,而不必遍历每个元素并将项目逐个添加到另一个列表？

有没有快速有效的方法来实现这一点,而不是实际迭代原始的元组列表,还是有一些内置的函数/方法来做到这一点？

Answer 1

lst = [(1,2), (1,3), (1,4), (1,5), (1,6)]

import itertools
list(itertools.chain(*lst))
# [1, 2, 1, 3, 1, 4, 1, 5, 1, 6]

或者:

[e for l in lst for e in l]
# [1, 2, 1, 3, 1, 4, 1, 5, 1, 6]

Answer 2

" 从根本上说,哪一个更快？使用"itertools"模块,还是使用列表解析？我基本上试图提高我的计算速度." - @davidadamojr

我一直在做一些测试,我发现下面的代码实际上更快.

list_ = [(1, 2), (1, 3), (1, 4), (1, 5), (1, 6)]
list(sum(list_, ()))

如果我错了,有人会纠正我.

以下是一些测试.

>>> list_ = [(1, 2), (1, 3), (1, 4), (1, 5), (1, 6)]
>>> 
>>> operation_1 = lambda: [tuple_item for tuple_ in list_ for tuple_item in tuple_]
>>> def operation_2 ():
        final_list = []
        for tuple_ in list_:
            for tuple_item in tuple_:
                final_list.append(tuple_item)
        return final_list

>>> operation_3 = lambda: reduce(list.__add__, map(list, list_))
>>> def operation_4 ():
        import itertools
        return list(itertools.chain(*list_))

>>> operation_5 = lambda: list(sum(list_, ()))
>>> 
>>> operation_1()
[1, 2, 1, 3, 1, 4, 1, 5, 1, 6]
>>> operation_2()
[1, 2, 1, 3, 1, 4, 1, 5, 1, 6]
>>> operation_3()
[1, 2, 1, 3, 1, 4, 1, 5, 1, 6]
>>> operation_4()
[1, 2, 1, 3, 1, 4, 1, 5, 1, 6]
>>> operation_5()
[1, 2, 1, 3, 1, 4, 1, 5, 1, 6]
>>> 
>>> import timeit
>>> 
>>> print('operation_1 completed in %s seconds.' % (timeit.timeit(operation_1)))
operation_1 completed in 1.57890490223 seconds.
>>> print('operation_2 completed in %s seconds.' % (timeit.timeit(operation_2)))
operation_2 completed in 2.90350501659 seconds.
>>> print('operation_3 completed in %s seconds.' % (timeit.timeit(operation_3)))
operation_3 completed in 5.08437990236 seconds.
>>> print('operation_4 completed in %s seconds.' % (timeit.timeit(operation_4)))
operation_4 completed in 3.85125378138 seconds.
>>> print('operation_5 completed in %s seconds.' % (timeit.timeit(operation_5)))
operation_5 completed in 1.2623826489 seconds.

Answer 3

使用chain.from_iterable,因为它通过懒洋洋地推进列表避免了不必要的一次性解包(导致冗余内存消耗):

>>> import itertools
>>> L = [(1,2), (1,3), (1,4), (1,5), (1,6)]
>>> list(itertools.chain.from_iterable(L))
[1, 2, 1, 3, 1, 4, 1, 5, 1, 6]