将元组列表转换为简单的平面列表

Question

假设我有一个像[(1,2), (1,3), (1,4), (1,5), (1,6)]这样的元组列表。我试图把它转换成一个简单的列表，比如[1,2,1,3,1,4,1,5,1,6]。

如何做到这一点，而不必迭代每个元素，并将项目逐个添加到另一个列表中？

是否有任何快速和有效的方法来做到这一点，而不实际迭代原来的元组列表？也许是内置的函数或方法？

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]

另一种选择是：

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

Answer 2

“Fundamentally，哪个更快？使用“迭代工具”模块，还是使用列表理解？我基本上是想提高我的计算速度，here.” - @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_, ()))

如果我错了谁来纠正我。

下面是一些测试。

>>> 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，因为它避免了不必要的一次性解压缩(这会导致冗余内存消耗)，因为它懒洋洋地推进列表：

>>> 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]