Question

可能重复：
How do I convert a tuple of tuples to a one-dimensional list using list comprehension?

我们说我有以下元组列表：

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

我正在尝试将其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，而无需遍历每个元素并将项目逐个添加到另一个列表中？

是否有任何快速有效的方法可以在不实际迭代原始元组列表的情况下执行此操作，或者是否有一些内置函数/方法来执行此操作？

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]

可替换地：

[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，因为它可以通过懒散地推进列表来避免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]

Answer 4

这是在itertools等特殊模块的性能和独立性方面做到这一点的最佳方法：

>>> l = [(1,2), (1,3), (1,4), (1,5), (1,6)] 
>>> reduce(list.__add__,map(list,l))
[1, 2, 1, 3, 1, 4, 1, 5, 1, 6]

将元组列表转换为简单的平面列表

4 个答案: