如何使用列表推导将元组的元组转换为一维列表？ [复制]

Question

我有一个元组 - 例如：

tupleOfTuples = ((1, 2), (3, 4), (5,))

我想将其转换为按顺序排列的所有元素的平面一维列表：

[1, 2, 3, 4, 5]

我一直在尝试通过列表理解来实现这一点。但我似乎无法弄清楚。我能够通过 for-each 循环来完成它：

myList = []
for tuple in tupleOfTuples:
   myList = myList + list(tuple)

但我觉得必须有一种方法可以通过列表理解来做到这一点。

一个简单的[list(tuple) for tuple in tupleOfTuples] 只是给你一个列表，而不是单个元素。我想我也许可以通过使用解包运算符来解包列表，如下所示：

[*list(tuple) for tuple in tupleOfTuples]

或

[*(list(tuple)) for tuple in tupleOfTuples]

...但这没有用。有任何想法吗？还是我应该坚持循环？

Answer 1

它通常被称为扁平化嵌套结构。

>>> tupleOfTuples = ((1, 2), (3, 4), (5,))
>>> [element for tupl in tupleOfTuples for element in tupl]
[1, 2, 3, 4, 5]

只是为了展示效率：

>>> import timeit
>>> it = lambda: list(chain(*tupleOfTuples))
>>> timeit.timeit(it)
2.1475738355700913
>>> lc = lambda: [element for tupl in tupleOfTuples for element in tupl]
>>> timeit.timeit(lc)
1.5745135182887857

ETA：请不要使用tuple 作为变量名，它会影响内置。

Answer 2

如果您没有很多元组，只需使用 sum。

>>> tupleOfTuples = ((1, 2), (3, 4), (5,))
>>> sum(tupleOfTuples, ())
(1, 2, 3, 4, 5)
>>> list(sum(tupleOfTuples, ())) # if you really need a list
[1, 2, 3, 4, 5]

如果您确实有很多元组，请使用list comprehension 或chain.from_iterable 来防止sum 的二次行为。

---

微基准测试：

• Python 2.6

  • 短元组的长元组

    $ python2.6 -m timeit -s 'tot = ((1, 2), )*500' '[element for tupl in tot for element in tupl]'
    10000 loops, best of 3: 134 usec per loop
    $ python2.6 -m timeit -s 'tot = ((1, 2), )*500' 'list(sum(tot, ()))'
    1000 loops, best of 3: 1.1 msec per loop
    $ python2.6 -m timeit -s 'tot = ((1, 2), )*500; from itertools import chain; ci = chain.from_iterable' 'list(ci(tot))'
    10000 loops, best of 3: 60.1 usec per loop
    $ python2.6 -m timeit -s 'tot = ((1, 2), )*500; from itertools import chain' 'list(chain(*tot))'
    10000 loops, best of 3: 64.8 usec per loop
    

  • 长元组的短元组

    $ python2.6 -m timeit -s 'tot = ((1, )*500, (2, )*500)' '[element for tupl in tot for element in tupl]'
    10000 loops, best of 3: 65.6 usec per loop
    $ python2.6 -m timeit -s 'tot = ((1, )*500, (2, )*500)' 'list(sum(tot, ()))'
    100000 loops, best of 3: 16.9 usec per loop
    $ python2.6 -m timeit -s 'tot = ((1, )*500, (2, )*500); from itertools import chain; ci = chain.from_iterable' 'list(ci(tot))'
    10000 loops, best of 3: 25.8 usec per loop
    $ python2.6 -m timeit -s 'tot = ((1, )*500, (2, )*500); from itertools import chain' 'list(chain(*tot))'
    10000 loops, best of 3: 26.5 usec per loop
    

• Python 3.1

  • 短元组的长元组

    $ python3.1 -m timeit -s 'tot = ((1, 2), )*500' '[element for tupl in tot for element in tupl]'
    10000 loops, best of 3: 121 usec per loop
    $ python3.1 -m timeit -s 'tot = ((1, 2), )*500' 'list(sum(tot, ()))'
    1000 loops, best of 3: 1.09 msec per loop
    $ python3.1 -m timeit -s 'tot = ((1, 2), )*500; from itertools import chain; ci = chain.from_iterable' 'list(ci(tot))'
    10000 loops, best of 3: 59.5 usec per loop
    $ python3.1 -m timeit -s 'tot = ((1, 2), )*500; from itertools import chain' 'list(chain(*tot))'
    10000 loops, best of 3: 63.2 usec per loop
    

  • 长元组的短元组

    $ python3.1 -m timeit -s 'tot = ((1, )*500, (2, )*500)' '[element for tupl in tot for element in tupl]'
    10000 loops, best of 3: 66.1 usec per loop
    $ python3.1 -m timeit -s 'tot = ((1, )*500, (2, )*500)' 'list(sum(tot, ()))'
    100000 loops, best of 3: 16.3 usec per loop
    $ python3.1 -m timeit -s 'tot = ((1, )*500, (2, )*500); from itertools import chain; ci = chain.from_iterable' 'list(ci(tot))'
    10000 loops, best of 3: 25.4 usec per loop
    $ python3.1 -m timeit -s 'tot = ((1, )*500, (2, )*500); from itertools import chain' 'list(chain(*tot))'
    10000 loops, best of 3: 25.6 usec per loop
    

观察：

• 如果外部元组很短，sum 会更快。
• 如果外部元组很长，list(chain.from_iterable(x)) 会更快。

Answer 3

您将元组链接在一起：

from itertools import chain
print list(chain(*listOfTuples))

如果您熟悉 itertools，则应该非常易读，如果没有明确的 list，您甚至可以生成生成器形式的结果。

Answer 4

我喜欢在这种情况下使用“reduce”（这就是 reduce 的用途！）

lot = ((1, 2), (3, 4), (5,))
print list(reduce(lambda t1, t2: t1 + t2, lot))

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

Answer 5

这些答案中的大多数仅适用于单一级别的展平。如需更全面的解决方案，请试试这个（来自http://rightfootin.blogspot.com/2006/09/more-on-python-flatten.html）：

def flatten(l, ltypes=(list, tuple)):
    ltype = type(l)
    l = list(l)
    i = 0
    while i < len(l):
        while isinstance(l[i], ltypes):
            if not l[i]:
                l.pop(i)
                i -= 1
                break
            else:
                l[i:i + 1] = l[i]
        i += 1
    return ltype(l)

Answer 6

使用itertools.chain的另一种解决方案

>>> tupleOfTuples = ((1, 2), (3, 4), (5,))
>>> from itertools import chain
>>> [x for x in chain.from_iterable(tupleOfTuples)]
[1, 2, 3, 4, 5]

Answer 7

对于多级且可读的代码：

def flatten(bla):
    output = []
    for item in bla:
        output += flatten(item) if hasattr (item, "__iter__") or hasattr (item, "__len__") else [item]
    return output

我无法将它放在一行中（并且仍然可读，即使到目前为止）