如何使用列表理解将元组的元组转换为一维列表[[副本]

它通常被称为展平嵌套结构。

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

>>> 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

预计到达时间 :请不要使用 tuple 作为变量名,它是内置的。

只是使用 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]

如果你有很多元组,使用 list comprehension chain.from_iterable 为了防止

微观基准:

• • 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
    

  • $ 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
    

• • 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
    

  • $ 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
    

• 总和
• list(chain.from_iterable(x)) 如果外部元组较长,则速度更快。

将元组链接在一起:

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

如果你熟悉 itertools list

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]

这些答案中的大多数只适用于单一水平的扁平化。要获得更全面的解决方案,请尝试以下方法(从 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)

itertools.chain

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

对于多级可读代码:

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

我无法把它放在一行中(即使到目前为止,也仍然可读)