Python列表转置和填充

我有一个列表列表，每个内部列表的长度为1或n（假设n>1）

我想转换列表，但不是截断较长的列表（如使用

zip

）或用

None

填充较短的列表，而是用它们自己的单数值填充较短的列表。换句话说，我想得到：

>>> [(1, 47, 3, 12), (1, 17, 3, 5), (1, 2, 3, 75), (1, 3, 3, 33)]

我可以通过几次迭代来实现这一点：

>>> maxlist = len(max(*uneven, key=len))
>>> maxlist
4
>>> from itertools import repeat
>>> uneven2 = [x if len(x) == maxlist else repeat(x[0], maxlist) for x in uneven]
>>> uneven2
[[1, 1, 1, 1], [47, 17, 2, 3], [3, 3, 3, 3], [12, 5, 75, 33]]
>>> zip(*uneven2)
[(1, 47, 3, 12), (1, 17, 3, 5), (1, 2, 3, 75), (1, 3, 3, 33)]

---

但有更好的方法吗？我真的需要事先知道

maxlist

才能完成此任务吗？

您可以永远重复一个元素列表：

uneven = [[1], [47, 17, 2, 3], [3], [12, 5, 75, 33]]

from itertools import repeat

print zip(*(repeat(*x) if len(x)==1 else x for x in uneven))

您可以改为使用：

---

这意味着你不需要提前知道

maxlist

。

我真的很喜欢@chris morgan模拟

itertools.izip_longest

的想法，所以当我最终得到灵感时，我写了一个

izip_cycle

函数

def izip_cycle(*iterables, **kwargs):
    """Make an iterator that aggregates elements from each of the iterables.
    If the iterables are of uneven length, missing values are filled-in by cycling the shorter iterables.
    If an iterable is empty, missing values are fillvalue or None if not specified.
    Iteration continues until the longest iterable is exhausted.
    """
    fillvalue = kwargs.get('fillvalue')
    counter = [len(iterables)]
    def cyclemost(iterable):
        """Cycle the given iterable like itertools.cycle, unless the counter has run out."""
        itb = iter(iterable)
        saved = []
        try:
            while True:
                element = itb.next()
                yield element
                saved.append(element)
        except StopIteration:
            counter[0] -= 1
            if counter[0] > 0:
                saved = saved or [fillvalue]
                while saved:
                    for element in saved:
                        yield element
    iterators = [cyclemost(iterable) for iterable in iterables]
    while iterators:
        yield tuple([next(iterator) for iterator in iterators])

print list(izip_cycle([], range(3), range(6), fillvalue='@'))

---

您可以模拟

itertools.izip_longest

，只需在序列用尽时使用序列的最后一个值，而不是

fillvalue

。@ChrisMorgan是对的，您看到了这个链接，可能还可以从那里使用一些其他函数。这非常有用，因为n=1内部列表是生成的。现在我已经进行了重构，所以列表是

[重复（1），[47,17,2,3]，重复（3），[12,5,75,33]]

，我可以只做

zip（*不均匀）

。

>>> from itertools import cycle
>>> uneven3 = [x if len(x) != 1 else cycle(x) for x in uneven]
>>> zip(*uneven3)
[(1, 47, 3, 12), (1, 17, 3, 5), (1, 2, 3, 75), (1, 3, 3, 33)]

def izip_cycle(*iterables, **kwargs):
    """Make an iterator that aggregates elements from each of the iterables.
    If the iterables are of uneven length, missing values are filled-in by cycling the shorter iterables.
    If an iterable is empty, missing values are fillvalue or None if not specified.
    Iteration continues until the longest iterable is exhausted.
    """
    fillvalue = kwargs.get('fillvalue')
    counter = [len(iterables)]
    def cyclemost(iterable):
        """Cycle the given iterable like itertools.cycle, unless the counter has run out."""
        itb = iter(iterable)
        saved = []
        try:
            while True:
                element = itb.next()
                yield element
                saved.append(element)
        except StopIteration:
            counter[0] -= 1
            if counter[0] > 0:
                saved = saved or [fillvalue]
                while saved:
                    for element in saved:
                        yield element
    iterators = [cyclemost(iterable) for iterable in iterables]
    while iterators:
        yield tuple([next(iterator) for iterator in iterators])

print list(izip_cycle([], range(3), range(6), fillvalue='@'))