试图用迭代工具的链来解决我的问题。我有一个迭代器列表,我想得到一个迭代器,它以无缝的方式遍历列表中迭代器的项。是否有办法这样做?也许另一种工具,而不是链条,会更合适?
我的代码的简化示例:
iter1 = iter([1,2])
iter2 = iter([3,4])
iter_list = [iter1, iter2]
chained_iter = chain(iter_list)预期结果:
chained_iter.next() --> 1
chained_iter.next() --> 2
chained_iter.next() --> 3
chained_iter.next() --> 4实际结果:
chained_iter.next() --> <listiterator at 0x10f374650>发布于 2017-08-02 07:26:49
您希望使用itertools.chain.from_iterable()来代替:
chained_iter = chain.from_iterable(iter_list)您将一个可迭代的元素传递给chain();它的设计目的是从多个可迭代性中获取元素,并将这些元素链接起来。单次迭代包含更多迭代器并不重要。
还可以使用*语法应用该列表:
chained_iter = chain(*iter_list)对于列表来说,这很好,但是如果iter_list本身就是一个没完没了的迭代器,那么您可能不希望Python试图将其扩展为单独的参数。
演示:
>>> from itertools import chain
>>> iter1 = iter([1, 2])
>>> iter2 = iter([3, 4])
>>> iter_list = [iter1, iter2]
>>> chained_iter = chain.from_iterable(iter_list)
>>> next(chained_iter)
1
>>> next(chained_iter)
2
>>> next(chained_iter)
3
>>> next(chained_iter)
4发布于 2017-08-02 07:33:15
你可以试试这个方法:
>>> from itertools import chain
>>> iter1 = iter([1,2])
>>> iter2 = iter([3,4])
>>> iter_list = [iter1, iter2]
>>> chained_iter = chain(*iter_list)
>>> next(chained_iter)
1
>>> next(chained_iter)
2
>>> next(chained_iter)
3
>>> next(chained_iter)
4
>>>https://stackoverflow.com/questions/45454169
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