python - 在Python中迭代一个带有未知子列表的列表？

Question

我正在研究一种聚类算法，当它对数据进行聚类后，它会返回包含单个文档（文档类）和文档组（类组）的列表，例如：

Group(Document(id='NSVcteD-5', name=u'1332410487000-2ed0728e9015028e7c41341011a1bd82'), Group(Document(id='NSVcteD-11', name=u'1332410485000-18ae371b18b3790874fb886085c770af'), Group(Document(id='NSVcteD-12', name=u'1332410484000-dc544efc146674289b126062b000a302'), Group(Document(id='NSVcteD-6', name=u'1332410487000-25e815a47779642df2a416495bd5174c'), Group(Document(id='NSVcteD-7', name=u'1332410485000-eb66881f5b1c633dd1609ad6fc18a45c'), Group(Document(id='NSVcteD-2', name=u'1332410487000-a39e2076ca4477e8a324081732bd36c0'), Group(Document(id='NSVcteD-9', name=u'1332410485000-db1acc63d72a63f65623610242394877'), Group(Group(Document(id='NSVcteD-13', name=u'1332410152000-13ea7da3c74917b86bb70e59ff356397'), Document(id='NSVcteD-3', name=u'1332410487000-6287c3d86e6416cb421b6f176a367e23')), Group(Document(id='NSVcteD-10', name=u'1332410485000-508937f6a4cae9ed79dbd54f016ca61c'), Group(Document(id='NSVcteD-4', name=u'1332410487000-4b16fa5633a9df1341690d9a32a4f06d'), Group(Document(id='NSVcteD-1', name=u'1332410487000-b6696b10ad4415c87e41e5367fd4bcfa'), Group(Document(id='NSVcteD-8', name=u'1332410485000-e3f77be9cddcb9efc07914654454d817'), Group(Document(id='NSVcteD-14', name=u'1332410151000-cc13783d0980106d686d64082121f6ac'), Document(id='NSVcteD-15', name=u'1332410151000-a91330e828e41ed3b8503f3133f61fc7'))))))))))))))

为了便于理解，只需带一个真实的 obj，由我的脚本生成。它是多列列表，我不知道如何迭代来操作它，例如转换为 json 样式字符串。

任何帮助深表感谢。

Answer 1

你的意思是这样的吗？请注意，这是完全低效的，并且会在一定深度后中断..

>>> def recursive_iterate(iterable):
...     iterated_object=[]
...     for elem in iterable:
...         if hasattr(elem,"__iter__"):
...             iterated_object.append(recursive_iterate(elem))
...         else:
...             iterated_object.append(elem)
...     return iterated_object
... 
>>> recursive_iterate([1,2,3,[4,5,6]])
[1, 2, 3, [4, 5, 6]]
>>> recursive_iterate([1,2,3,xrange(10)])
[1, 2, 3, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]]
>>> recursive_iterate([1,2,3,[4,5,6,[xrange(10)]]])
[1, 2, 3, [4, 5, 6, [[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]]]]