如何扁平化列表,而留下一些嵌套(在python)
问题描述:
我有一个字符串列表的列表清单,这样的事情(代表章节,段落和文本的句子)):如何扁平化列表,而留下一些嵌套(在python)
[ [[ ['chp1p1s1'], ['chp1p1s2'], ['chp1p1s3'] ],
[ ['chp1p2s1'], ['chp1p2s2'], ['chp1p2s3'] ]],
[[ ['chp2p1s1'], ['chp2p1s2'], ['chp2p1s3'] ],
[ ['chp2p2s1'], ['chp2p2s2'], ['chp2p2s3'] ]] ]
我知道如何(通过[x for y in z for x in y]
例如)完全地压平这个名单,但我希望做的是部分变平,到最后是这样的:
[ [ ['chp1p1s1'], ['chp1p1s2'], ['chp1p1s3'],
['chp1p2s1'], ['chp1p2s2'], ['chp1p2s3'] ],
[ ['chp2p1s1'], ['chp2p1s2'], ['chp2p1s3'],
['chp2p2s1'], ['chp2p2s2'], ['chp2p2s3'] ] ]
我设法通过一些for循环来解决这个问题:
semiflattend_list=list()
for chapter in chapters:
senlist=list()
for paragraph in chapter:
for sentences in paragraph:
senlist.append(sentences)
semiflattend_list.append(senlist)
但我想知道是否有更好,更短的解决方案? (我不认为,zip
是很长的路要走,因为我的列表的大小不同。)
答
我可以使用itertools.chain
方法看到的最简单的变化:
q = [
[[ ['chp1p1s1'], ['chp1p1s2'], ['chp1p1s3'] ],
[ ['chp1p2s1'], ['chp1p2s2'], ['chp1p2s3'] ]],
[[ ['chp2p1s1'], ['chp2p1s2'], ['chp2p1s3'] ],
[ ['chp2p2s1'], ['chp2p2s2'], ['chp2p2s3'] ]]
]
r = [list(itertools.chain(*g)) for g in q]
print(r)
[[['chp1p1s1'], ['chp1p1s2'], ['chp1p1s3'], ['chp1p2s1'], ['chp1p2s2'], ['chp1p2s3']],
[['chp2p1s1'], ['chp2p1s2'], ['chp2p1s3'], ['chp2p2s1'], ['chp2p2s2'], ['chp2p2s3']]]
那么,是什么[list(itertools.chain(*g)) for g in q]
的意思是:
# If I only had this
[g for g in q]
# I would get the same I started with.
# What I really want is to expand the nested lists
# * before an iterable (basically) converts the iterable into its parts.
func foo(bar, baz):
print(bar + " " + baz)
lst = ["cat", "dog"]
foo(*lst) # prints "cat dog"
# itertools.chain accepts an arbitrary number of lists, and then outputs
# a generator of the results:
c = itertools.chain([1],[2])
# c is now <itertools.chain object at 0x10e1fce10>
# You don't want an generator though, you want a list. Calling `list` converts that:
o = list(c)
# o is now [1,2]
# Now, together:
myList = [[2],[3]]
flattened = list(itertools.chain(*myList))
# flattened is now [2,3]
+0
这实际上解决了它!你可能请解释一下这个星号/图示操作符?在[doc](https://docs.python.org/3/tutorial/controlflow.html#unpacking-argument-lists)中,它并没有向我解释这可能对此有何帮助。 – dia
你给的例子实际上是两个不同列表的元组,我不认为这是你的意思。你可能弄乱了括号或者逗号,但是我们需要一个可再生的例子 –