如何展平列表/嵌套列表列表？

Question

我有这样的python代码:

newlist =[[52, None, None], [129, None, None], [56, None, None], [111, None, None],  
          [22, None, None], [33, None, None], [28, None, None], [52, None, None],  
          [52, None, None], [52, None, None], [129, None, None], [56, None, None],  
          [111, None, None], [22, None, None], [33, None, None], [28, None, None]]

我想要的newlist是:

newlist =[52, None, None,129, None, None,56, None, None,111, None, None,22, 
          None, None,33, None, None,28, None, None,52, None, None,52, None,  
          None,52, None, None,129, None, None,56, None, None, 111, None,  
          None,22, None, None,33, None, None,28, None, None]

有办法解决吗？

Answer 1

您要做的是将列表展平.根据Python的Zen,你正在努力做正确的事情.引用那个

Flat优于嵌套.

1. 所以你可以像这样使用列表理解

   newlist = [item for items in newlist for item in items]
   

2. 或者您可以使用chain从itertools这样的

   from itertools import chain
   newlist = list(chain(*newlist))
   

3. 或者您可以使用chain.from_iterable,无需解压缩列表

   from itertools import chain
   newlist = list(chain.from_iterable(newlist))
   

4. 使用sum功能

   newlist = sum(newlist, [])
   

5. 使用reduce功能

   newlist = reduce(lambda x,y: x+y, newlist)
   

6. 用operator.add.这将是比快reduce用lambda版本.

   import operator
   newlist = reduce(operator.add, newlist)
   

编辑:为了完整起见,包括在Python中列出列表列表中的答案.

我尝试在Python 2.7中对所有这些进行计时,就像这样

from timeit import timeit
print(timeit("[item for items in newlist for item in items]", "from __main__ import newlist"))
print(timeit("sum(newlist, [])", "from __main__ import newlist"))
print(timeit("reduce(lambda x,y: x+y, newlist)", "from __main__ import newlist"))
print(timeit("reduce(add, newlist)", "from __main__ import newlist; from operator import add"))
print(timeit("list(chain(*newlist))", "from __main__ import newlist; from itertools import chain"))
print(timeit("list(chain.from_iterable(newlist))", "from __main__ import newlist; from itertools import chain"))

在我的机器上输出

2.26074504852
2.45047688484
3.50180387497
2.56596302986
1.78825688362
1.61612296104

因此,最有效的方法是list(chain.from_iterable(newlist))在Python 2.7中使用.在Python 3.3中进行相同的测试

from timeit import timeit
print(timeit("[item for items in newlist for item in items]", "from __main__ import newlist"))
print(timeit("sum(newlist, [])", "from __main__ import newlist"))
print(timeit("reduce(lambda x,y: x+y, newlist)", "from __main__ import newlist; from functools import reduce"))
print(timeit("reduce(add, newlist)", "from __main__ import newlist; from operator import add; from functools import reduce"))
print(timeit("list(chain(*newlist))", "from __main__ import newlist; from itertools import chain"))
print(timeit("list(chain.from_iterable(newlist))", "from __main__ import newlist; from itertools import chain"))

在我的机器上输出

2.26074504852
2.45047688484
3.50180387497
2.56596302986
1.78825688362
1.61612296104

因此,无论是Python 2.7还是3.3,都可以list(chain.from_iterable(newlist))用来展平嵌套列表.

Answer 2

最简单的一个：

newlist = sum(newlist, [])
print newlist

Answer 3

尝试：

newlist = [j for i in newlist for j in i]

Answer 4

temp = []
for small_list in newlist:
    temp += small_list
newlist = temp

这应该可以做到。