如何展平列表/嵌套列表的列表？ [复制]

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newlist 喜欢：

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]

有什么办法可以解决吗？

Answer 1

您正在尝试做的事情称为展平列表。根据Zen of Python，您正在尝试做正确的事情。引用自那

平面比嵌套好。

1. 所以你可以像这样使用列表推导

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

2. 或者你可以像这样使用itertools中的chain

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

3. 或者你可以使用chain.from_iterable，这里不需要解包列表

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

4. 使用sum函数

   newlist = sum(newlist, [])
   

5. 使用reduce函数

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

6. 使用operator.add。这将比带有lambda 版本的reduce 更快。

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

编辑：为了完整起见，也包括在Making a flat list out of list of lists in Python 中找到的答案。

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，最有效的方法是在 Python 2.7 中使用 list(chain.from_iterable(newlist))。在 Python 3.3

中运行相同的测试

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

所以，无论是 Python 2.7 还是 3.3，都可以使用 list(chain.from_iterable(newlist)) 来展平嵌套列表。

Answer 2

只是最简单的一个：

newlist = sum(newlist, [])
print newlist

Answer 3

试试：

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

Answer 4

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

应该这样做。