对于无限嵌套的元素，这是递归工作的:

def iterFlatten(root):
    if isinstance(root, (list, tuple)):
        for element in root:
            for e in iterFlatten(element):
                yield e
    else:
        yield root

结果:

>>> b = [["a", ("b", "c")], "d"]
>>> list(iterFlatten(b))
['a', 'b', 'c', 'd']

这就是所谓的扁平化，有很多实现。

这个怎么样，尽管它只适用于1级深嵌套:

>>> x = [["a","b"], ["c"]]
>>> for el in sum(x, []):
...     print el
...
a
b
c

从这些链接中，显然最完整的-fast-elegant-etc实现如下:

def flatten(l, ltypes=(list, tuple)):
    ltype = type(l)
    l = list(l)
    i = 0
    while i < len(l):
        while isinstance(l[i], ltypes):
            if not l[i]:
                l.pop(i)
                i -= 1
                break
            else:
                l[i:i + 1] = l[i]
        i += 1
    return ltype(l)

来晚了，但是…

我是python的新手，有lisp的背景。这是我想出的(检查lulz的var名称):

def flatten(lst):
    if lst:
        car,*cdr=lst
        if isinstance(car,(list,tuple)):
            if cdr: return flatten(car) + flatten(cdr)
            return flatten(car)
        if cdr: return [car] + flatten(cdr)
        return [car]

似乎有用。测试:

flatten((1,2,3,(4,5,6,(7,8,(((1,2)))))))

返回:

[1, 2, 3, 4, 5, 6, 7, 8, 1, 2]

I had a similar problem when I had to create a dictionary that contained the elements of an array and their count. The answer is relevant because, I flatten a list of lists, get the elements I need and then do a group and count. I used Python's map function to produce a tuple of element and it's count and groupby over the array. Note that the groupby takes the array element itself as the keyfunc. As a relatively new Python coder, I find it to me more easier to comprehend, while being Pythonic as well.

在我讨论代码之前，这里有一个我必须首先平化的数据示例:

{ "_id" : ObjectId("4fe3a90783157d765d000011"), "status" : [ "opencalais" ],
  "content_length" : 688, "open_calais_extract" : { "entities" : [
  {"type" :"Person","name" : "Iman Samdura","rel_score" : 0.223 }, 
  {"type" : "Company",  "name" : "Associated Press",    "rel_score" : 0.321 },          
  {"type" : "Country",  "name" : "Indonesia",   "rel_score" : 0.321 }, ... ]},
  "title" : "Indonesia Police Arrest Bali Bomb Planner", "time" : "06:42  ET",         
  "filename" : "021121bn.01", "month" : "November", "utctime" : 1037836800,
  "date" : "November 21, 2002", "news_type" : "bn", "day" : "21" }

是来自Mongo的查询结果。下面的代码将这样的列表集合平铺开来。

def flatten_list(items):
  return sorted([entity['name'] for entity in [entities for sublist in  
   [item['open_calais_extract']['entities'] for item in items] 
   for entities in sublist])

首先，我将提取所有的“实体”集合，然后对于每个实体集合，遍历字典并提取name属性。

对于无限嵌套的元素，这是递归工作的:

def iterFlatten(root):
    if isinstance(root, (list, tuple)):
        for element in root:
            for e in iterFlatten(element):
                yield e
    else:
        yield root

结果:

>>> b = [["a", ("b", "c")], "d"]
>>> list(iterFlatten(b))
['a', 'b', 'c', 'd']

总是有reduce(不推荐使用functools):

>>> x = [ [ 'a', 'b'], ['c'] ]
>>> for el in reduce(lambda a,b: a+b, x, []):
...  print el
...
__main__:1: DeprecationWarning: reduce() not supported in 3.x; use functools.reduce()
a
b
c
>>> import functools
>>> for el in functools.reduce(lambda a,b: a+b, x, []):
...   print el
...
a
b
c
>>>

不幸的是，用于列表连接的加号操作符不能用作函数——或者幸运的是，如果为了提高可见性，您希望lambdas变得更丑一些。