是一个简短的语法连接列表列表到一个单一的列表(或迭代器)在python?
例如,我有一个列表,如下所示,我想迭代a,b和c。
x = [["a","b"], ["c"]]
我能想到的最好的是如下。
result = []
[ result.extend(el) for el in x]
for el in result:
print el
是一个简短的语法连接列表列表到一个单一的列表(或迭代器)在python?
例如,我有一个列表,如下所示,我想迭代a,b和c。
x = [["a","b"], ["c"]]
我能想到的最好的是如下。
result = []
[ result.extend(el) for el in x]
for el in result:
print el
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)
您所描述的就是所谓的将列表扁平化,有了这些新知识,您将能够在谷歌上找到许多解决方案(没有内置的扁平化方法)。以下是其中一个,来自http://www.daniel-lemire.com/blog/archives/2006/05/10/flattening-lists-in-python/:
def flatten(x):
flat = True
ans = []
for i in x:
if ( i.__class__ is list):
ans = flatten(i)
else:
ans.append(i)
return ans
import itertools
a = [['a','b'], ['c']]
print(list(itertools.chain.from_iterable(a)))
这给了
['a', 'b', 'c']
遗憾的是,Python没有一个简单的方法来平展列表。试试这个:
def flatten(some_list):
for element in some_list:
if type(element) in (tuple, list):
for item in flatten(element):
yield item
else:
yield element
它将递归地将列表平展;你可以这样做
result = []
[ result.extend(el) for el in x]
for el in flatten(result):
print el
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。
如果你只深入一层,一个嵌套的理解也可以:
>>> x = [["a","b"], ["c"]]
>>> [inner
... for outer in x
... for inner in outer]
['a', 'b', 'c']
在一行上,它变成:
>>> [j for i in x for j in i]
['a', 'b', 'c']
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']
或者递归操作:
def flatten(input):
ret = []
if not isinstance(input, (list, tuple)):
return [input]
for i in input:
if isinstance(i, (list, tuple)):
ret.extend(flatten(i))
else:
ret.append(i)
return ret
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]
对于一级扁平化,如果你关心速度,在我尝试过的所有条件下,这比之前的任何答案都快。(也就是说,如果您需要结果作为列表。如果你只需要在运行中迭代它,那么链的例子可能更好。)它的工作原理是预先分配一个最终大小的列表,并按片复制部分(这是一种比任何迭代器方法都低级别的块复制):
def join(a):
"""Joins a sequence of sequences into a single sequence. (One-level flattening.)
E.g., join([(1,2,3), [4, 5], [6, (7, 8, 9), 10]]) = [1,2,3,4,5,6,(7,8,9),10]
This is very efficient, especially when the subsequences are long.
"""
n = sum([len(b) for b in a])
l = [None]*n
i = 0
for b in a:
j = i+len(b)
l[i:j] = b
i = j
return l
带注释的排序时间列表:
[(0.5391559600830078, 'flatten4b'), # join() above.
(0.5400412082672119, 'flatten4c'), # Same, with sum(len(b) for b in a)
(0.5419249534606934, 'flatten4a'), # Similar, using zip()
(0.7351131439208984, 'flatten1b'), # list(itertools.chain.from_iterable(a))
(0.7472689151763916, 'flatten1'), # list(itertools.chain(*a))
(1.5468521118164062, 'flatten3'), # [i for j in a for i in j]
(26.696547985076904, 'flatten2')] # sum(a, [])
如果你需要一个列表,而不是一个生成器,使用list():
from itertools import chain
x = [["a","b"], ["c"]]
y = list(chain(*x))
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。
性能比较:
import itertools
import timeit
big_list = [[0]*1000 for i in range(1000)]
timeit.repeat(lambda: list(itertools.chain.from_iterable(big_list)), number=100)
timeit.repeat(lambda: list(itertools.chain(*big_list)), number=100)
timeit.repeat(lambda: (lambda b: map(b.extend, big_list))([]), number=100)
timeit.repeat(lambda: [el for list_ in big_list for el in list_], number=100)
[100*x for x in timeit.repeat(lambda: sum(big_list, []), number=1)]
生产:
>>> import itertools
>>> import timeit
>>> big_list = [[0]*1000 for i in range(1000)]
>>> timeit.repeat(lambda: list(itertools.chain.from_iterable(big_list)), number=100)
[3.016212113769325, 3.0148865239060227, 3.0126415732791028]
>>> timeit.repeat(lambda: list(itertools.chain(*big_list)), number=100)
[3.019953987082083, 3.528754223385439, 3.02181439266457]
>>> timeit.repeat(lambda: (lambda b: map(b.extend, big_list))([]), number=100)
[1.812084445152557, 1.7702404451095965, 1.7722977998725362]
>>> timeit.repeat(lambda: [el for list_ in big_list for el in list_], number=100)
[5.409658160700605, 5.477502077679354, 5.444318360412744]
>>> [100*x for x in timeit.repeat(lambda: sum(big_list, []), number=1)]
[399.27587954973444, 400.9240571138051, 403.7521153804846]
这是在Windows XP 32位的Python 2.7.1上,但上面评论中的@temoto得到from_iterable比map+extend更快,所以它相当依赖于平台和输入。
不要使用sum(big_list, [])