加入python中的列表列表

Question

在python中将列表列表加入单个列表(或迭代器)的简短语法是什么？

例如,我有一个如下列表,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

Answer 1

import itertools
a = [['a','b'], ['c']]
print(list(itertools.chain.from_iterable(a)))

Answer 2

如果你只是深入一级,嵌套的理解也会起作用:

>>> x = [["a","b"], ["c"]]
>>> [inner
...     for outer in x
...         for inner in outer]
['a', 'b', 'c']

在一行,这成为:

>>> [j for i in x for j in i]
['a', 'b', 'c']

Answer 3

x = [["a","b"], ["c"]]

result = sum(x, [])

Answer 4

这被称为展平,并且有很多实现:

• 更多关于python flatten
• Python技巧
• 在Python中展平列表

这个怎么样,虽然它只适用于1级深度嵌套:

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

从这些链接,显然最完整,快速优雅等实现如下:

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)

Answer 5

l = []
map(l.extend, list_of_lists)

最短!

Answer 6

这对于无限嵌套元素递归工作:

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']

Answer 7

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位上的Python 2.7.1,但上面的评论中的@temoto from_iterable必须更快map+extend,因此它非常依赖于平台和输入.

保持距离 sum(big_list, [])

Answer 8

迟到了,但......

我是python的新手,来自lisp背景.这就是我提出的(查看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]

似乎工作.测试:

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

收益:

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

Answer 9

您所描述的内容被称为展平列表,利用这些新知识,您将能够在Google上找到许多解决方案(没有内置的展平方法).以下是其中之一,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

Answer 10

如果需要列表而不是生成器，请使用 list():

from itertools import chain
x = [["a","b"], ["c"]]
y = list(chain(*x))