join list of lists in python [duplicate]

2019-01-01 01:35发布

This question already has an answer here:

Is the a short syntax for joining a list of lists into a single list( or iterator) in python?

For example I have a list as follows and I want to iterate over a,b and c.

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

The best I can come up with is as follows.

result = []
[ result.extend(el) for el in x] 

for el in result:
  print el

标签: python
15条回答
孤独总比滥情好
2楼-- · 2019-01-01 02:14

What you're describing is known as flattening a list, and with this new knowledge you'll be able to find many solutions to this on Google (there is no built-in flatten method). Here is one of them, from 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
查看更多
不流泪的眼
3楼-- · 2019-01-01 02:17
import itertools
a = [["a","b"], ["c"]]
print list(itertools.chain.from_iterable(a))
查看更多
初与友歌
4楼-- · 2019-01-01 02:18

For one-level flatten, if you care about speed, this is faster than any of the previous answers under all conditions I tried. (That is, if you need the result as a list. If you only need to iterate through it on the fly then the chain example is probably better.) It works by pre-allocating a list of the final size and copying the parts in by slice (which is a lower-level block copy than any of the iterator methods):

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

Sorted times list with comments:

[(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, [])
查看更多
墨雨无痕
5楼-- · 2019-01-01 02:18

Late to the party but ...

I'm new to python and come from a lisp background. This is what I came up with (check out the var names for lulz):

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]

Seems to work. Test:

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

returns:

[1, 2, 3, 4, 5, 6, 7, 8, 1, 2]
查看更多
素衣白纱
6楼-- · 2019-01-01 02:23

If you need a list, not a generator, use list():

from itertools import chain
x = [["a","b"], ["c"]]
y = list(chain(*x))
查看更多
皆成旧梦
7楼-- · 2019-01-01 02:24

Or a recursive operation:

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
查看更多
登录 后发表回答