join list of lists in python [duplicate]

Question

This question already has an answer here:

• How to make a flat list out of list of lists? 39 answers

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

Answer 1

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

result = sum(x, [])

Answer 2

This is known as flattening, and there are a LOT of implementations out there:

• More on python flatten
• Python tricks
• Flattening lists in Python

How about this, although it will only work for 1 level deep nesting:

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

From those links, apparently the most complete-fast-elegant-etc implementation is the following:

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 3

There's always reduce (being deprecated to 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
>>>

Unfortunately the plus operator for list concatenation can't be used as a function -- or fortunate, if you prefer lambdas to be ugly for improved visibility.

Answer 4

Sadly, Python doesn't have a simple way to flatten lists. Try this:

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

Which will recursively flatten a list; you can then do

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

for el in flatten(result):
      print el

Answer 5

If you're only going one level deep, a nested comprehension will also work:

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

On one line, that becomes:

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

Answer 6

A performance comparison:

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)]

Producing:

>>> 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]

This is with Python 2.7.1 on Windows XP 32-bit, but @temoto in the comments above got from_iterable to be faster than map+extend, so it's quite platform and input dependent.

Stay away from sum(big_list, [])