python list comprehensions; compressing a list of lists?

Question

guys. I\'m trying to find the most elegant solution to a problem and wondered if python has anything built-in for what I\'m trying to do.

What I\'m doing is this. I

Answer 1

The question proposed flatmap. Some implementations are proposed but they may unnecessary creating intermediate lists. Here is one implementation that's based on iterators.

def flatmap(func, *iterable):
    return itertools.chain.from_iterable(map(func, *iterable))

In [148]: list(flatmap(os.listdir, ['c:/mfg','c:/Intel']))
Out[148]: ['SPEC.pdf', 'W7ADD64EN006.cdr', 'W7ADD64EN006.pdf', 'ExtremeGraphics', 'Logs']

In Python 2.x, use itertools.map in place of map.

Answer 2

You can find a good answer in itertools' recipes:

def flatten(listOfLists):
    return list(chain.from_iterable(listOfLists))

(Note: requires Python 2.6+)

Answer 3

If listA=[list1,list2,list3]
flattened_list=reduce(lambda x,y:x+y,listA)

This will do.

Answer 4

You can have nested iterations in a single list comprehension:

[filename for path in dirs for filename in os.listdir(path)]

which is equivalent (at least functionally) to:

filenames = []
for path in dirs:
    for filename in os.listdir(path):
        filenames.append(filename)

Answer 5

Google brought me next solution:

def flatten(l):
   if isinstance(l,list):
      return sum(map(flatten,l))
   else:
      return l

Answer 6

subs = []
map(subs.extend, (os.listdir(d) for d in dirs))

(but Ants's answer is better; +1 for him)