Given a multidimensional (nested) Python list, how can I easily get a single list that contains all the elements of its sub-lists?
For example, given [[1,2,3,4,5]] I want to get [1,2,3,4,5]; similarly [[1,2], [3,4]] should become [1,2,3,4].
4 Answers
Use itertools.chain:
itertools.chain(*iterables):Make an iterator that returns elements from the first iterable until it is exhausted, then proceeds to the next iterable, until all of the iterables are exhausted. Used for treating consecutive sequences as a single sequence.
from itertools import chain
A = [[1,2], [3,4]]
print list(chain(*A))
# or better: (available since Python 2.6)
print list(chain.from_iterable(A))
The output is:
[1, 2, 3, 4]
[1, 2, 3, 4]
4 Comments
msh855
It useful to clarify that all objects within the list must be lists themselves. For example, if 'A= ['year', [1] ]' then 'chain' will not produce the expected results. It requires 'A = [['year'], [1]]'.
moooeeeep
@msh855 A string is an iterable too. So, in this case, it seems to be more an issue with the expectation, rather than with the result.
msh855
Have you tried it, or is indeed a matter of expectation?. I tried this: A= ['year', [1,2,'bad']] list(chain(*A)) list(chain.from_iterable(A)) and produced: Out[153]: ['y', 'e', 'a', 'r', 1, 2, 'bad']
moooeeeep
You pass two iterables, both of which are iterated. The first produces the characters in the string, the second produces the list elements. If you need a different behavior, you might need to write specific code.
Use reduce function
reduce(lambda x, y: x + y, A, [])
Or sum
sum(A, [])
itertools provides the chain function for that:
From http://docs.python.org/library/itertools.html#recipes:
def flatten(listOfLists):
"Flatten one level of nesting"
return chain.from_iterable(listOfLists)
Note that the result is an iterable, so you may need list(flatten(...)).
[item for sublist in l for item in sublist]. This is a duplicate of a number of other questions. See, for example, stackoverflow.com/q/952914/623518, stackoverflow.com/q/406121/623518 and stackoverflow.com/q/457215/623518.