Converting a list of tuples into a simple flat list [duplicate]

Answer 1

lst = [(1,2), (1,3), (1,4), (1,5), (1,6)]

import itertools
list(itertools.chain(*lst))
# [1, 2, 1, 3, 1, 4, 1, 5, 1, 6]

Alternatively:

[e for l in lst for e in l]
# [1, 2, 1, 3, 1, 4, 1, 5, 1, 6]

Answer 2

Here is the best way to do it in terms of performance and independence of special modules like itertools:

>>> l = [(1,2), (1,3), (1,4), (1,5), (1,6)] 
>>> reduce(list.__add__,map(list,l))
[1, 2, 1, 3, 1, 4, 1, 5, 1, 6]

Answer 3

Use chain.from_iterable, as it avoids unnecessary all-at-once unpacking (which leads to redundant memory consumption) by lazily advancing through list:

>>> import itertools
>>> L = [(1,2), (1,3), (1,4), (1,5), (1,6)]
>>> list(itertools.chain.from_iterable(L))
[1, 2, 1, 3, 1, 4, 1, 5, 1, 6]

Answer 4

“Fundamentally, which one is faster? Using the "itertools" module, or using a list comprehension? I'm basically trying to improve my computation speed here.” - @davidadamojr

I've been doing some tests and I find that the code below is actually faster.

list_ = [(1, 2), (1, 3), (1, 4), (1, 5), (1, 6)]
list(sum(list_, ()))

Someone correct me if I'm wrong.

Here are some tests below.

>>> list_ = [(1, 2), (1, 3), (1, 4), (1, 5), (1, 6)]
>>> 
>>> operation_1 = lambda: [tuple_item for tuple_ in list_ for tuple_item in tuple_]
>>> def operation_2 ():
        final_list = []
        for tuple_ in list_:
            for tuple_item in tuple_:
                final_list.append(tuple_item)
        return final_list

>>> operation_3 = lambda: reduce(list.__add__, map(list, list_))
>>> def operation_4 ():
        import itertools
        return list(itertools.chain(*list_))

>>> operation_5 = lambda: list(sum(list_, ()))
>>> 
>>> operation_1()
[1, 2, 1, 3, 1, 4, 1, 5, 1, 6]
>>> operation_2()
[1, 2, 1, 3, 1, 4, 1, 5, 1, 6]
>>> operation_3()
[1, 2, 1, 3, 1, 4, 1, 5, 1, 6]
>>> operation_4()
[1, 2, 1, 3, 1, 4, 1, 5, 1, 6]
>>> operation_5()
[1, 2, 1, 3, 1, 4, 1, 5, 1, 6]
>>> 
>>> import timeit
>>> 
>>> print('operation_1 completed in %s seconds.' % (timeit.timeit(operation_1)))
operation_1 completed in 1.57890490223 seconds.
>>> print('operation_2 completed in %s seconds.' % (timeit.timeit(operation_2)))
operation_2 completed in 2.90350501659 seconds.
>>> print('operation_3 completed in %s seconds.' % (timeit.timeit(operation_3)))
operation_3 completed in 5.08437990236 seconds.
>>> print('operation_4 completed in %s seconds.' % (timeit.timeit(operation_4)))
operation_4 completed in 3.85125378138 seconds.
>>> print('operation_5 completed in %s seconds.' % (timeit.timeit(operation_5)))
operation_5 completed in 1.2623826489 seconds.