How do I convert a tuple of tuples to a one-dimensional list using list comprehension? [duplicate]

Answer 1

You're chaining the tuples together:

from itertools import chain
print list(chain(*listOfTuples))

Should be pretty readable if you're familiar with itertools, and without the explicit list you even have your result in generator form.

Answer 2

Just use sum if you don't have a lot of tuples.

>>> tupleOfTuples = ((1, 2), (3, 4), (5,))
>>> sum(tupleOfTuples, ())
(1, 2, 3, 4, 5)
>>> list(sum(tupleOfTuples, ())) # if you really need a list
[1, 2, 3, 4, 5]

If you do have a lot of tuples, use list comprehension or chain.from_iterable to prevent the quadratic behavior of sum.

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Micro-benchmarks:

• Python 2.6

  • Long tuple of short tuples

    $ python2.6 -m timeit -s 'tot = ((1, 2), )*500' '[element for tupl in tot for element in tupl]'
    10000 loops, best of 3: 134 usec per loop
    $ python2.6 -m timeit -s 'tot = ((1, 2), )*500' 'list(sum(tot, ()))'
    1000 loops, best of 3: 1.1 msec per loop
    $ python2.6 -m timeit -s 'tot = ((1, 2), )*500; from itertools import chain; ci = chain.from_iterable' 'list(ci(tot))'
    10000 loops, best of 3: 60.1 usec per loop
    $ python2.6 -m timeit -s 'tot = ((1, 2), )*500; from itertools import chain' 'list(chain(*tot))'
    10000 loops, best of 3: 64.8 usec per loop
    

  • Short tuple of long tuples

    $ python2.6 -m timeit -s 'tot = ((1, )*500, (2, )*500)' '[element for tupl in tot for element in tupl]'
    10000 loops, best of 3: 65.6 usec per loop
    $ python2.6 -m timeit -s 'tot = ((1, )*500, (2, )*500)' 'list(sum(tot, ()))'
    100000 loops, best of 3: 16.9 usec per loop
    $ python2.6 -m timeit -s 'tot = ((1, )*500, (2, )*500); from itertools import chain; ci = chain.from_iterable' 'list(ci(tot))'
    10000 loops, best of 3: 25.8 usec per loop
    $ python2.6 -m timeit -s 'tot = ((1, )*500, (2, )*500); from itertools import chain' 'list(chain(*tot))'
    10000 loops, best of 3: 26.5 usec per loop
    

• Python 3.1

  • Long tuple of short tuples

    $ python3.1 -m timeit -s 'tot = ((1, 2), )*500' '[element for tupl in tot for element in tupl]'
    10000 loops, best of 3: 121 usec per loop
    $ python3.1 -m timeit -s 'tot = ((1, 2), )*500' 'list(sum(tot, ()))'
    1000 loops, best of 3: 1.09 msec per loop
    $ python3.1 -m timeit -s 'tot = ((1, 2), )*500; from itertools import chain; ci = chain.from_iterable' 'list(ci(tot))'
    10000 loops, best of 3: 59.5 usec per loop
    $ python3.1 -m timeit -s 'tot = ((1, 2), )*500; from itertools import chain' 'list(chain(*tot))'
    10000 loops, best of 3: 63.2 usec per loop
    

  • Short tuple of long tuples

    $ python3.1 -m timeit -s 'tot = ((1, )*500, (2, )*500)' '[element for tupl in tot for element in tupl]'
    10000 loops, best of 3: 66.1 usec per loop
    $ python3.1 -m timeit -s 'tot = ((1, )*500, (2, )*500)' 'list(sum(tot, ()))'
    100000 loops, best of 3: 16.3 usec per loop
    $ python3.1 -m timeit -s 'tot = ((1, )*500, (2, )*500); from itertools import chain; ci = chain.from_iterable' 'list(ci(tot))'
    10000 loops, best of 3: 25.4 usec per loop
    $ python3.1 -m timeit -s 'tot = ((1, )*500, (2, )*500); from itertools import chain' 'list(chain(*tot))'
    10000 loops, best of 3: 25.6 usec per loop
    

Observation:

• sum is faster if the outer tuple is short.
• list(chain.from_iterable(x)) is faster if the outer tuple is long.

Answer 3

Another solution using itertools.chain

>>> tupleOfTuples = ((1, 2), (3, 4), (5,))
>>> from itertools import chain
>>> [x for x in chain.from_iterable(tupleOfTuples)]
[1, 2, 3, 4, 5]

Answer 4

For multilevel, and readable code:

def flatten(bla):
    output = []
    for item in bla:
        output += flatten(item) if hasattr (item, "__iter__") or hasattr (item, "__len__") else [item]
    return output

I could not get this to fit in one line (and remain readable, even by far)

Answer 5

Most of these answers will only work for a single level of flattening. For a more comprehensive solution, try this (from http://rightfootin.blogspot.com/2006/09/more-on-python-flatten.html):

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 6

it's typically referred to as flattening a nested structure.

>>> tupleOfTuples = ((1, 2), (3, 4), (5,))
>>> [element for tupl in tupleOfTuples for element in tupl]
[1, 2, 3, 4, 5]

Just to demonstrate efficiency:

>>> import timeit
>>> it = lambda: list(chain(*tupleOfTuples))
>>> timeit.timeit(it)
2.1475738355700913
>>> lc = lambda: [element for tupl in tupleOfTuples for element in tupl]
>>> timeit.timeit(lc)
1.5745135182887857

ETA: Please don't use tuple as a variable name, it shadows built-in.