[FIXED] Concatenation of inner lists or ints

Issue

I feel like I'm missing something obvious, but there it is... I would like to go from:

lst = [[0, 1, 3, 7, 8, 11, 12], [8, 0, 1, 2, 3, 14], 2]

to:

output = [0, 1, 3, 7, 8, 11, 12, 8, 0, 1, 2, 3, 14, 2]

I can do this with a for loop such as:

output = []
for l in lst:
    if hasattr(l, '__iter__'):
        output.extend(l)
    else:
        output.append(l)

Maybe the for-loop is fine, but it feels like there should be a more elegant way to do this... Trying to do this with numpy seems even more convoluted because ragged arrays aren't easily handled... so you can't (for example):

output = np.asanyarray(lst).flatten().tolist()

Thanks in advance.

Update:

Here's my comparison between the two methods provided by @T.J and @Ashwini - thanks to both!

In [5]: %paste
from itertools import chain
from collections import Iterable
lis = [[0, 1, 3, 7, 8, 11, 12], [8, 0, 1, 2, 3, 14], 2]
def solve(lis):
    for x in lis:
        if isinstance(x, Iterable) and not isinstance(x, basestring):
            yield x
        else:
            yield [x]

%timeit list(chain.from_iterable(solve(lis)))

%timeit [a for x in lis for a in (x if isinstance(x, Iterable) and not isinstance(x,basestring) else [x])]
## -- End pasted text --
100000 loops, best of 3: 10.1 us per loop
100000 loops, best of 3: 8.12 us per loop

Update2:

...
lis = lis *10**5
%timeit list(chain.from_iterable(solve(lis)))

%timeit [a for x in lis for a in (x if isinstance(x, Iterable) and not isinstance(x,basestring) else [x])]
## -- End pasted text --
1 loops, best of 3: 699 ms per loop
1 loops, best of 3: 698 ms per loop

Solution

Here is a pretty straightforward approach that uses a list comprehension:

>>> data = [[0, 1, 3, 7, 8, 11, 12], [8, 0, 1, 2, 3, 14], 2]
>>> [a for x in data for a in (x if isinstance(x, list) else [x])]
[0, 1, 3, 7, 8, 11, 12, 8, 0, 1, 2, 3, 14, 2]

Here are timing comparisons, it looks like my version is slightly faster (note that I modified my code to use collections.Iterable as well to make sure the comparison is fair):

In [9]: %timeit list(chain.from_iterable(solve(data)))
100000 loops, best of 3: 9.22 us per loop

In [10]: %timeit [a for x in data for a in (x if isinstance(x, Iterable) else [x])]
100000 loops, best of 3: 6.45 us per loop

Answered By - Andrew Clark