For one-level flatten, if you care about speed, this is faster than any of the previous answers under all conditions I tried. (That is, if you need the result as a list. If you only need to iterate through it on the fly then the chain example is probably better.) It works by pre-allocating a list of the final size and copying the parts in by slice (which is a lower-level block copy than any of the iterator methods):
def join(a):
"""Joins a sequence of sequences into a single sequence. (One-level flattening.)
E.g., join([(1,2,3), [4, 5], [6, (7, 8, 9), 10]]) = [1,2,3,4,5,6,(7,8,9),10]
This is very efficient, especially when the subsequences are long.
"""
n = sum([len(b) for b in a])
l = [None]*n
i = 0
for b in a:
j = i+len(b)
l[i:j] = b
i = j
return l
Sorted times list with comments:
[(0.5391559600830078, 'flatten4b'), # join() above.
(0.5400412082672119, 'flatten4c'), # Same, with sum(len(b) for b in a)
(0.5419249534606934, 'flatten4a'), # Similar, using zip()
(0.7351131439208984, 'flatten1b'), # list(itertools.chain.from_iterable(a))
(0.7472689151763916, 'flatten1'), # list(itertools.chain(*a))
(1.5468521118164062, 'flatten3'), # [i for j in a for i in j]
(26.696547985076904, 'flatten2')] # sum(a, [])