I have a list of tuples that looks like this:
[('a', 'b'), ('c', 'd'), (('e', 'f'), ('h', 'i'))]
I want to turn it into this:
[('a', 'b'), ('c', 'd'), ('e', 'f'), ('h', 'i')]
What is the most Pythonic way to do this?
I have a list of tuples that looks like this:
[('a', 'b'), ('c', 'd'), (('e', 'f'), ('h', 'i'))]
I want to turn it into this:
[('a', 'b'), ('c', 'd'), ('e', 'f'), ('h', 'i')]
What is the most Pythonic way to do this?
one-line, using list comprehension:
l = [('a', 'b'), ('c', 'd'), (('e', 'f'), ('h', 'i'))]
result = [z for y in (x if isinstance(x[0],tuple) else [x] for x in l) for z in y]
print(result)
yields:
[('a', 'b'), ('c', 'd'), ('e', 'f'), ('h', 'i')]
this is artificially creating a list if the element is not a tuple of tuples, then flattening all does the job. To avoid creating a single element list [x]
, (x for _ in range(1))
can also do the job (although it appears clunky)
Limitation: doesn't handle more than 1 level of nesting. In which case, a more complex/recursive solution must be coded.
Adjust the canonical un-flatten recipe to only unflatten when there are tuples in the value:
def flatten(l):
for el in l:
if isinstance(el, tuple) and any(isinstance(sub, tuple) for sub in el):
for sub in flatten(el):
yield sub
else:
yield el
This will only unwrap tuples, and only if there are other tuples in it:
>>> sample = [('a', 'b'), ('c', 'd'), (('e', 'f'), ('h', 'i'))]
>>> list(flatten(sample))
[('a', 'b'), ('c', 'd'), ('e', 'f'), ('h', 'i')]
A one-line solution would be using itertools.chain
:
>>> l = [('a', 'b'), ('c', 'd'), (('e', 'f'), ('h', 'i'))]
>>> from itertools import chain
>>> [*chain.from_iterable(x if isinstance(x[0], tuple) else [x] for x in l)]
[('a', 'b'), ('c', 'd'), ('e', 'f'), ('h', 'i')]