Question

With a structure like this

hapts = [('1|2', '1|2'), ('3|4', '3|4')]

I need to zip it (sort of…) to get the following:

end = ['1|1', '2|2', '3|3', '4|4']

I started working with the following code:

zipped=[] 
for i in hapts:    
    tete = zip(i[0][0], i[1][0]) 
    zipped.extend(tete)
    some = zip(i[0][2], i[1][2])
    zipped.extend(some)        

… and got it zipped like this:

zipped = [('1', '1'), ('2', '2'), ('3', '3'), ('4', '4')]

Any suggestions on how to continue? Furthermore i’m sure there should a more elegant way to do this, but is hard to pass to Google an accurate definition of the question 😉
Thx!

Answer 1

You are very close to solving this, I would argue the best solution here is a simple str.join() in a list comprehension:

["|".join(values) for values in zipped]

This also has the bonus of working nicely with (potentially) more values, without modification.

If you wanted tuples (which is not what your requested output shows, as brackets don’t make a tuple, a comma does), then it is trivial to add that in:

[("|".join(values), ) for values in zipped]

Also note that zipped can be produced more effectively too:

>>> zipped = itertools.chain.from_iterable(zip(*[part.split("|") for part in group]) for group in hapts)
>>> ["|".join(values) for values in zipped]
['1|1', '2|2', '3|3', '4|4']

And to show what I meant before about handling more values elegantly:

>>> hapts = [('1|2|3', '1|2|3', '1|2|3'), ('3|4|5', '3|4|5', '3|4|5')]
>>> zipped = itertools.chain.from_iterable(zip(*[part.split("|") for part in group]) for group in hapts)
>>> ["|".join(values) for values in zipped]
['1|1|1', '2|2|2', '3|3|3', '3|3|3', '4|4|4', '5|5|5']