How can I combine a tuple elements into a list in python

Question

Consider the following code below:

list1 = ['1x', '2x']
list2 = ['x18', 'x74']
list3 = [('100p1', '100p2'), ('300p1', '300p2')]

gen_list = [[a,b] for a in list1 for b in list2]

for new_list in gen_list:
    for c in list3:
        print(new_list.extend(c))

My target result is like this:

[['1x','x18, '100p1', '100p2'],
 ['1x','x74, '100p1', '100p2'],
 ['1x','x18, '300p1', '300p2'],
 ['1x','x74, '300p1', '300p2'],
 ['2x','x18, '100p1', '100p2'],   
 ['2x','x74, '100p1', '100p2'],
 ['2x','x18, '300p1', '300p2'],
 ['2x','x74, '300p1', '300p2']]

but the result of the above code is this:

None
None
None
None
None
None
None
None

What necessary correction do I need to apply on my code? Thanks in advance.

print(new_list.extend(c)) prints None because extends returns None. Just print new_listafter the loops. — Jean-François Fabre
– Jean-François Fabre ♦, Commented Jul 5, 2017 at 9:16

Maarten Fabré · Accepted Answer · 2017-07-06 14:49:08Z

3

using itertools.product, unpacking and a list comprehension

[[l[0], l[2], *l[1]] for l in itertools.product(list1, list3, list2)]

or

[[l1, l2, *l3] for l1, l3, l2 in itertools.product(list1, list3, list2)]

before Python 3.5

For versions before Python 3.5 you could do something like this

[[l1, l2] + list(l3) for l1, l3, l2 in itertools.product(list1, list3, list2)]

If you know l3 only contains 2 elements you can use nested unpacking, as mentioned by @ShadowRanger

[[a, b, c1, c2] for a, (c1, c2), b in itertools.product(list1, list3, list2)]

edited Jul 6, 2017 at 14:49

answered Jul 5, 2017 at 10:42

Maarten Fabré

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5 Comments

Meeyaw Over a year ago

I tried this line here but still returns an syntax error like this: Syntax Error: can use starred expression only as assignment target My system is in python 3.4. I don't really know why this is a syntax error.

Maarten Fabré Over a year ago

It's a new feature in 3.5 docs.python.org/3/whatsnew/… I 'll see for a way that works in 3.4 when I'm at my pc

ShadowRanger Over a year ago

@MaartenFabré: Note: To get the exact ordering desired by the OP, you need to change the argument ordering for itertools.product from list1, list2, list3 to list1, list3, list2, so that the value derived from list2 is alternating most rapidly; similarly, you'd change l1, l2, l3 to l1, l3, l2 so unpacking matches. Demo (w/plain loop to match OP's code)

ShadowRanger Over a year ago

@MaartenFabré: Also, given list3 is a list of fixed length tuples, even pre-3.5 you can avoid creating/concatenating temporary lists by using nested unpacking syntax; it's faster, as it avoids the generic function call lookup and dispatch required by the list constructor: [[a, b, c1, c2] for a, (c1, c2), b in itertools.product(list1, list3, list2)] (Try it online!)

Meeyaw Over a year ago

Thanks for the effort. I actually used the one suggested by @ShadowRanger as it is simple and straightforward, not worrying about using the * or + operators for I am not well-trained to use those techniques. Though the explanations are good and informative. I might as well include it on my personal knowledge base. Thanks everyone!

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