combinations between two lists?

Question

It’s been a while and I’m having trouble wrapping my head around a algorithm I’m try to make. Basically, I have two lists and want to get all the combinations of the two lists.

I might not be explaining it correct so here’s a example.

name = 'a', 'b'
number = 1, 2

the output in this case would be:

1.  A1 B2
2.  B1 A2

The tricky part is I might have more items in the “name” variable than items in the “number” variable(number will always be equal to or less than the name variable).

I’m confused how to do all the combinations (nested for loop?) and even more confused on the logic to shift the items in the name variable in the event that there are more items in name than they are in the number list.

I’m not the best programmer but think I can give it a shot if someone can help me clarify the logic/algoriythm to achieve this. So I have just been stuck on nested for loops.

Update:

Here's the output with 3 variables and 2 numbers:

name = 'a', 'b', 'c'
number = 1, 2

output:

1.  A1 B2
2.  B1 A2
3.  A1 C2
4.  C1 A2
5.  B1 C2
6.  C1 B2

Answer 1

May be simpler than the simplest one above:

>>> a = ["foo", "bar"]
>>> b = [1, 2, 3]
>>> [(x,y) for x in a for y in b]
[('foo', 1), ('foo', 2), ('foo', 3), ('bar', 1), ('bar', 2), ('bar', 3)]

without any import

Answer 2

You might want to try a one line list comprehension:

>>> [name+number for name in 'ab' for number in '12']
['a1', 'a2', 'b1', 'b2']
>>> [name+number for name in 'abc' for number in '12']
['a1', 'a2', 'b1', 'b2', 'c1', 'c2']

Answer 3

a tiny improvement for the answer from interjay, to make the result as a flatten list.

>>> list3 = [zip(x,list2) for x in itertools.permutations(list1,len(list2))]
>>> import itertools
>>> chain = itertools.chain(*list3)
>>> list4 = list(chain)
[('a', 1), ('b', 2), ('a', 1), ('c', 2), ('b', 1), ('a', 2), ('b', 1), ('c', 2), ('c', 1), ('a', 2), ('c', 1), ('b', 2)]

reference from this link

Answer 4

Without itertools

[(list1[i], list2[j]) for i in xrange(len(list1)) for j in xrange(len(list2))]

Answer 5

The simplest way is to use itertools.product:

a = ["foo", "melon"]
b = [True, False]
c = list(itertools.product(a, b))
>> [("foo", True), ("foo", False), ("melon", True), ("melon", False)]

Answer 6

I was looking for a list multiplied by itself with only unique combinations, which is provided as this function.

import itertools
itertools.combinations(list, n_times)

---

Here as an excerpt from the Python docs on itertools That might help you find what your looking for.

Combinatoric generators:

Iterator                                 | Results
-----------------------------------------+----------------------------------------
product(p, q, ... [repeat=1])            | cartesian product, equivalent to a 
                                         |   nested for-loop
-----------------------------------------+----------------------------------------
permutations(p[, r])                     | r-length tuples, all possible 
                                         |   orderings, no repeated elements
-----------------------------------------+----------------------------------------
combinations(p, r)                       | r-length tuples, in sorted order, no 
                                         |   repeated elements
-----------------------------------------+----------------------------------------
combinations_with_replacement(p, r)      | r-length tuples, in sorted order, 
                                         | with repeated elements
-----------------------------------------+----------------------------------------
product('ABCD', repeat=2)                | AA AB AC AD BA BB BC BD CA CB CC CD DA DB DC DD
permutations('ABCD', 2)                  | AB AC AD BA BC BD CA CB CD DA DB DC
combinations('ABCD', 2)                  | AB AC AD BC BD CD
combinations_with_replacement('ABCD', 2) | AA AB AC AD BB BC BD CC CD DD