Combinations between two lists?

Question

It has been a while, and I am having trouble wrapping around the algorithm I'm trying to do. Basically, I have two lists and you want to get all combinations of these two lists.

Maybe I won’t explain it correctly, so here is an example.

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

the output in this case will be:

 1. A1 B2 2. B1 A2 

The hard part: I can have more elements in the variable "name" than in the variable "number" (the number will always be equal to or less than the variable name).

I got confused how to make all combinations (nested in a loop?) And even more confused with logic to move elements in a name variable in case there are more elements in the list than in the list of numbers.

I am not the best programmer, but I think I can give him a chance if someone can help me clarify the logic / algorithm to achieve this. So I was just stuck on nested loops.

Update:

Here is the output with three variables and two numbers:

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

exit:

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

Answer 1

Suppose len(list1) >= len(list2) . Then you need to take all permutations of length len(list2) from list1 and map them to elements from list2. In python:

 import itertools list1=['a','b','c'] list2=[1,2] [zip(x,list2) for x in itertools.permutations(list1,len(list2))] 

Returns

 [[('a', 1), ('b', 2)], [('a', 1), ('c', 2)], [('b', 1), ('a', 2)], [('b', 1), ('c', 2)], [('c', 1), ('a', 2)], [('c', 1), ('b', 2)]] 

Answer 2

The easiest 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 3

Could be simpler than simple 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 4

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 is an excerpt from the Python documentation for itertools This may help you find what you are 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 

Answer 5

You might want to try a one-line schedule:

 >>> [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 6

a slight improvement in the response from interjay to make the result a smoothing 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)] 

link from this link

Answer 7

Without itertools

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

Answer 8

Answering the question “given two lists, find all possible permutations of pairs of one element from each list” and using the basic Python functionality (i.e. without itertools) and, therefore, simplify its replication for other programming languages:

 def rec(a, b, ll, size): ret = [] for i,e in enumerate(a): for j,f in enumerate(b): l = [e+f] new_l = rec(a[i+1:], b[:j]+b[j+1:], ll, size) if not new_l: ret.append(l) for k in new_l: l_k = l + k ret.append(l_k) if len(l_k) == size: ll.append(l_k) return ret a = ['a','b','c'] b = ['1','2'] ll = [] rec(a,b,ll, min(len(a),len(b))) print(ll) 

Returns

 [['a1', 'b2'], ['a1', 'c2'], ['a2', 'b1'], ['a2', 'c1'], ['b1', 'c2'], ['b2', 'c1']] 

Answer 9

Or KISS answer for short lists:

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

Not as productive as itertools, but you are using python, so performance is no longer your main concern ...

I like all the other answers too!

Answer 10

The best way to find out all the combinations for a large number of lists:

 import itertools from pprint import pprint inputdata = [ ['a', 'b', 'c'], ['d'], ['e', 'f'], ] result = list(itertools.product(*inputdata)) pprint(result) 

the result will be:

 [('a', 'd', 'e'), ('a', 'd', 'f'), ('b', 'd', 'e'), ('b', 'd', 'f'), ('c', 'd', 'e'), ('c', 'd', 'f')] 

Answer 11

The best answers to this question are only suitable for the specific lengths of the lists that are provided.

Here is a version that works for any input length. It also makes the algorithm understandable in terms of mathematical concepts of combination and permutation.

 from itertools import combinations, permutations list1 = ['1', '2'] list2 = ['A', 'B', 'C'] num_elements = min(len(list1), len(list2)) list1_combs = list(combinations(list1, num_elements)) list2_perms = list(permutations(list2, num_elements)) result = [ tuple(zip(perm, comb)) for comb in list1_combs for perm in list2_perms ] for idx, ((l11, l12), (l21, l22)) in enumerate(result): print(f'{idx}: {l11}{l12} {l21}{l22}') 

This outputs:

 0: A1 B2 1: A1 C2 2: B1 A2 3: B1 C2 4: C1 A2 5: C1 B2