How do I create a recursion of multiple for loops of lists in Python to get combination?
Question:
I am currently using python to execute my code. My input into this function is a list within a list like [ [1,2,3],[4,5],[6,7,8,9] ] The goal is to iterate and create all the possible combinations for each list.
The only way I have it in my mind is to do multiple for loops on each sublist within the master list. Each combination must have at least one element from each sublist so since there are 3 sublists, all combinations must have 3 elements.
for a in [1,2,3]:
for b in [4,5]:
for c in [6,7,8,9]:
l.append([a,b,c])
So one combination would be [1,4,6],[1,4,7],[1,4,8],[1,4,9]. And the next loop with be [1,5,6]..[1,5,7]…and so forth.
This works but my problem is that I don’t know how many sublists will be in the master list (input) so I cant just keep writing for loops indefinitely. I know there must be a way to write a recursive function but I have never done this and don’t know how it works. I don’t think it should be too difficult but can someone show me the algorithm to accomplish this please?
Thank you so much!!
Answers:
Try using the itertools library:
import itertools
arr = [[1,2], [3,4], [5,6]]
list(itertools.product(*arr))
# => [(1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6)]
If you are trying to learn about recursion, think this way: The combinations of all the lists consist of taking one item at a time from the first list and prepending this item to all the combinations of the remaining lists.
This way you get
def combinations(rest, prefix = [], result = []):
if rest:
first = rest.pop()
for item in first:
prefix.append(item)
combinations(rest, prefix, result)
prefix.pop()
rest.append(first)
else:
result.append(list(reversed(prefix)))
return result
NB I am far from a Python expert. It’s likely there is a more elegant way to code this. Note that because I’m pushing and popping at the ends of lists I need to reverse the final results. The advantage of this is that it’s fast and should produce no garbage.
In Idle:
>>> combinations([ [1,2,3],[4,5],[6,7,8,9] ] )
[[1, 4, 6], [2, 4, 6], [3, 4, 6], [1, 5, 6], [2, 5, 6], [3, 5, 6],
[1, 4, 7], [2, 4, 7], [3, 4, 7], [1, 5, 7], [2, 5, 7], [3, 5, 7],
[1, 4, 8], [2, 4, 8], [3, 4, 8], [1, 5, 8], [2, 5, 8], [3, 5, 8],
[1, 4, 9], [2, 4, 9], [3, 4, 9], [1, 5, 9], [2, 5, 9], [3, 5, 9]]
I am currently using python to execute my code. My input into this function is a list within a list like [ [1,2,3],[4,5],[6,7,8,9] ] The goal is to iterate and create all the possible combinations for each list.
The only way I have it in my mind is to do multiple for loops on each sublist within the master list. Each combination must have at least one element from each sublist so since there are 3 sublists, all combinations must have 3 elements.
for a in [1,2,3]:
for b in [4,5]:
for c in [6,7,8,9]:
l.append([a,b,c])
So one combination would be [1,4,6],[1,4,7],[1,4,8],[1,4,9]. And the next loop with be [1,5,6]..[1,5,7]…and so forth.
This works but my problem is that I don’t know how many sublists will be in the master list (input) so I cant just keep writing for loops indefinitely. I know there must be a way to write a recursive function but I have never done this and don’t know how it works. I don’t think it should be too difficult but can someone show me the algorithm to accomplish this please?
Thank you so much!!
Try using the itertools library:
import itertools
arr = [[1,2], [3,4], [5,6]]
list(itertools.product(*arr))
# => [(1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6)]
If you are trying to learn about recursion, think this way: The combinations of all the lists consist of taking one item at a time from the first list and prepending this item to all the combinations of the remaining lists.
This way you get
def combinations(rest, prefix = [], result = []):
if rest:
first = rest.pop()
for item in first:
prefix.append(item)
combinations(rest, prefix, result)
prefix.pop()
rest.append(first)
else:
result.append(list(reversed(prefix)))
return result
NB I am far from a Python expert. It’s likely there is a more elegant way to code this. Note that because I’m pushing and popping at the ends of lists I need to reverse the final results. The advantage of this is that it’s fast and should produce no garbage.
In Idle:
>>> combinations([ [1,2,3],[4,5],[6,7,8,9] ] )
[[1, 4, 6], [2, 4, 6], [3, 4, 6], [1, 5, 6], [2, 5, 6], [3, 5, 6],
[1, 4, 7], [2, 4, 7], [3, 4, 7], [1, 5, 7], [2, 5, 7], [3, 5, 7],
[1, 4, 8], [2, 4, 8], [3, 4, 8], [1, 5, 8], [2, 5, 8], [3, 5, 8],
[1, 4, 9], [2, 4, 9], [3, 4, 9], [1, 5, 9], [2, 5, 9], [3, 5, 9]]