Python provides direct methods to find permutations and combinations of a sequence. These methods are present in itertools package.
Permutation
First import itertools package to implement the permutations method in python. This method takes a list as an input and returns an object list of tuples that contain all permutations in a list form.
Python3
from itertools import permutations
perm = permutations[[1, 2, 3]]
for i in list[perm]:
print [i]
Output:
[1, 2, 3] [1, 3, 2] [2, 1, 3] [2, 3, 1] [3, 1, 2] [3, 2, 1]
It generates n! permutations if the length of the input sequence is n.
If want to get permutations of length L then implement it in this way.
Python3
from itertools import permutations
perm =
permutations[[1, 2, 3], 2]
for i in list[perm]:
print [i]
Output:
[1, 2] [1, 3] [2, 1] [2, 3] [3, 1] [3, 2]
It generates nCr * r! permutations if the length of the input sequence is n and the input parameter is r.
Combination
This method takes a list and an input r as an input and return an object list of tuples which contain all possible combination of
length r in a list form.
Python3
from itertools import combinations
comb = combinations[[1, 2, 3], 2]
for i in list[comb]:
print [i]
Output:
[1, 2] [1, 3] [2, 3]
1. Combinations are emitted in lexicographic sort order of input. So, if the input list is sorted, the combination tuples will be
produced in sorted order.
Python3
from itertools import combinations
comb = combinations[[1, 2, 3], 2]
for i in list[comb]:
print [i]
Output:
[1, 2] [1, 3] [2, 3]
2. Elements are treated as unique based on their position, not on their value. So if the input elements are unique, there will
be no repeat values in each combination.
Python3
from itertools import combinations
comb = combinations[[2, 1, 3], 2]
for i in list[comb]:
print [i]
Output:
[2, 1] [2, 3] [1, 3]
3. If we want to make a combination of the same element to the same element then we use
combinations_with_replacement.
Python3
from itertools import combinations_with_replacement
comb = combinations_with_replacement[[1, 2, 3], 2]
for i in list[comb]:
print [i]
Output:
[1, 1] [1, 2] [1, 3] [2, 2] [2, 3] [3, 3]
Given an array of size n, generate and print all possible combinations of r elements in array.
Examples:
Input : arr[] = [1, 2, 3, 4],
r = 2
Output : [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
This problem has existing recursive solution please refer Print all possible combinations of r elements in a given array of size n link. We will solve this problem in python using itertools.combinations[] module.
What does itertools.combinations[] do ?
It returns r length subsequences of elements from the input iterable. Combinations are emitted in lexicographic sort order. So, if the input iterable is sorted, the combination tuples will be produced in sorted order.
- itertools.combinations[iterable, r] :
It return r-length tuples in sorted order with no repeated elements. For Example, combinations[‘ABCD’, 2] ==> [AB, AC, AD, BC, BD, CD]. - itertools.combinations_with_replacement[iterable, r] :
It return r-length tuples in sorted order with repeated elements. For Example, combinations_with_replacement[‘ABCD’, 2] ==> [AA, AB, AC, AD, BB, BC, BD, CC, CD, DD].
from itertools import combinations
def rSubset[arr, r]:
return list[combinations[arr, r]]
if __name__ == "__main__":
arr = [1, 2, 3, 4]
r = 2
print [rSubset[arr, r]]
Output:
[[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]
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