Given a list. The task is to write a Python program to get all pairwise combinations from the list.
Finding all Pairs (No uniqueness)
Example:
Input: [1,”Mallika”,2,”Yash”]
Output: [(1, ‘Mallika’), (1, 2), (1, ‘Yash’), (‘Mallika’, 1), (‘Mallika’, 2), (‘Mallika’, ‘Yash’), (2, 1), (2, ‘Mallika’), (2, ‘Yash’), (‘Yash’, 1), (‘Yash’, ‘Mallika’), (‘Yash’, 2)]
Method 1: Using simple loops
We can access all combinations of the list using two loops to iterate over list indexes. If both the index counters are on the same index value, we skip it, else we print the element at index i followed by the element at index j in order.
The time complexity of this method is O(n2) since we require two loops to iterate over lists.
Python3
# declaring a listlst = [1,"Mallika",2,"Yash"]# outputoutput = []# looping over listfor i in range(0,len(lst)): for j in range(0,len(lst)): # checking if i and j indexes are not on same element if (i!=j): # add to output output.append((lst[i],lst[j]))# view outputprint(output) |
Output:
[(1, ‘Mallika’), (1, 2), (1, ‘Yash’), (‘Mallika’, 1), (‘Mallika’, 2), (‘Mallika’, ‘Yash’), (2, 1), (2, ‘Mallika’), (2, ‘Yash’), (‘Yash’, 1), (‘Yash’, ‘Mallika’), (‘Yash’, 2)]
Time Complexity: O(n*n)
Auxiliary Space: O(n)
Method 2: Using itertools
Python provides support of itertools standard library which is used to create iterators for efficient looping. The library provides support for various kinds of iterations, in groups, sorted order, etc. The permutations() functions of this library are used to get through all possible orderings of the list of elements, without any repetitions. The permutations() functions have the following syntax:
itertools.permutations(lst,r)
Where r depicts the r-length tuples, that is, 2 depicts a pair,3 depicts a triplet. The first argument is the specified list.
The function returns the list of groups of elements returned after forming the permutations. The output contains n x (n-1) number of elements, where n is the size of the list since each element is subsequently is multiplied with all others. The time required to compute permutations is roughly exponential in the order of the size of the list.
Python3
# importing required libraryimport itertools# creating a list of elements belonging to integers and stringslst = [1, "Mallika", 2, "Yash"]# simulating permutations of the list in a group of 2pair_order_list = itertools.permutations(lst, 2)# printing the elements belonging to permutationsprint(list(pair_order_list)) |
Time Complexity: O(n)
Auxiliary Space: O(n)
Output:
[(1, ‘Mallika’), (1, 2), (1, ‘Yash’), (‘Mallika’, 1), (‘Mallika’, 2), (‘Mallika’, ‘Yash’), (2, 1), (2, ‘Mallika’), (2, ‘Yash’), (‘Yash’, 1), (‘Yash’, ‘Mallika’), (‘Yash’, 2)]
Note:
- The pairs are printed in the order of the sequence of arrival of elements in the list.
- In the case of all same elements, the method still continues to form pairs and return them, even if they are duplicates.
Python3
# importing required libraryimport itertools# declaring a listlst = [2,2,2]# creating a list of pairs of the listordered_list = itertools.permutations(lst,2)# looping over the elements belonging to the# pair of permutationsfor i in ordered_list: # printing ith element print(i) |
Output :
(2, 2) (2, 2) (2, 2) (2, 2) (2, 2) (2, 2)
Time Complexity: O(n)
Auxiliary Space: O(1)
Finding all Unique Pairs (Uniqueness)
However, the permutations’ method doesn’t distinguish between (a, b) and (b, a) pairs and returns them both. The itertools library also supports a combinations() method that prints either of the (a, b) or (b, a) pairs and not both. The output number of elements is equivalent to (n-1)! where n is the length of the list. The time required to compute combinations is roughly polynomial.
Example:
Input: [1,”Mallika”,2,”Yash”]
Output: [(1, ‘Mallika’), (1, 2), (1, ‘Yash’), (‘Mallika’, 2), (‘Mallika’, ‘Yash’), (2, ‘Yash’)]
Python3
# importing required libraryimport itertools# creating a list of elements belonging 3 to integers and stringslst = [1,"Mallika",2,"Yash"]# simulating permutations of the list in # a group of 2pair_order_list = itertools.combinations(lst,2)# printing the elements belonging to permutationsprint (list(pair_order_list)) |
Output:
[(1, ‘Mallika’), (1, 2), (1, ‘Yash’), (‘Mallika’, 2), (‘Mallika’, ‘Yash’), (2, ‘Yash’)]
Time Complexity: O(n)
Auxiliary Space: O(n)
