Combinations in Python without using itertools

Last Updated : 27 Feb, 2020

Itertools in Python is a module that produces complex iterators with the help of methods that work on iterators. This module works as a fast, memory-efficient tool that is used either by themselves or in combination to form iterator algebra.

Printing Combinations Using itertools

Using Itertools we can display all the possible combinations of the string in a quite optimized way. To display the combination, it requires 2 parameters. First is the string and second is the length of substrings needed. The following example makes all combinations for the string ‘abc’ using itertools.

Example:

# Import combinations from itertools 
from itertools import combinations  
    
      
def n_length_combo(arr, n):  
      
    # using set to deal  
    # with duplicates   
    return list(combinations(arr, n))  
    
# Driver Function  
if __name__ == "__main__":  
    arr = 'abc' 
    n = 2
    print (n_length_combo([x for x in arr], n) ) 

Output

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

Printing Combinations Without using itertools

By using recursion.

To create combinations without using itertools, iterate the list one by one and fix the first element of the list and make combinations with the remaining list. Similarly, iterate with all the list elements one by one by recursion of the remaining list.

# Function to create combinations  
# without itertools 
def n_length_combo(lst, n): 
      
    if n == 0: 
        return [[]] 
      
    l =[] 
    for i in range(0, len(lst)): 
          
        m = lst[i] 
        remLst = lst[i + 1:] 
          
        for p in n_length_combo(remLst, n-1): 
            l.append([m]+p) 
              
    return l 
  
# Driver code 
if __name__=="__main__": 
    arr ="abc"
    print(n_length_combo([x for x in arr], 2)) 

Output

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

By using iterations

In this, return the first combination of n elements from the string as it is, then other combinations are made by considering each element by its position. Each element is treated as unique based on its position, not on its value. So if the input elements are unique, there will be no repeat values in each combination.

import numpy 
  
  
def n_length_combo(iterable, r): 
      
    char = tuple(iterable) 
    n = len(char) 
      
    if r > n: 
        return
      
    index = numpy.arange(r) 
      
    # retruns the first sequence  
    yield tuple(char[i] for i in index) 
      
    while True: 
          
        for i in reversed(range(r)): 
            if index[i] != i + n - r: 
                break
        else: 
            return
          
        index[i] += 1
          
        for j in range(i + 1, r): 
              
            index[j] = index[j-1] + 1
              
        yield tuple(char[i] for i in index) 
          
# Driver code 
print([x for x in n_length_combo("abc", 2)]) 