Combinations in Python without using itertools

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:

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)])

  Output

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

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