# Combinations in Python without using itertools

• Difficulty Level : Expert
• Last Updated : 21 Jun, 2022

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 itself 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 requires 2 parameters. First is the string and the second is the length of substrings needed. The following example makes all combinations for the string ‘abc’ using itertools.

Example:

## Python3

 `# 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')]`

### A. 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.

## Python3

 `# 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``:]``        ` `        ``remainlst_combo ``=` `n_length_combo(remLst, n``-``1``)``        ``for` `p ``in` `remainlst_combo:``             ``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', 'a'], ['b', 'c'], ['c', 'a'], ['c', 'b']]`

### B. 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.

## Python3

 `import` `numpy`  `def` `n_length_combo(iterable, r):``    ` `    ``char ``=` `tuple``(iterable)``    ``n ``=` `len``(char)``    ` `    ``if` `r > n:``        ``return``    ` `    ``index ``=` `numpy.arange(r)``    ` `    ``# returns 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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