Python itertools permutations with repetition
You are looking for the Cartesian Product. Show
In your case, this would be
To get a random dice roll (in a totally inefficient way): (The following information and more can be found from my upcoming, to-be-announced book on recursive algorithms. I'll update this post when the book is released in 2022.) If you a list, dictionary, or other iterable object of values you need to generate combinations and permutations from, Python has the built-in
You can also reuse the values multiple times, which is called permutations with repetition and combinations with repetition (also called replacement):
(Note that permutations with repetition is effectively the same as a password cracker that tries every possible combination of characters to brute force a password.) The number of permutations and combinations quickly grows when more values are added to the iterable object. The total number of permutations and combinations is given in the following:
But to have Python generate permutations, you can use >>> import itertools >>> for v in itertools.permutations(['A', 'B', 'C']): ... print(v) ... ('A', 'B', 'C') ('A', 'C', 'B') ('B', 'A', 'C') ('B', 'C', 'A') ('C', 'A', 'B') ('C', 'B', 'A') >>> >>> list(itertools.permutations(['A', 'B', 'C'])) [('A', 'B', 'C'), ('A', 'C', 'B'), ('B', 'A', 'C'), ('B', 'C', 'A'), ('C', 'A', 'B'), ('C', 'B', 'A')] >>> list(itertools.permutations(['A', 'B', 'C'], 2)) [('A', 'B'), ('A', 'C'), ('B', 'A'), ('B', 'C'), ('C', 'A'), ('C', 'B')] >>> list(itertools.permutations(['A', 'B', 'C'], 1)) [('A',), ('B',), ('C',)] To have Python generate combinations, you can use >>> import itertools >>> for v in itertools.combinations(['A', 'B', 'C'], 2): ... print(v) ... ('A', 'B') ('A', 'C') ('B', 'C') >>> list(itertools.combinations(['A', 'B', 'C'], 1)) [('A',), ('B',), ('C',)] >>> list(itertools.combinations(['A', 'B', 'C'], 2)) [('A', 'B'), ('A', 'C'), ('B', 'C')] >>> list(itertools.combinations(['A', 'B', 'C'], 3)) [('A', 'B', 'C')] Note that the >>> powerSet = [] >>> import itertools >>> for k in range(4): ... powerSet.extend(itertools.combinations(['A', 'B', 'C'], k)) ... >>> powerSet [(), ('A',), ('B',), ('C',), ('A', 'B'), ('A', 'C'), ('B', 'C'), ('A', 'B', 'C')] To get permutations with repetition/replacement, call >>> import itertools >>> for v in itertools.product(['A', 'B', 'C'], repeat=3): ... print(v) ... ('A', 'A', 'A') ('A', 'A', 'B') ('A', 'A', 'C') ('A', 'B', 'A') ('A', 'B', 'B') ('A', 'B', 'C') ('A', 'C', 'A') ('A', 'C', 'B') ('A', 'C', 'C') ('B', 'A', 'A') ('B', 'A', 'B') ('B', 'A', 'C') ('B', 'B', 'A') ('B', 'B', 'B') ('B', 'B', 'C') ('B', 'C', 'A') ('B', 'C', 'B') ('B', 'C', 'C') ('C', 'A', 'A') ('C', 'A', 'B') ('C', 'A', 'C') ('C', 'B', 'A') ('C', 'B', 'B') ('C', 'B', 'C') ('C', 'C', 'A') ('C', 'C', 'B') ('C', 'C', 'C') >>> list(itertools.product(['A', 'B', 'C'], repeat=3)) [('A', 'A', 'A'), ('A', 'A', 'B'), ('A', 'A', 'C'), ('A', 'B', 'A'), ('A', 'B', 'B'), ('A', 'B', 'C'), ('A', 'C', 'A'), ('A', 'C', 'B'), ('A', 'C', 'C'), ('B', 'A', 'A'), ('B', 'A', 'B'), ('B', 'A', 'C'), ('B', 'B', 'A'), ('B', 'B', 'B'), ('B', 'B', 'C'), ('B', 'C', 'A'), ('B', 'C', 'B'), ('B', 'C', 'C'), ('C', 'A', 'A'), ('C', 'A', 'B'), ('C', 'A', 'C'), ('C', 'B', 'A'), ('C', 'B', 'B'), ('C', 'B', 'C'), ('C', 'C', 'A'), ('C', 'C', 'B'), ('C', 'C', 'C')] To get combinations with repetition/replacement, call >>> import itertools >>> for v in itertools.combinations_with_replacement(['A', 'B', 'C'], 3): ... print(v) ... ('A', 'A', 'A') ('A', 'A', 'B') ('A', 'A', 'C') ('A', 'B', 'B') ('A', 'B', 'C') ('A', 'C', 'C') ('B', 'B', 'B') ('B', 'B', 'C') ('B', 'C', 'C') ('C', 'C', 'C') >>> list(itertools.combinations_with_replacement(['A', 'B', 'C'], 3)) [('A', 'A', 'A'), ('A', 'A', 'B'), ('A', 'A', 'C'), ('A', 'B', 'B'), ('A', 'B', 'C'), ('A', 'C', 'C'), ('B', 'B', 'B'), ('B', 'B', 'C'), ('B', 'C', 'C'), ('C', 'C', 'C')] If you're like me and you had trouble remembering the differences between permutations and combinations, with and without repetition, and which Python functions implement them, bookmark this page to have easy access in the future. How do you do permutations with repetition in Python?Python String: Exercise-52 with Solution. Sample Solution:-. Python Code: from itertools import product def all_repeat(str1, rno): chars = list(str1) results = [] for c in product(chars, repeat = rno): results.append(c) return results print(all_repeat('xyz', 3)) print(all_repeat('xyz', 2)) print(all_repeat('abcd', 4)). How do you solve permutations with repetition?Permutation with repetition: This method is used when we are asked to make different choices each time and with different objects.
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Formulas for Permutations.. How do you do permutations in Python without Itertools?A.
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.
How do you find the permutation of a list in Python?The number of permutations on a set of n elements is given by n!. For example, there are 2! = 2*1 = 2 permutations of {1, 2}, namely {1, 2} and {2, 1}, and 3! = 3*2*1 = 6 permutations of {1, 2, 3}, namely {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2} and {3, 2, 1}.
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