Hướng dẫn permutations with replacement python

Python provides direct methods to find permutations and combinations of a sequence. These methods are present in itertools package.

Permutation 

First import itertools package to implement the permutations method in python. This method takes a list as an input and returns an object list of tuples that contain all permutations in a list form. 
 

Python3

from itertools import permutations 

perm = permutations([1, 2, 3]) 

for i in list(perm): 

    print (i) 

Output: 

(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)

It generates n! permutations if the length of the input sequence is n. 
If want  to get permutations of length L then implement it in this way. 
 

Python3

from itertools import permutations 

perm = permutations([1, 2, 3], 2) 

for i in list(perm): 

    print (i) 

Output: 

(1, 2)
(1, 3)
(2, 1)
(2, 3)
(3, 1)
(3, 2)

It generates nCr * r! permutations if the length of the input sequence is n and the input parameter is r.

Combination 

This method takes a list and an input r as an input and return an object list of tuples which contain all possible combination of length r in a list form. 
 

Python3

from itertools import combinations

comb = combinations([1, 2, 3], 2)

for i in list(comb):

    print (i)

Output: 

(1, 2)
(1, 3)
(2, 3)

1. Combinations are emitted in lexicographic sort order of input. So, if the input list is sorted, the combination tuples will be produced in sorted order. 
 

Python3

from itertools import combinations 

comb = combinations([1, 2, 3], 2) 

for i in list(comb): 

    print (i)

Output: 

(1, 2)
(1, 3)
(2, 3)

2. Elements are treated as unique based on their position, not on their value. So if the input elements are unique, there will be no repeat values in each combination. 
 

Python3

from itertools import combinations 

comb = combinations([2, 1, 3], 2) 

for i in list(comb): 

    print (i)

Output: 

(2, 1)
(2, 3)
(1, 3)

3. If we want to make a combination of the same element to the same element then we use combinations_with_replacement. 
 

Python3

from itertools import combinations_with_replacement 

comb = combinations_with_replacement([1, 2, 3], 2) 

for i in list(comb): 

    print (i) 

Output:

(1, 1)
(1, 2)
(1, 3)
(2, 2)
(2, 3)
(3, 3) 

You are looking for the Cartesian Product.

Nội dung chính

• How do you do permutations with repetition in Python?
• How do you solve permutations with repetition?
• How do you do permutations in Python without Itertools?
• How do you find the permutation of a list in Python?

Nội dung chính

• How do you do permutations with repetition in Python?
• How do you solve permutations with repetition?
• How do you do permutations in Python without Itertools?
• How do you find the permutation of a list in Python?

In mathematics, a Cartesian product (or product set) is the direct product of two sets.

In your case, this would be {1, 2, 3, 4, 5, 6} x {1, 2, 3, 4, 5, 6}. itertools can help you there:

import itertools
x = [1, 2, 3, 4, 5, 6]
[p for p in itertools.product(x, repeat=2)]
[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), 
 (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), 
 (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), 
 (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)]

To get a random dice roll (in a totally inefficient way):

import random
random.choice([p for p in itertools.product(x, repeat=2)])
(6, 3)

(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 itertools module as part of its standard library. The permutations of an iterable are every possible ordering of all of the values, while the combinations are every possible selection of some, none, or all of the values. For example, the permutations and combinations of the set {'A', 'B', 'C'} are:

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 itertools.permutations():

>>> 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 itertools.combinations():

>>> 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 combinations() function takes a second argument for the number of values to select. To get all combinations (also called the power set), you'll need to make multiple calls to combinations():

>>> 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 itertools.product() and pass the size of the iterable object for its repeat argument:

>>> 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 itertools.combinations_with_replacement():

>>> 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. ... 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}.