I'm a bit late on this topic, but think I can help someone.
You can use product from itertools:
from itertools import product
n = [1, 2, 3]
result = product[n, repeat=3] # You can change the repeat more then n length
print[list[result]]
Output:
[[1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 2, 1], [1, 2, 2], [1, 2, 3], [1, 3, 1],
[1, 3, 2], [1, 3, 3], [2, 1, 1], [2, 1, 2], [2, 1, 3], [2, 2, 1], [2, 2, 2],
[2, 2, 3], [2, 3, 1], [2, 3, 2], [2, 3, 3], [3, 1, 1], [3, 1, 2], [3, 1, 3],
[3, 2, 1], [3, 2, 2], [3, 2, 3], [3, 3, 1], [3, 3, 2], [3, 3, 3]]
Another example, but changing repeat arguement:
from itertools import product
n = [1, 2, 3]
result = product[n, repeat=4] # Changing repeat to 4
print[list[result]]
Output:
[1, 1, 2, 3], [1, 1, 3, 1], [1, 1, 3, 2], [1, 1, 3, 3], [1, 2, 1, 1],
[1, 2, 1, 2], [1, 2, 1, 3], [1, 2, 2, 1], [1, 2, 2, 2], [1, 2, 2, 3],
[1, 2, 3, 1], [1, 2, 3, 2], [1, 2, 3, 3], [1, 3, 1, 1], [1, 3, 1, 2],
[1, 3, 1, 3], [1, 3, 2, 1], [1, 3, 2, 2], [1, 3, 2, 3], [1, 3, 3, 1],
[1, 3, 3, 2], [1, 3, 3, 3], [2, 1, 1, 1], [2, 1, 1, 2], [2, 1, 1, 3],
[2, 1, 2, 1], [2, 1, 2, 2], [2, 1, 2, 3], [2, 1, 3, 1], [2, 1, 3, 2],
[2, 1, 3, 3], [2, 2, 1, 1], [2, 2, 1, 2], [2, 2, 1, 3], [2, 2, 2, 1],
[2, 2, 2, 2], [2, 2, 2, 3], [2, 2, 3, 1], [2, 2, 3, 2], [2, 2, 3, 3],
[2, 3, 1, 1], [2, 3, 1, 2], [2, 3, 1, 3], [2, 3, 2, 1], [2, 3, 2, 2],
[2, 3, 2, 3], [2, 3, 3, 1], [2, 3, 3, 2], [2, 3, 3, 3], [3, 1, 1, 1],
[3, 1, 1, 2], [3, 1, 1, 3], [3, 1, 2, 1], [3, 1, 2, 2], [3, 1, 2, 3],
[3, 1, 3, 1], [3, 1, 3, 2], [3, 1, 3, 3], [3, 2, 1, 1], [3, 2, 1, 2],
[3, 2, 1, 3], [3, 2, 2, 1], [3, 2, 2, 2], [3, 2, 2, 3], [3, 2, 3, 1],
[3, 2, 3, 2], [3, 2, 3, 3], [3, 3, 1, 1], [3, 3, 1, 2], [3, 3, 1, 3],
[3, 3, 2, 1], [3, 3, 2, 2], [3, 3, 2, 3], [3, 3, 3, 1], [3, 3, 3, 2],
[3, 3, 3, 3]]```
Given 3 digits a, b, and c. The task is to find all the possible combinations from these digits.
Examples:
Input: [1, 2, 3] Output: 1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1 Input: [0, 9, 5] Output: 0 9 5 0 5 9 9 0 5 9 5 0 5 0 9 5 9 0
Method 1: Brute force or Naive approach
The naive approach is to run 3 loops from 0 to 3 and print all the numbers from the list if the indexes are not equal to each other.
Example:
Python3
def comb[L]:
for i in
range[3]:
for j in range[3]:
for k in range[3]:
if [i!=j and j!=k and i!=k]:
print[L[i], L[j], L[k]]
comb[[1, 2, 3]]
Output:
1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1
Method 2: Using itertools.permutations[]
This method takes a list as an input and returns an object list of tuples that contain all permutation in a list form.
Example:
Python3
from itertools import permutations
comb = permutations[[1, 2, 3], 3]
for i in
comb:
print[i]
Output:
[1, 2, 3] [1, 3, 2] [2, 1, 3] [2, 3, 1] [3, 1, 2] [3, 2, 1]
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]