从组合列表中获取唯一组

Question

combinations=[[0, 1, 2],[0, 2, 3], [0, 2, 4],[5, 6, 8], [5, 6, 9],[6, 7, 10], [6, 8, 9],[3,4,11],[7,9,10],[1,4,7],[9,10,11]]

output needed=
1) [[0,1,2],[5, 6, 8],[3,4,11],[7,9,10]],
2) [[0, 2, 3],[1,4,7],[5, 6, 8],[9,10,11]]

So I have gotten a list of all possible combination in sets of three, and I wanted to know if I could get a list of combinations in which all elements are there. ie [0,1,2,3,4,5,6,7,8,9,10,11] but in set of 3 Thanks in advance

Answer 1

This combination is a mathematical techniques with calculates the number of possible arrangements in a collection of list or items. The unique combination of two lists in python can be performed by pairing each element of the first list with the elements of the second list.
It can be done by using the methods namely permutation() of itertools package and zip() function and the other method is by using product() of itertools and zip() function.

Here is the example:

import itertools
from itertools import product

list_1 = ["b","c","d"]
list_2 = [1,4,9]

unique_combinations = []

unique_combinations = list(list(zip(list_1, element))
                           for element in product(list_2, repeat = len(list_1)))

Answer 2

If you are using only numeric data types in your array, you might be better off using numpy .

From what I can glean from your question, you need an array containing all the unique elements in combinations , and group them in threes in any random fashion.

import numpy as np

# Extracting the unique elements.
combinations = np.unique(combinations)

# Shuffling the elements.
combinations = np.random.permutation(combinations)

# Reshaping the unique elements in groups of three.
combinations = combinations.reshape((-1, 3))

Refer: NumPy docs