Let's say I have a list of four values. I want to find all combinations of two of the values. For example, I would like to get an output like:
[[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]]
As you can see, I do not want repetitions, for example [0, 1] and [1, 0]
This needs to be able to be used with larger numbers, not just 4, and I will have to iterate through all of the combos
I am using Python 3 and Windows, and this would ideally be an inbuilt function, a simple bit of list comprehension code, or something I can import. I have tried making this with range, but I do not know how to exclude the numbers that I have already done from it.
The combination is a mathematical technique which calculates the number of possible arrangements in a collection of items or list. In combination order of selection doesn’t matter. The unique combination of two lists in Python can be formed by pairing each element of the first list with the elements of the second list.
Example:
List_1 = ["a","b"] List_2 = [1,2] Unique_combination = [[['a',1],['b',2]],[['a',2],['b',1]]]
Method 1 : Using permutation[] of itertools package and zip[] function.
Approach :
- Import itertools package and initialize list_1 and list_2.
- Create an empty list of ‘unique_combinations’ to store the resulting combinations so obtained.
- Call itertools.permutations[ ] which will return permutations of list_1 with length of list_2. Generally, the length of the shorter list is taken and if both lists are equal, use either.
- For loop is used and zip[] function is called to pair each permutation and shorter list element into the combination.
- Then each combination is converted into a list and append to the combination list.
Below is the implementation.
Python3
import itertools
from itertools import permutations
list_1 = ["a", "b", "c","d"]
list_2 = [1,4,9]
unique_combinations = []
permut = itertools.permutations[list_1, len[list_2]]
for comb in permut:
zipped = zip[comb, list_2]
unique_combinations.append[list[zipped]]
print[unique_combinations]
Output :
[[[‘a’, 1], [‘b’, 4], [‘c’, 9]], [[‘a’, 1], [‘b’, 4], [‘d’, 9]], [[‘a’, 1], [‘c’, 4], [‘b’, 9]], [[‘a’, 1], [‘c’, 4], [‘d’, 9]], [[‘a’, 1], [‘d’, 4], [‘b’, 9]], [[‘a’, 1], [‘d’, 4], [‘c’, 9]], [[‘b’, 1], [‘a’, 4], [‘c’, 9]], [[‘b’, 1], [‘a’, 4], [‘d’, 9]], [[‘b’, 1], [‘c’, 4], [‘a’, 9]], [[‘b’, 1], [‘c’, 4], [‘d’, 9]], [[‘b’, 1], [‘d’, 4], [‘a’, 9]], [[‘b’, 1], [‘d’, 4], [‘c’, 9]], [[‘c’, 1], [‘a’, 4], [‘b’, 9]], [[‘c’, 1], [‘a’, 4], [‘d’, 9]], [[‘c’, 1], [‘b’, 4], [‘a’, 9]], [[‘c’, 1], [‘b’, 4], [‘d’, 9]], [[‘c’, 1], [‘d’, 4], [‘a’, 9]], [[‘c’, 1], [‘d’, 4], [‘b’, 9]], [[‘d’, 1], [‘a’, 4], [‘b’, 9]], [[‘d’, 1], [‘a’, 4], [‘c’, 9]], [[‘d’, 1], [‘b’, 4], [‘a’, 9]], [[‘d’, 1], [‘b’, 4], [‘c’, 9]], [[‘d’, 1], [‘c’, 4], [‘a’, 9]], [[‘d’, 1], [‘c’, 4], [‘b’, 9]]]
Method 2 : Using product[] of itertools package and zip[] function.
Approach :
- Import itertools package and initialize list_1 and list_2.
- Create an empty list of ‘unique_combinations’ to store the resulting combinations so obtained.
- product[] is called to find all possible combinations of elements.
- And zip[] is used to pair up all these combinations, converting each element into a list and append them to the desired combination list.
Below is the implementation.
Python3
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]]]
print[unique_combinations]
Output :
[[[‘b’, 1], [‘c’, 1], [‘d’, 1]], [[‘b’, 1], [‘c’, 1], [‘d’, 4]], [[‘b’, 1], [‘c’, 1], [‘d’, 9]], [[‘b’, 1], [‘c’, 4], [‘d’, 1]], [[‘b’, 1], [‘c’, 4], [‘d’, 4]], [[‘b’, 1], [‘c’, 4], [‘d’, 9]], [[‘b’, 1], [‘c’, 9], [‘d’, 1]], [[‘b’, 1], [‘c’, 9], [‘d’, 4]], [[‘b’, 1], [‘c’, 9], [‘d’, 9]], [[‘b’, 4], [‘c’, 1], [‘d’, 1]], [[‘b’, 4], [‘c’, 1], [‘d’, 4]], [[‘b’, 4], [‘c’, 1], [‘d’, 9]], [[‘b’, 4], [‘c’, 4], [‘d’, 1]], [[‘b’, 4], [‘c’, 4], [‘d’, 4]], [[‘b’, 4], [‘c’, 4], [‘d’, 9]], [[‘b’, 4], [‘c’, 9], [‘d’, 1]], [[‘b’, 4], [‘c’, 9], [‘d’, 4]], [[‘b’, 4], [‘c’, 9], [‘d’, 9]], [[‘b’, 9], [‘c’, 1], [‘d’, 1]], [[‘b’, 9], [‘c’, 1], [‘d’, 4]], [[‘b’, 9], [‘c’, 1], [‘d’, 9]], [[‘b’, 9], [‘c’, 4], [‘d’, 1]], [[‘b’, 9], [‘c’, 4], [‘d’, 4]], [[‘b’, 9], [‘c’, 4], [‘d’, 9]], [[‘b’, 9], [‘c’, 9], [‘d’, 1]], [[‘b’, 9], [‘c’, 9], [‘d’, 4]], [[‘b’, 9], [‘c’, 9], [‘d’, 9]]]