A round-robin algorithm implementation written in Python. #round-robin #scheduling #algorithm #python

roundRobin.py

#!/usr/bin/python

div1 = ["Lions", "Tigers", "Jaguars", "Cougars"]
div2 = ["Whales", "Sharks", "Piranhas", "Alligators"]
div3 = ["Cubs", "Kittens", "Puppies", "Calfs"]

def create_schedule(list):
    """ Create a schedule for the teams in the list and return it"""
    s = []

    if len(list) % 2 == 1: list = list + ["BYE"]

    for i in range(len(list)-1):

        mid = len(list) / 2
        l1 = list[:mid]
        l2 = list[mid:]
        l2.reverse()	

        # Switch sides after each round
        if(i % 2 == 1):
            s = s + [ zip(l1, l2) ]
        else:
            s = s + [ zip(l2, l1) ]

        list.insert(1, list.pop())

    return s


def main():
    for round in create_schedule(div1):
        for match in round:
            print match[0] + " - " + match[1]
    print
    for round in create_schedule(div2):
        for match in round:
            print match[0] + " - " + match[1]
    print
    for round in create_schedule(div3): 
        for match in round:
            print match[0] + " - " + match[1]
    print
    for round in create_schedule(div1+div2+div3): 
        for match in round:
            print match[0] + " - " + match[1]
        print


if __name__ == "__main__":
    main()

Un mes entero buscando la forma de desarrollar un algoritmo que permita realizar emparejamientos y no encontraba, hasta que me dí cuenta que ya existían.
Gracias por el método

from itertools import cycle
A = [[1,2,3],[4,5,6],[7]]
B = [[8],[9,10,11],[12,13]]
i = len(A)
j = 0
C = [ ] #for results
list_num = cycle(k for k in range ( i ))
for x in list_num:
j += 1
if j == i*3:
break

if A[x]:
    C.append(A[x].pop(0))
    
if B[x]:
    C.append(B[x].pop(0))        

Output=========>[1, 8, 4, 9, 7, 12, 2, 5, 10, 13, 3, 6, 11]