锦标赛冠军 python while 循环

Question

I made a function 'Match' which takes two country and return a winner according to my algorithm.
EX)

def Match(England, Brazil) ----> England, 
def Match(Mexico, France) ---> France  

I need to write function Winner which takes list of 2^n nations and run a tournament and find a winner.
EX)

def Winner([England, Brazil, Mexico, France]) ---> France 

England---Brazil,        Mexico---France  (Semifinal)

England---France (final)

France (winner) Return value

The length of nation list differs and I cannot make a winner choosing algorithm. The match-up is It would be great if the code uses while loop rather than for loop ^^.

Answer 1

Your tournament method should make matches between consecutive pairs of players, for ["a", "b", "c", "d", "e", "f", "g", "h"] it should be a/b , c/d , e/f and g/h

You can achieve these with slicing and zip

• countries[::2] takes 1 on 2 so ["a", "c", "e", "g"]

• countries[1::2] same but starting at 1 so ["b", "d", "f", "h"]

• zip pairs these 2 lists to create pairs of opponents

Keep the winner of each match, and call tournament recursivly with the next round of players which contains half of players

# FOR DEMO PURPOSER
def match(country_a, country_b):
    return random.choice([country_a, country_b])

def tournament(countries):
    n = len(countries)
    if not ((n & (n - 1) == 0) and n != 0):
        raise Exception("Size isn't power of 2")

    if n == 2:
        return match(*countries)

    next_round = []
    for player1, player2 in zip(countries[::2], countries[1::2]):
        winner = match(player1, player2)
        next_round.append(winner)

    return tournament(next_round)

Using list-comprehension the for-loop and return can be replaced with

return tournament([match(p1, p2) for p1, p2 in zip(countries[::2], countries[1::2])])

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Improvement full code

After some time and discussion with the OP here's a major improvement of the full code with the major rules:

• don't call the exact same method in a loop, do it outside
• don't call a method again and again if it does the same, store the data somewhere