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evYpe
Asked:2020-02-20 13:08:14 +0000 UTC2020-02-20 13:08:14 +0000 UTC

Python中解决问题的代码优化

  • 772

我是初学者程序员。我正在上课。我在课程的框架内解决了这个问题，但我想让我的解决方案更加美观和优化。如何改进和简化我的代码？

一个任务：

输入是数字 n 然后 n 行，格式如下： 'Zenith;3;Spartak;1' 你需要编写一个程序来计算所有匹配的结果，并以下列格式显示结果：

球队：Total_games 胜 平 负 Total_points

我的决定：

n = int(input())
x = [input().split(';') for i in range(n)]
res = {}
g = 1
wp = 3
for i in x:
    if int(i[1]) > int(i[3]):
        if i[0] in res:
            z = res.get(i[0])
            z[0] += g
            z[1] += g
            z[4] += wp
        else:
            res[i[0]] = [1, 1, 0, 0, 3]
        if i[2] in res:
            z = res.get(i[2])
            z[0] += g
            z[3] += g
        else:
            res[i[2]] = [1, 0, 0, 1, 0]
    if int(i[1]) < int(i[3]):
        if i[2] in res:
            z = res.get(i[2])
            z[0] += g
            z[1] += g
            z[4] += wp
        else:
            res[i[2]] = [1, 1, 0, 0, 3]
        if i[0] in res:
            z = res.get(i[0])
            z[0] += g
            z[3] += g
        else:
            res[i[0]] = [1, 0, 0, 1, 0]
    if int(i[1]) == int(i[3]):
        if i[0] in res:
            z = res.get(i[0])
            z[0] += g
            z[2] += g
            z[4] += g
        else:
            res[i[0]] = [1, 0, 1, 0, 1]
        if i[2] in res:
            z = res.get(i[2])
            z[0] += g
            z[2] += g
            z[4] += g
        else:
            res[i[2]] = [1, 0, 1, 0, 1]
for i in res:
    print(i, ' '.join(map(str, res.get(i))), sep=':')

感谢您的关注！

python

3 个回答

  1. Best Answer

    引入函数和数据结构是合理的。看看这个选项。

    game_result={
    'WIN':3,
    'LOSE':0,
    'DRAW':1
}

def get_result(a):
    if a[1]>a[3]:
        return 'WIN','LOSE'
    if a[1]==a[3]:
        return 'DRAW','DRAW'
    if a[1]<a[3]:
        return 'LOSE','WIN'

n = int(input())
lst = [input().split(';') for i in range(n)]
res = {}
data={'games':1, 'WIN':0,'DRAW':0,'LOSE':0,'score':0}
for x in lst:
    r1,r2=get_result(x)
    if x[0] not in res.keys():
        res[x[0]]=data.copy()
    else:
        res[x[0]]['games']+=1
    if x[2] not in res.keys():
        res[x[2]]=data.copy()
    else:
        res[x[2]]['games']+=1
    res[x[0]][r1]+=1
    res[x[0]]['score']+=game_result[r1]
    res[x[2]][r2]+=1
    res[x[2]]['score']+=game_result[r2]

for x in res:
    print(x,res[x])
    • 0

  2. 删除 if i[0] in res: else:使用defaultdict

    res=defaultdict(lambda: list((0,0,0,0)))

    您可以从输出中删除 get 和 map。

    for k, v in result.items():
    print(k, end=": ")
    print(*v, sep=' ')

    if < , if >, if == 可以改成 if, elif, else

    • 0

  3. 基于@becouse 的想法。希望以更惯用的风格。需要 python 3.7

    from dataclasses import dataclass
from collections import defaultdict

@dataclass
class Results:
    games: int = 0
    win: int = 0
    draw: int = 0
    lose: int = 0

    def add_match(self, result):
        self.games += 1
        value = getattr(self, result)
        setattr(self, result, value + 1)

    @property
    def points(self):
        return self.win*3 + self.draw*1

    def __repr__(self):
        return f'{self.games}, {self.win}, {self.draw}, {self.lose}, {self.points}'


def get_result(score):
    first, second = score
    if first > second:
        return 'win', 'lose'
    elif first == second:
        return 'draw', 'draw'
    else:
        return 'lose', 'win'


#n = int(input())
#matches = [input().split(';') for i in range(n)]
matches = [['Зенит', 3, 'Спартак', 1], ['Зенит', 1, 'Динамо', 2]]
results = defaultdict(Results)

for match in matches:
    teams = match[::2]
    score = match[1::2]
    for team, result in zip(teams, get_result(score)):
        results[team].add_match(result)

for k, v in results.items():
    print(k, v)

    Зенит 2, 1, 0, 1, 3
Спартак 1, 0, 0, 1, 0
Динамо 1, 1, 0, 0, 3
    • 0

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