Question

我被要求设计一个排行榜，

这就是我试过的

def leader():

    file = open ("res.txt","r")
    reader = csv.reader(file)
    print (("{:20s}{:20s}{:20s}{:20s}".format("\nPlayer","Matches played","Won","Lost")))
    won = 100

    for r in reader:
        won = won-1
        if r[2] == str(won):
           print (("{:20s}{:20s}{:20s}{:20s}".format(r[0],r[1],r[2],r[3])))



    file.close()

我的csv文件看起来像这样

Leeroy,19,7,12
Jenkins,19,8,11
Tyler,19,0,19
Napoleon Wilson,19,7,12
Big Boss,19,7,12
Game Dude,19,5,14
Macho Man,19,3,16
Space Pirate,19,6,13
Billy Casper,19,7,12
Otacon,19,7,12
Big Brother,19,7,12
Ingsoc,19,5,14
Ripley,19,5,14
M'lady,19,4,15
Einstein100,19,8,11
Dennis,19,5,14
Esports,19,8,11
RNGesus,19,7,12
Kes,19,9,10
Magnitude,19,6,13

我希望首先展示获胜最多的人，你能帮忙吗？

Answer 1

尝试将文件啜饮到列表中：

all_rows = list(reader)

然后对大多数胜利的关键字上的所有行进行排序：

sorted_rows = sorted(all_rows, key=(lambda row: row[2]), reverse=True)

您可能希望堆叠多种排序，或者定义更复杂的密钥，以便您可以在相同数量的胜利中强制执行排序（例如，8场比赛中的7场比100场比赛中的7场胜利更好）。

Answer 2

keys = ["Player","Matches played","Won","Lost"]

# read file
with open("res.txt") as f:
    content = f.readlines()
# you may also want to remove whitespace characters like `\n` at the end of each line and 
# split it by ',' then form a dictionary out of it 
content = [dict(zip(keys,x.strip().split(','))) for x in content]
# sort the above list of dicts in decreasing orde.
final_result = sorted(content, key=lambda k: k['Won'], reverse=True)

print final_result# sorted By score
# first element is the one with highest Won value
print final_result[0] # highest Won

Python：排行榜

2 个答案: