保持名稱和得分一起排序時

Question

，所以我需要一些高分到進行排序，這裏是我的代碼已經有：

def sortscores(): 
    namelist = [] 
    scorelist = [] 
    hs = open("hst.txt", "r") 
    hscounter = 0 
    for line in hs: 
     if counter%2 !=0: 
      name = line 
      othername = name[0:len(name)-1] 
      namelist.append(othername) 
     else: 
      scorelist.append(int(line)) 

這使姓名和分數到列表所以現在我需要有他們排序，但我不能使用.sort（）函數，因爲我必須自己寫這種排序，所以任何人都可以告訴我，我會怎麼做？ （同時保持名稱與正確的得分的分值爲降序排列）

Answer 1

，你可以讓你的字典的副本，發現最大的價值，關鍵保存到列表，從字典中刪除鍵，然後做直到複製的字典爲空。

import copy 

scores = {'hawks': 23, 'eagles': 42, 'knights': 33, 'rabbits': 44} #this or read from .txt 
scorescopy = copy.deepcopy(scores) #makes a copy of the dict, so you don't change the dict when deleting keys from the copy 
rank = [] #the list in which we want the keys ranked by value 

def keywithmaxval(scores): #finde the key with the highest value (stolen from another stackoverflow question) 
    values = list(scores.values()) 
    keys = list(scores.keys()) 
    return keys[values.index(max(values))] 

while len(scorescopy) > 0: #repeats until copy of dict is empty 
    maxkey = keywithmaxval(scorescopy) 
    scorescopy.pop(maxkey) #deletes key from copy of dict 
    rank.append(maxkey) #puts key in the ranked list 

print 'rank', rank #list of keys ranked by value 
print 'copy of dict', scorescopy #copy of dict, should be empty after we looped trough 
print 'original dict',scores #original dict, should be unchanged 

print '\nRank:' 
for key in rank: print key,':',scores[key] #pretty list of keys and vals 

Answer 2

如果您保存您的高分在(name, score)元組，那麼你可以很容易地讓他們在一起。由於您需要自己編寫排序函數，因此查看在其他問題中使用元組的示例可能會有所幫助。這裏有一個例子，簡單地找到最高分，同時保持名稱和分數。

首先設置數據。您可以使用zip這個

names = ['John', 'Jane', 'Tim', 'Sara'] 
scores = [100, 120, 80, 90] 
data = list(zip(names, scores)) # For Python 2.X you don't need the 'list' constructor 
print(data) 

輸出：

[('John', 100), ('Jane', 120), ('Tim', 80), ('Sara', 90)] 

現在發現的最大項：

max_entry = ('', 0) 
for entry in data: 
    if entry[1] > max_entry[1]: 
     max_entry = entry 

print(max_entry) 

輸出：

('Jane', 120)