在while循環中計數輸入？

Question

我需要一個功能我必須寫，其中包括一個while循環將繼續下去，直到用戶輸入一個空的輸入，一旦出現這種情況，該函數將返回一個名稱已進入

的次數幫助到目前爲止，我的代碼是：

while True: 
    name = input('Enter a name:') 

    lst = name.split() 
    count={} 
    for n in lst: 
     if n in count: 
      count[n]+=1 

    for n in count: 
     if count[n] == 1: 
      print('There is {} student named {}'.format(count[n],\ 
                n)) 
     else: 

      print('There are {} students named {}'.format(count[n],\ 
                 n)) 

此不贅述，只要求用戶而返回1

輸出應該是這樣的：

Enter next name:Bob 
Enter next name:Mike 
Enter next name:Bob 
Enter next name:Sam 
Enter next name:Mike 
Enter next name:Bob 
Enter next name: 

There is 1 student named Sam 
There are 2 students named Mike 
There are 3 students named Bob 

Answer 1

for n in lst: 
     if n in count: 
      count[n]+=1 

在上述代碼中，n永遠不會添加到您的count字典中。即count仍然在循環後空..

Answer 2

什麼@zzk是說每個人的使用建議是

for n in lst: 
    if n in count: 
     count[n]+=1 
    else: 
     count[n]=1 

對於大部分功能答案

count= {} 

while True: 

    name = raw_input('Enter a name:') 
    lst = name.split() 

    for n in lst: 
     count[n] = count.get(n, 0) + 1 

    if not lst: 
     for n in count: 
      if count[n] == 1: 
       print('There is {} student named {}'.format(count[n],n)) 
      else: 
       print('There are {} students named {}'.format(count[n],n)) 
     break 

Answer 3

除了事實，你絕不添加n到您的count字典，您可以在while循環的每次迭代中一次又一次地初始化此字典。你必須把它放在循環之外。

count= {} 

while True: 
    name = input('Enter a name:') 

    lst = name.split() 
    for n in lst: 
     if n in count: 
      count[n] += 1 
     else: 
      count[n] = 1 

    for n in count: 
     if count[n] == 1: 
      print('There is {} student named {}'.format(count[n],\ 
                n)) 
     else: 

      print('There are {} students named {}'.format(count[n],\ 
                 n)) 

Answer 4

以下是有點矯枉過正。這只是要知道有標準模塊collections，它包含Counter類。無論如何，我更喜歡問題中使用的簡單解決方案（在刪除錯誤之後）。第一個函數讀取輸入並在輸入空名稱時中斷。第二個函數顯示結果：

#!python3 

import collections 


def enterStudents(names=None): 

    # Initialize the collection of counted names if it was not passed 
    # as the argument. 
    if names is None: 
     names = collections.Counter() 

    # Ask for names until the empty input. 
    while True: 
     name = input('Enter a name: ') 

     # Users are beasts. They may enter whitespaces. 
     # Strip it first and if the result is empty string, break the loop. 
     name = name.strip() 
     if len(name) == 0: 
      break 

     # The alternative is to split the given string to the first 
     # name and the other names. In the case, the strip is done 
     # automatically. The end-of-the-loop test can be based 
     # on testing the list. 
     # 
     # lst = name.split() 
     # if not lst: 
     #  break 
     # 
     # name = lst[0] 
     #     (my thanks to johnthexiii ;) 


     # New name entered -- update the collection. The update 
     # uses the argument as iterable and adds the elements. Because 
     # of this the name must be wrapped in a list (or some other 
     # iterable container). Otherwise, the letters of the name would 
     # be added to the collection. 
     # 
     # The collections.Counter() can be considered a bit overkill. 
     # Anyway, it may be handy in more complex cases.  
     names.update([name])  

    # Return the collection of counted names.  
    return names 


def printStudents(names): 
    print(names) # just for debugging 

    # Use .most_common() without the integer argument to iterate over 
    # all elements of the collection. 
    for name, cnt in names.most_common(): 
     if cnt == 1: 
      print('There is one student named', name) 
     else: 
      print('There are {} students named {}'.format(cnt, name)) 


# The body of a program. 
if __name__ == '__main__': 
    names = enterStudents() 
    printStudents(names) 

有部分代碼可以刪除。 enterStudents()中的name參數允許調用該函數爲現有集合添加名稱。初始化爲None用於將空初始集合作爲默認集合。

如果要收集包括空格在內的所有內容，則name.strip()不是必需的。

它打印出我的控制檯上

c:\tmp\___python\user\so15350021>py a.py 
Enter a name: a 
Enter a name: b 
Enter a name: c 
Enter a name: a 
Enter a name: a 
Enter a name: a 
Enter a name: 
Counter({'a': 4, 'b': 1, 'c': 1}) 
There are 4 students named a 
There is one student named b 
There is one student named c