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Question

分組串輸入我正在試圖做的一個問題了我的書，並問：

實現功能名稱，它沒有任何輸入，反覆詢問 用戶輸入一個學生的名字。當用戶輸入一個空白 字符串時，該功能應打印每個名稱，即具有該名稱的 學生的編號。

用法示例：

Usage: 
names() 
Enter next name: Valerie 
Enter next name: Bob 
Enter next name: Valerie 
Enter next name: John 
Enter next name: Amelia 
Enter next name: Bob 
Enter next name: 
There is 1 student named Amelia 
There are 2 students named Bob 
There is 1 student named John 
There are 2 students named Valerie 

到目前爲止，我有這樣的代碼：

def names(): 
    names = [] 
    namecount = {a:name.count(a) for a in names} 
    while input != (''): 
     name = input('Enter next name: ') 
     names = name 
     if input == ('') 
      for x in names.split(): 
       print ('There is', x ,'named', names[x]) 

我真的在這裏失去了任何的投入將助陣噸。另外，如果可能的話，請解釋如何修復我的代碼

Answer 1

在你的函數中namings有很多問題，你正在使用'names'這樣的變量，用於函數名以及'input'是一個用於讀取用戶輸入的python函數名 - 所以你必須避免使用這個。你也定義一個namecount變量作爲一個字典，並嘗試在填充前初始化它。因此，嘗試檢查下面的解決方案：

def myFunc(): 
    names = [] 
    name = '' 
    while True: #bad stuff you can think on your own condition 
     name = raw_input('press space(or Q) to exit or enter next name: ') 
     if name.strip() in ('', 'q', 'Q'): 
      for x in set(names): 
       print '{0} is mentioned {1} times'.format(x, names.count(x)) 
      break 
     else: 
      names.append(name) 

myFunc() 

OR：

from collections import defaultdict 

def myFunc(): 
    names = defaultdict(int) 
    name = '' 
    while True: #bad stuff you can think on your own condition 
     name = raw_input('press space(or Q) to exit or enter next name: ') 
     if name.strip() in ('', 'q', 'Q'): 
      for x in set(names): 
       print '{0} is mentioned {1} times'.format(x, names[x]) 
      break 
     else: 
      names[name] += 1 

Answer 2

我改寫了你的功能爲您提供：

def names(): 
    names = {} # Creates an empty dictionary called names 
    name = 'cabbage' # Creates a variable, name, so when we do our while loop, 
        # it won't immediately break 
        # It can be anything really. I just like to use cabbage 
    while name != '': # While name is not an empty string 
     name = input('Enter a name! ') # We get an input 
     if name in names: # Checks to see if the name is already in the dictionary 
      names[name] += 1 # Adds one to the value 
     else: # Otherwise 
      names[name] = 1 # We add a new key/value to the dictionary 
    del names[''] # Deleted the key '' from the dictionary 
    for i in names: # For every key in the dictionary 
     if names[i] > 1: # Checks to see if the value is greater for 1. Just for the grammar :D 
      print("There are", names[i], "students named", i) # Prints your expected output 
     else: # This runs if the value is 1 
      print("There is", names[i], "student named", i) # Prints your expected output 

在做names()：

Enter a name! bob 
Enter a name! bill 
Enter a name! ben 
Enter a name! bob 
Enter a name! bill 
Enter a name! bob 
Enter a name! 
There are 3 students named bob 
There are 2 students named bill 
There is 1 student named ben 

Answer 3

讓我們來分析你的代碼：

def names(): 
    names = [] 
    namecount = {a:name.count(a) for a in names} 
    while input != (''): 
     name = input('Enter next name: ') 
     names = name 
     if input == ('') 
      for x in names.split(): 
       print ('There is', x ,'named', names[x]) 

似乎有是幾個問題，讓我們來一一列舉了

1. 的while循環的條件
   
   • 你想要做的檢查，如果從用戶輸入''（什麼）什麼..
   • input是獲取用戶輸入的內置函數，所以它永遠不會是('')。
2. 的names = name聲明
   
   • 你想要做什麼是添加name到列表names。
   • 這裏你將names更改爲一個字符串，這不是你想要的。
3. 的if的條件
   • 相同1。
4. 的for環
   • 讓我們忽略..只是無效..在這裏..

我們解決這些問題如下（解決方案具有相同的編號以上問題，它解決）

1. 將條件更改爲像name != ''。 另外，在循環開始之前，您需要輸入一次才能正常工作，在這種情況下有獎金，第一個輸入可以有不同的提示。
2. 使用names.append(name)添加name到names。
3. 同1
4. 只要看看下面循環...

試試這個

def names(): 
    names = [] 
    name = input('Enter a name: ').strip() # get first name 
    while name != '': 
     names.append(name) 
     name = raw_input('Enter next name: ').strip() # get next name 

    for n in set(names): # in a set, no values are repeated 
     print '%s is mentioned %s times' % (n, names.count(n)) # print output 

Answer 4

def names(): 
     counters = {} 

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

      if name == ' ': 
        break 

      if name in counters: 
        counters[name] += 1 
      else: 
        counters[name] = 1 
     for name in counters: 
      if counters[name] == 1: 
        print('There is {} student named {}'.format(counters[name],name)) 
      else: 
        print('There are {} student named {}'.format(counters[name],name)) 


names()