Question

我正试图从我的书中提出一个问题并且问：

实现不接受任何输入的函数名称并重复询问用户输入学生的名字。当用户输入空白时字符串，该函数应该打印每个名称，数量有这个名字的学生。

使用示例：

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

到目前为止，我有这段代码：

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])

我真的迷失在这里，任何投入都会有所帮助。如果可能的话，请解释如何修复我的代码

Answer 1

你的函数中有很多关于namings的问题，你正在使用诸如'names'这样的变量用于函数名称以及'input'这是一个用于读取用户输入的python函数名 - 所以你有避免使用这个。您还将namecount变量定义为dict并尝试在填充之前初始化它。因此，请尝试检查以下解决方案：

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

我为你重写了你的功能：

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

执行names()时：

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

让我们分析您的代码：

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])

似乎有一些问题，让我们列出它们

while循环的条件
- 您要做的是检查来自用户的输入是''（无）......
- input是一个内置函数，用于从用户那里获取输入，因此它永远不会是('')。
names = name声明
- 您要做的是将name添加到列表names。
- 此处您正在将names更改为字符串，这不是您想要的。
if的条件
- 与1相同。
for循环
- 让我们忽略..只是无效......这里..

我们将这些问题解决如下（解决方案与上面解决的问题具有相同的编号）

将条件更改为name != ''之类的内容。此外，在循环开始之前，您需要输入一次才能使其工作，在这种情况下有一个奖励，第一个输入可以有不同的提示。
使用names.append(name)将name添加到names。
与1相同。
请看下面的for循环......

试试这个

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()

按计数对字符串输入进行分组

4 个答案: