Question

我正在尝试编写一个不接受输入的函数，而是要求输入一个名称。如果输入空字符串，则该函数应为每个名称打印具有相同名称的人数。

例如：

>>>name()
Enter a name: Paul
Enter a name: Bill
Enter a name: John
Enter a name: Paul
Enter a name: Nick
Enter a name: Bill
Enter a name: Bill
Enter a name:    
There is 1 person named John
There is 1 person named Nick
There are 2 people named Paul
There are 3 people named Bill

到目前为止，我有：

def name():
    name = input ('Enter a name: ')
    count = 0

    while name:
        if name == input ('Enter a name: '):
            count = count + 1
        else:
            print (count)

我很确定我没有正确计算。你怎么能正确地完成这个功能呢？当你不知道有多少不同的名字时，你如何区分不同的输入并计算它们？

另外，如果可能的话，即使效率不高，我仍然希望学习基本代码。

Answer 1

只是为了一个单行的乐趣：

>>> collections.Counter(iter(functools.partial(input, "Enter a name: "), ""))
Enter a name: Paul
Enter a name: Bill
Enter a name: John
Enter a name: Paul
Enter a name: Nick
Enter a name: Bill
Enter a name: Bill
Enter a name: 
Counter({'Bill': 3, 'Paul': 2, 'Nick': 1, 'John': 1})

这可能不是在实际代码中执行此操作的方法。

Answer 2

from collections import defaultdict
dic = defaultdict(int)
while True:
    name = input ('Enter a name: ')
    if name:
        dic[name] += 1
    else:
        for k,v in sorted(dic.items(), key = lambda x: (x[1],x[0])):
            print ("There is {} person named {}".format(v,k))
        break

演示：

$ python3 so.py
Enter a name: Paul
Enter a name: Bill
Enter a name: John
Enter a name: Paul
Enter a name: Nick
Enter a name: Bill
Enter a name: Bill
Enter a name: 
There is 1 person named John
There is 1 person named Nick
There is 2 person named Paul
There is 3 person named Bill

Answer 3

Python附带了一个专门用于计数的集合，名为Counter。

import collections

counts = collections.Counter()

while True:
    name = input('Enter a name: ')
    if not name:
        break
    counts[name] += 1

for name, count in counts.items():
    print('There is {} person named {}'.format(count, name))

如果您希望结果的顺序从最不常见到最常见，Counter有一个按计数排序的函数most_common。不幸的是，它是向后的...但你可以通过调用reversed：

来解决这个问题

for name, count in reversed(counts.most_common()):
    print('There is {} person named {}'.format(count, name))

或者，如果您更喜欢按名称排序：

for name, count in sorted(counts.items()):
    print('There is {} person named {}'.format(count, name))

Answer 4

你可以使用一个简单的python地图。这个功能：

names = {}

while True:
    name = raw_input('Enter a name: ')
    if not name:
        break
    if name not in names:
        names[name] = 0
    names[name] = names[name] + 1

for name in sorted(names, key=names.get):
    if names[name] < 2:
        print 'There is %d person named %s' % (names[name], name)
    else:
        print 'There are %d people named %s' % (names[name], name)

有输入时如何计算

4 个答案: