Question

我试图让这个代码：

from collections import Counter
a = input('Votes: ')
a = a.split(',')
count = Counter(a)
print (count)

当我输入这样的内容时：

One,One,Two,Two,Three,Four,Five,Five

打印出来：

One: 2
Two: 2
Three: 1
Four: 1
Five:2

而不是：

Counter({'One': 2, 'Two': 2, 'Five': 2, 'Three': 1, 'Four': 1})

Answer 1

循环Counter.most_common()输出：

for key, c in count.most_common():
    print("{}: {}".format(key, c))

most_common()方法为您提供排序顺序的项目，从最常见到最不常见：

如果您需要先按照他们的顺序订购。在a中排序{One之前Two只是因为先提到One，然后按照str.index()的字符串索引对它们进行排序。

for key, c in sorted(count.items(), key=lambda i: a.index(i[0])):
    print("{}: {}".format(key, c))

如果您需要通过数字的序数解释来订购它们，请使用将字词翻译成数字的字典：

numbers = {'one': 1, 'two': 2, 'three': 3, 'four': 4, 'five': 5} # expand as needed
for key, c in sorted(count.items(), key=lambda i: numbers[i[0].lower()]):
    print("{}: {}".format(key, c))

演示：

>>> from collections import Counter
>>> a = 'One,One,Two,Two,Three,Four,Five,Five'.split(',')
>>> count = Counter(a)
>>> for key, c in count.most_common():
...     print("{}: {}".format(key, c))
... 
Five: 2
Two: 2
One: 2
Three: 1
Four: 1
>>> for key, c in sorted(count.items(), key=lambda i: a.index(i[0])):
...     print("{}: {}".format(key, c))
... 
One: 2
Two: 2
Three: 1
Four: 1
Five: 2
>>> numbers = {'one': 1, 'two': 2, 'three': 3, 'four': 4, 'five': 5}
>>> for key, c in sorted(count.items(), key=lambda i: numbers[i[0].lower()]):
...     print("{}: {}".format(key, c))
... 
One: 2
Two: 2
Three: 1
Four: 1
Five: 2

Answer 2

from collections import Counter
a = input('Votes: ')
a = a.split(',')
count = Counter(a)

for key in sorted(set(a),key=lambda x: a.index(x)):
     print ("{}: {}".format(key,count[key]))

In [13]: for key in sorted(set(a),key=lambda x: a.index(x)):
   ....:          print ("{}: {}".format(key,count[key]))
   ....:     

One: 2
Two: 2
Three: 1
Four: 1
Five: 2

创建set以删除重复项，并使用sorted与lambda进行排序，以根据a列表中与{匹配的input列表中相应值的索引进行排序{1}}订单

Answer 3

一个班轮，分类：

a = ['One','One','Two','Two','Three','Four','Five','Five']
counts = Counter(a)
print('\n'.join('{}: {}'.format(*x) for x in sorted(counts.items(), key=lambda x: a.index(x[0]))))

在单独的行上显示“计数器”结果（Python）

3 个答案: