如何计算列表中的唯一值 [英] How do I count unique values inside a list

问题描述

所以我正在尝试制作这个程序，它会要求用户输入并将值存储在数组/列表中.
然后当输入一个空行时，它会告诉用户这些值中有多少是唯一的.
我是出于现实生活的原因而不是为了解决问题而构建它.

So I'm trying to make this program that will ask the user for input and store the values in an array / list.
Then when a blank line is entered it will tell the user how many of those values are unique.
I'm building this for real life reasons and not as a problem set.

enter: happy
enter: rofl
enter: happy
enter: mpg8
enter: Cpp
enter: Cpp
enter:
There are 4 unique words!

我的代码如下:

# ask for input
ipta = raw_input("Word: ")

# create list 
uniquewords = [] 
counter = 0
uniquewords.append(ipta)

a = 0   # loop thingy
# while loop to ask for input and append in list
while ipta: 
  ipta = raw_input("Word: ")
  new_words.append(input1)
  counter = counter + 1

for p in uniquewords:

..这就是我到目前为止所获得的全部内容.
我不确定如何计算列表中唯一的单词数?
如果有人可以发布解决方案，以便我可以从中学习，或者至少向我展示它会很棒，谢谢！

..and that's about all I've gotten so far.
I'm not sure how to count the unique number of words in a list?
If someone can post the solution so I can learn from it, or at least show me how it would be great, thanks!

推荐答案

另外，使用 collections.Counter 重构你的代码:

In addition, use collections.Counter to refactor your code:

from collections import Counter

words = ['a', 'b', 'c', 'a']

Counter(words).keys() # equals to list(set(words))
Counter(words).values() # counts the elements' frequency

输出:

['a', 'c', 'b']
[2, 1, 1]

这篇关于如何计算列表中的唯一值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！