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

因此，我正在尝试制作一个程序，该程序将要求用户输入并将值存储在数组/列表中.
然后，当输入空白行时，它将告诉用户这些值中有多少是唯一的.
我出于现实原因而不是问题集来构建它.

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掌握的一切.
我不确定如何计算列表中单词的唯一数量?
如果有人可以发布解决方案，以便我可以从中学习，或者至少告诉我它会是多么棒，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

此外，使用

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]

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