如何找到列表中最常见的元素? [英] How to find most common elements of a list?

问题描述

提供以下列表

['Jellicle', 'Cats', 'are', 'black', 'and', 'white,', 'Jellicle', 'Cats', 
 'are', 'rather', 'small;', 'Jellicle', 'Cats', 'are', 'merry', 'and', 
 'bright,', 'And', 'pleasant', 'to', 'hear', 'when', 'they', 'caterwaul.', 
 'Jellicle', 'Cats', 'have', 'cheerful', 'faces,', 'Jellicle', 'Cats', 
 'have', 'bright', 'black', 'eyes;', 'They', 'like', 'to', 'practise', 
 'their', 'airs', 'and', 'graces', 'And', 'wait', 'for', 'the', 'Jellicle', 
 'Moon', 'to', 'rise.', '']

我试图计算每个单词出现的次数并显示前3个.

I am trying to count how many times each word appears and display the top 3.

但是，我只想查找首字母大写的前三位，而忽略所有首字母大写的单词.

However I am only looking to find the top three that have the first letter capitalized and ignore all words that do not have the first letter capitalized.

我敢肯定有比这更好的方法，但是我的想法是做以下事情:

I am sure there is a better way than this, but my idea was to do the following:

1. 将列表中的第一个单词放入另一个名为uniquewords的列表中
2. 从原始列表中删除第一个单词及其所有重复单词
3. 将新的第一个单词添加到唯一单词中
4. 删除第一个单词，并从原始列表中删除所有重复的单词.
5. 等...
6. 直到原始列表为空....
7. 计算唯一单词中每个单词出现在原始列表中的次数
8. 找到前3名并打印

推荐答案

如果您使用的是Python的早期版本，或者您有充分的理由推出自己的单词计数器(我想听听！)，您可以使用dict尝试以下方法.

If you are using an earlier version of Python or you have a very good reason to roll your own word counter (I'd like to hear it!), you could try the following approach using a dict.

Python 2.6.1 (r261:67515, Feb 11 2010, 00:51:29) 
[GCC 4.2.1 (Apple Inc. build 5646)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> word_list = ['Jellicle', 'Cats', 'are', 'black', 'and', 'white,', 'Jellicle', 'Cats', 'are', 'rather', 'small;', 'Jellicle', 'Cats', 'are', 'merry', 'and', 'bright,', 'And', 'pleasant', 'to', 'hear', 'when', 'they', 'caterwaul.', 'Jellicle', 'Cats', 'have', 'cheerful', 'faces,', 'Jellicle', 'Cats', 'have', 'bright', 'black', 'eyes;', 'They', 'like', 'to', 'practise', 'their', 'airs', 'and', 'graces', 'And', 'wait', 'for', 'the', 'Jellicle', 'Moon', 'to', 'rise.', '']
>>> word_counter = {}
>>> for word in word_list:
...     if word in word_counter:
...         word_counter[word] += 1
...     else:
...         word_counter[word] = 1
... 
>>> popular_words = sorted(word_counter, key = word_counter.get, reverse = True)
>>> 
>>> top_3 = popular_words[:3]
>>> 
>>> top_3
['Jellicle', 'Cats', 'and']

最重要的提示:只要您想使用这种算法，交互式Python解释器就是您的朋友.只需键入它，然后观察它，然后检查整个过程中的元素即可.

Top Tip: The interactive Python interpretor is your friend whenever you want to play with an algorithm like this. Just type it in and watch it go, inspecting elements along the way.

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