从字符串中删除常见单词(及其复数形式)的技术 [英] Technique to remove common words(and their plural versions) from a string

问题描述

我正在尝试通过解析一长串文本来查找食谱的标签(关键字).文字包含配方成分，说明和简短说明.

I am attempting to find tags(keywords) for a recipe by parsing a long string of text. The text contains the recipe ingredients, directions and a short blurb.

您认为从标签列表中删除常用字词最有效的方法是什么?

通常的意思是"the"，"at"，"there"，"their"等.

我有2种方法可以使用，您认为在速度方面更有效吗?您知道我可以采用哪种更有效的方法吗?

I have 2 methodologies I can use, which do you think is more efficient in terms of speed and do you know of a more efficient way I could do this?

方法1:
-确定每个单词出现的次数(使用库集合)
-列出常用词，并通过尝试从Collection对象中删除该键(如果存在)来从Collection对象中删除所有常用词".
-因此，速度将取决于变量delims的长度

Methodology 1:
- Determine the number of times each word occurs(using the library Collections)
- Have a list of common words and remove all 'Common Words' from the Collection object by attempting to delete that key from the Collection object if it exists.
- Therefore the speed will be determined by the length of the variable delims

import collections from Counter
delim     = ['there','there\'s','theres','they','they\'re'] 
# the above will end up being a really long list!
word_freq = Counter(recipe_str.lower().split())
for delim in set(delims):
    del word_freq[delim]
return freq.most_common()

方法2:
-对于可以为复数形式的常见单词，请查看配方字符串中的每个单词，并检查其是否部分包含了普通单词的非复数形式.例如;对于字符串"There's a test"，检查每个单词以查看它是否包含"there"，如果存在，则将其删除.

Methodology 2:
- For common words that can be plural, look at each word in the recipe string, and check if it partially contains the non-plural version of a common word. Eg; For the string "There's a test" check each word to see if it contains "there" and delete it if it does.

delim         = ['this','at','them'] # words that cant be plural
partial_delim = ['there','they',] # words that could occur in many forms
word_freq     = Counter(recipe_str.lower().split())
for delim in set(delims):
    del word_freq[delim]
# really slow 
for delim in set(partial_delims):
    for word in word_freq:
        if word.find(delim) != -1:
           del word_freq[delim]
return freq.most_common()

推荐答案

我只会做这样的事情:

from nltk.corpus import stopwords
s=set(stopwords.words('english'))

txt="a long string of text about him and her"
print filter(lambda w: not w in s,txt.split())

可打印

['long', 'string', 'text']

如果您认为散列集查找为O(1)，则在复杂度方面，字符串中的单词数应为O(n).

and in terms of complexity should be O(n) in number of words in the string, if you believe the hashed set lookup is O(1).

FWIW，我的NLTK版本定义了127个停用词:

FWIW, my version of NLTK defines 127 stopwords:

'all', 'just', 'being', 'over', 'both', 'through', 'yourselves', 'its', 'before', 'herself', 'had', 'should', 'to', 'only', 'under', 'ours', 'has', 'do', 'them', 'his', 'very', 'they', 'not', 'during', 'now', 'him', 'nor', 'did', 'this', 'she', 'each', 'further', 'where', 'few', 'because', 'doing', 'some', 'are', 'our', 'ourselves', 'out', 'what', 'for', 'while', 'does', 'above', 'between', 't', 'be', 'we', 'who', 'were', 'here', 'hers', 'by', 'on', 'about', 'of', 'against', 's', 'or', 'own', 'into', 'yourself', 'down', 'your', 'from', 'her', 'their', 'there', 'been', 'whom', 'too', 'themselves', 'was', 'until', 'more', 'himself', 'that', 'but', 'don', 'with', 'than', 'those', 'he', 'me', 'myself', 'these', 'up', 'will', 'below', 'can', 'theirs', 'my', 'and', 'then', 'is', 'am', 'it', 'an', 'as', 'itself', 'at', 'have', 'in', 'any', 'if', 'again', 'no', 'when', 'same', 'how', 'other', 'which', 'you', 'after', 'most', 'such', 'why', 'a', 'off', 'i', 'yours', 'so', 'the', 'having', 'once'

显然，您可以提供自己的一套；我同意对您的问题的评论，即可能最容易(也是最快)的是仅提供您要消除的所有变体，除非您想消除比这更多的单词，但这会成为一个问题发现有趣的事物比消除虚假的事物更重要.

obviously you can provide your own set; I'm in agreement with the comment on your question that it's probably easiest (and fastest) to just provide all the variations you want to eliminate up front, unless you want to eliminate a lot more words than this but then it becomes more a question of spotting interesting ones than eliminating spurious ones.

这篇关于从字符串中删除常见单词(及其复数形式)的技术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！