从python中的词袋中搜索文本答案

【问题标题】：Search text from bag of words in python从python中的词袋中搜索文本
【发布时间】：2020-02-14 00:19:08
【问题描述】：

假设我有一袋关键字。例如：

['profit low', 'loss increased', 'profit lowered']

我有一个 pdf 文档，我从中解析了整个文本，现在我想得到与词袋匹配的句子。

让我们说一句话是：

'The profit in the month of November lowered from 5% to 3%.'

这应该匹配在 bag of words 'profit lowered' 匹配这个句子。

在 python 中解决这个问题的最佳方法是什么？

【问题讨论】：

标签： python nlp text-processing text-parsing

【解决方案1】：

# input
checking_words = ['profit low', 'loss increased', 'profit lowered']
checking_string = 'The profit in the month of November lowered from 5% to 3%.'

trans_check_words = checking_string.split()
# output
for word_bug in [st.split() for st in checking_words]:
    if word_bug[0] in trans_check_words and word_bug[1] in trans_check_words:
        print(word_bug)

【讨论】：

【解决方案2】：

您想检查所有 Check Words 列表元素是否在长句中

sentence = 'The profit in the month of November lowered from 5% to 3%.'

words = ['profit','month','5%']

for element in words:
    if element in sentence:
        #do something with it
        print(element)

如果你想更简洁，可以使用这个单行循环将匹配的单词收集到一个列表中：

sentence = 'The profit in the month of November lowered from 5% to 3%.'

words = ['profit','month','5%']

matched_words = [] # Will collect the matched words in the next life loop:

[matched_words.append(word) for word in words if word in sentence]

print(matched_words)

如果您的列表中的每个元素上都有“间隔”单词，您想通过使用 split() 方法来处理它。

sentence = 'The profit in the month of November lowered from 5% to 3%.'

words = ['profit low','month high','5% 3%']

single_words = []
for w in words:
    for s in range(len(w.split(' '))):
        single_words.append(w.split(' ')[s])

matched_words = [] # Will collect the matched words in the next life loop:
[matched_words.append(word) for word in single_words if word in sentence]

print(matched_words)

【讨论】：

【解决方案3】：

您可以尝试以下方法：

将词袋转化为句子：

bag_of_words = ['profit low', 'loss increased', 'profit lowered']    
bag_of_word_sent =  ' '.join(bag_of_words)

然后是句子列表：

list_sents = ['The profit in the month of November lowered from 5% to 3%.']

使用 Levenshtein 距离：

import distance
for sent in list_sents:
    dist = distance.levenshtein(bag_of_word_sent, sent)
    if dist > len(bag_of_word_sent):
        # do something
        print(dist)

【讨论】：