从字符串列表中获取拆分句子的索引

Question

所需的结果是一个函数或一种查找字符串列表中句子位置的方法。

sentence = 'The cat went to the pool yesterday'

structure = ['The cat went,', 'to the pool yesterday.','I wonder if you realize the effect you are having on me. It hurts. A lot.']

例如

def findsentence(sentence, list of strings):
      # do something to get the output, vec of positions to find the sentence in hte string list
return output

findsentence(sentence, structure)
> (0,1) # beacuse the phrase is splitted in the list...

注意！！

挑战在于不能准确地找到句子。看例子，这句话是句子位置0的一部分，结构位置1的一部分。

所以这不是一个简单的字符串操作问题。

Answer 1

使用以下内容：

sentence = "foo sam bar go"
structure = ["rq", "foo sam", "bar go", "ca", "da"]

def findsentencelist(sentence, list_of_strings):
    l = []
    for item in list_of_strings:
        if item in sentence:
            l.append(list_of_strings.index(item))
    return l

print str(findsentencelist(sentence, structure))

Yahli，希望这对你有所帮助。

编辑：

你的变量有问题。你的句子必须是一个字符串——而不是一个列表。编辑您的变量并再次尝试此函数:)

第二次编辑：我想我终于明白你想做什么了。如果这个效果更好，请告诉我。

第三次编辑：耶稣，希望这个能解决你的问题。让我知道它是否成功了:)

Answer 2

我只是删除了 structure 上的标点符号以使其正常工作：

sentence = 'The cat went to the pool yesterday'

structure = ['The cat went,', 'to the pool yesterday.','I wonder if you realize the effect you are having on me. It hurts. A lot.','Life is too short as it is. In short, she had a cushion job.']

import string

def findsentence(sentence, list_of_strings):
    return tuple(i for i, s in enumerate(list_of_strings) if s.translate(None, string.punctuation) in sentence)


print findsentence(sentence, structure)
# (0, 1)

Answer 3

去掉标点符号后。您可以使用此代码获取索引，

for i,j in enumerate(structure):
     if j in sentence:
          print(i)

希望这能解决您的问题。 python 很灵活，还有很多其他解决方案。

从字符串列表中获取拆分句子的索引

get index of splited sentences from a string list

python

string

text-mining

注意！！