python - 如何根据词对的存在来选择子串？ Python

Question

我有大量的句子，我想从中提取以某些单词组合开头的子句子。例如，我想提取以“what does”或“what is”等开头的句子片段(实质上是从句子中删除出现在单词对之前的单词)。句子和单词对都是存储在 DataFrame 中:

'Sentence'                                    'First2'                                    
0  If this is a string what does it say?      0 can I    
1  And this is a string, should it say more?  1 should it    
2  This is yet another string.                2 what does
3  etc. etc.                                  3 etc. etc

我想从上面的例子中得到的结果是:

0 what does it say?
1 should it say more?
2

下面最明显的解决方案(至少对我而言)不起作用。它只使用第一个单词对 b 遍历所有句子 r，而不使用其他 b。

a = df['Sentence']
b = df['First2'] 

#The function seems to loop over all r's but only over the first b:
def func(z): 
    for x in b:
        if x in r:
            s = z[z.index(x):] 
            return s
        else:
            return ‘’

df['Segments'] = a.apply(func)

似乎以这种方式同时循环两个 DataFrame 是行不通的。有没有更有效的方法来做到这一点？

Answer 1

我认为您的代码中存在错误。

else:
    return ''

这意味着如果第一次比较不匹配，'func' 将立即返回。这可能就是代码不返回任何匹配项的原因。

示例工作代码如下:

# The function seems to loop over all r's but only over the first b:
def func(sentence, first_twos=b):
    for first_two in first_twos:
        if first_two in sentence:
            s = sentence[sentence.index(first_two):]
            return s
    return ''

df['Segments'] = a.apply(func)

输出:

df:   
{   
'First2': ['can I', 'should it', 'what does'],   
'Segments': ['what does it say? ', 'should it say more?', ''],   
'Sentence': ['If this is a string what does it say? ', 'And this is a string, should it say more?', 'This is yet another string.  '  ]  
}