Python dataframe在满足条件时选择连续跨距

假设我有停止词列表：

STOP = ['under', 'its', 'agreement', 'financed'] 

对于给定的数据帧：

lst = ['Kan.-based National', 'Kan.-based National Pizza', 'stock market', 
   'Pittsburg Kan.-based National Pizza', 'the stock market', 'revolving credit',
   'revolving credit agreement', 'its revolving credit agreement', 'under its revolving credit agreement', 
   'financed under its revolving credit agreement']

df = pd.DataFrame(lst)

即：

0   Kan.-based National
1   Kan.-based National Pizza
2   stock market
3   Pittsburg Kan.-based National Pizza
4   the stock market
5   revolving credit
6   revolving credit agreement
7   its revolving credit agreement
8   under its revolving credit agreement
9   financed under its revolving credit agreement

0   Pittsburg Kan.-based National Pizza
1   the stock market
2   revolving credit
3   revolving credit agreement
4   its revolving credit agreement
5   under its revolving credit agreement
6   financed under its revolving credit agreement

我想获得：

out = ['Pittsburg Kan.-based National Pizza', 'the stock market', 'revolving credit',
       'revolving credit agreement', 'its revolving credit agreement', 'under its revolving credit agreement', 
       'financed under its revolving credit agreement']

df_out = pd.DataFrame(out)

即：

0   Kan.-based National
1   Kan.-based National Pizza
2   stock market
3   Pittsburg Kan.-based National Pizza
4   the stock market
5   revolving credit
6   revolving credit agreement
7   its revolving credit agreement
8   under its revolving credit agreement
9   financed under its revolving credit agreement

0   Pittsburg Kan.-based National Pizza
1   the stock market
2   revolving credit
3   revolving credit agreement
4   its revolving credit agreement
5   under its revolving credit agreement
6   financed under its revolving credit agreement

注意：行的顺序并不重要

说明：

由于，

'Kan.-based National'

和

'Kan.-based National Pizza'

仅相差一个单词

'Pizza'

，并且

停止

列表中没有单词，因此我们希望选择最长的跨度，即

'Kan.-based National Pizza'

。 但是，

'Pittsburg Kan.-based National Pizza'

与

'Pittsburg.-based National Pizza'

也只有一个词不同

'Pittsburg'

，并且在

停止

列表中没有词，我们想选择最长的跨度，即

'Pittsburg Kan.-based National Pizza'

我们不能选择

“根据其循环信贷协议融资”

作为从

“循环信贷”

开始的最长期限，因为单词出现在

停止

列表中。因此，我们不会删除它的较小跨度

或者，在旁道上，如果字符串以（a | an | the）开头，其常用跨距之间的差异仅为一个单词。对于例如-

“股票市场”

和

“股票市场”

，我们希望选择最长的跨度，即

“股票市场”

我试着做：

delete_from_best_constituents = []
for u in best_parse_constituents:
    for v in best_parse_constituents:
        if u.lower().startswith('the') or v.lower().startswith('the'):
            u_part = u.lower().split('the')[-1].strip()
            v_part =  v.lower().split('the')[-1].strip()
            cond1 = all([w.lower() not in STOP for w in u_part.split()])
            cond2 = all([w.lower() not in STOP for w in v_part.split()])
            if u_part == v.lower() or v_part == u.lower() and cond1 and cond2:
                if not u.lower().startswith('the'):
                    delete_from_best_constituents.append(u)

---

假设XS是一组字符串。让YS是XS中X的集合，这样对于XS中的所有X-hat，X-hat不是X的子字符串。例如，如果XS是

{“blue”、“blueegg”、“hello”、“hello world”}

，那么

YS

是

{“blueegg”、“hello world”}

？你是在计算y吗？您正在尝试删除所有短子字符串吗？我不知道你想做什么。是否只保留较长的超级字符串？是否在任何时候找到另一个字符串的子字符串时，删除短字符串并保留较长字符串？例如，如果您发现

“hello world”

和

“hello”

这两个词，您保留

“hello world”

并丢弃

“hello”

？您写道，我们不能选择“根据其循环信贷协议融资”作为以“循环信贷”开头的最长字符串，这一点都没有意义。字符串

“根据其循环信贷协议融资”

并不是以字符串“循环信贷”开头的。您可以这样写：“我们不能做[BLAH]，因为单词出现在

停止列表中。”确切的“单词”是什么
STOP
列表中的哪些词确实会导致问题？您能否提供“span”的定义？你一直在谈论“这个跨度”、“那个跨度”等等，但你从来没有说过“跨度”到底是什么。你能在你的问题中加入一个指向“跨度”定义的超链接吗？