Pandas.apply 在 spacy doc 列上返回 none 值

Question

我在我的熊猫df 'sp500news3'上运行以下内容，它返回一个 None 值

def extract_ticker(title):
    for word in title:
        if word in constituents['Symbol']:
            return word
sp500news3['tickers'] = sp500news3['title'].apply(extract_ticker)
#sp500news3 sample:

  index date_publish    title   tickers
0   79944   2007-01-29 19:08:35 (MSFT, Vista, corporate, sales, go, very, well) None
1   181781  2007-12-14 19:39:06 (WMB, No, Anglican, consensus, on, Episcopal, Church)   None
2   213175  2008-01-22 11:17:19 (CSX, quarterly, profit, rises) None
3   93554   2008-01-22 18:52:56 (C, says, 30, bln, capital, helps, exceed, target)  None

成分["符号"]：样本

0      TWX  
1      C  
2      MSFT  
3      WMB ...

从以下位置复制空间文档：

constituents =  pd.DataFrame({"Symbol":["TWX","C","MSFT","WMB"]})
sp500news3 = pd.DataFrame({"title":["MSFT Vista corporate sales go very well","WMB No Anglican consensus on Episcopal Church","CSX quarterly profit rises",'C says 30 bln capital helps exceed target','TWX plans cable spinoff']})
import spacy
nlp = spacy.load('en_core_web_sm')
sp500news3['title'] = sp500news3['title'].apply(nlp)

Answer 1

你必须使用word.text，因为在迭代spacy.tokens.doc.Doc时，它会迭代Token，这不会为字符串实现__eq__：

for word in title:
    if word.text in constituents['Symbol'].values:
        return word

---

以您的示例为例：

In [11]: sp500news3['title'].apply(extract_ticker)
Out[11]:
0    MSFT
1     WMB
2    None
3       C
4     TWX
Name: title, dtype: object