包含单词列表的列的单词分数总和

Question

我有一个单词栏：

> print(df['words'])
0       [awww, thats, bummer, shoulda, got, david, car...   
1       [upset, that, he, cant, update, his, facebook,...   
2       [dived, many, time, ball, managed, save, rest,...   
3       [whole, body, feel, itchy, like, it, on, fire]   
4       [no, it, not, behaving, at, all, im, mad, why,...   
5       [not, whole, crew]

另一个情感栏是每个单词的“情感”值：

> print(sentiment) 
           abandon  -2
0        abandoned  -2
1         abandons  -2
2         abducted  -2
3        abduction  -2
4       abductions  -2
5            abhor  -3
6         abhorred  -3
7        abhorrent  -3
8           abhors  -3
9        abilities   2
...

对于df['words']中的每一行单词，我想总结它们各自的情感值。对于感情中不存在的词语，等于0。你知道吗

到目前为止，我的情况是：

df['sentiment_value'] = Sum(df['words'].apply(lambda x: ''.join(x+x for x in sentiment))

预期结果

print(df['sentiment_value'])
0        -5   
1         2   
2        15  
3        -6   
4        -8   
...

Answer 1

如果将分数作为一个系列，以单词作为标签：

In [11]: s  # e.g. sentiment.set_index("word")["score"]
Out[11]:
abandon     -2
abandoned   -2
abandons    -2
abducted    -2
abduction   -2
Name: score, dtype: int64

然后您可以查看列表的分数：

In [12]: s.loc[["abandon", "abducted"]].sum()
Out[12]: -4

所以应用的方法是：

df['words'].apply(lambda ls: s.loc[ls])

---

如果需要支持缺少的单词（不在s中），可以使用reindex:

In [21]: s.reindex(["abandon", "abducted", "missing_word"]).sum()
Out[21]: -4.0

df['words'].apply(lambda ls: s.reindex(ls))