python - 列表元素与 pandas 列的关键字匹配

Question

我有如下元素列表:

 A=  ['loans','s-class','veyron','trump','rihana','drake','election']

我还有另一个 pandas 数据框 B，其中包含列 category 和 words，它是逗号分隔的字符串:-

category              words
audi                  a4, a6
bugatti               veyron, chiron
mercedez              s-class, e-class
dslr                  canon, nikon
apple                 iphone,macbook,ipod
finance               sales,loans,sales price
politics              trump, election, votes
entertainment         spiderman,thor, ironmen
music                 beiber, rihana,drake
........              ..............
.........             .........

我只想将列表 A 的元素映射到列 words 并将相应的 category 分配到一个新列表中。因此，预期输出将是。

matched_categories=['finance','mercedez','bugatti','politics','music','music','politics']

Answer 1

按 boolean indexing 过滤与 iat选择第一个匹配值:

#if always matched all values
matched_categories = [df.loc[df['words'].str.contains(x), 'category'].iat[0] for x in A]
print (matched_categories)
['finance', 'mercedez', 'bugatti', 'politics', 'music', 'music', 'politics']

如果某些值不匹配，则更通用的解决方案 - 然后返回 不匹配 值:

#added last aaa value
A = ['loans','s-class','veyron','trump','rihana','drake','election','aaa']

matched_categories = [next(iter(df.loc[df['words'].str.contains(x),'category']),'not matched')
                      for x in A]
print (matched_categories)
['finance', 'mercedez', 'bugatti', 'politics', 'music', 'music', 'politics', 'not matched']