列表元素与 pandas 列的关键字匹配

Question

我有元素列表：

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

我也有另一只大熊猫数据框B与列category和words是逗号分隔字符串： -

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A用列映射列表元素words并将相应的分配category到新列表中。所以，预期的输出是。

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

Answer 1

通过过滤器boolean indexing与iat用于选择第一个匹配的值：

#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not matched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']