我有如下元素列表：
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
........ ..............
......... .........
所有我要地图列表的元素A与列words，并指定相应的category进入一个新的list.So，预计产量会。
matched_categories=['finance','mercedez','bugatti','politics','music','music','politics']
用筛选boolean indexing以iat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，则为更通用的解决方案-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']
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