python - 使用字符串列表过滤数据框

Question

我有以下数据框，称为 pop_df:

pop_df
Out[33]: 
            2014       2015       2016       2017       2018       2019
Geo                                                                    
AL     2892394.0  2885796.0  2875592.0  2876591.0  2870324.0  2862427.0
AL0    2892394.0  2885796.0  2875592.0  2876591.0  2870324.0  2862427.0
AL01    844921.0   836448.0   830981.0   826904.0   819793.0   813758.0
AL011   134332.0   131054.0   129056.0   125579.0   120978.0   118948.0
AL012   276058.0   277989.0   280205.0   284823.0   289626.0   290126.0
         ...        ...        ...        ...        ...        ...
UKN12   142028.0   142756.0   143363.0   143746.0   144105.0   144367.0
UKN13   139774.0   140222.0   140752.0   141368.0   141994.0   142565.0
UKN14   137722.0   139426.0   140691.0   141917.0   143286.0   144771.0
UKN15   136332.0   136904.0   137492.0   138000.0   138441.0   138948.0
UKN16   114696.0   115171.0   115581.0   116057.0   116612.0   117051.0

[2034 rows x 6 columns]

我还有一个欧盟国家代码列表:

EuropeanUnion=["BE","BG","CZ","DK","DE","EE","IE","EL","ES","FR","HR","IT","CY","LV","LT","LU","HU","MT","NL","AT","PL","PT","RO","SI","SK","FI","SE"]

我想过滤数据框以仅保留列表中国家代码开始的行。我尝试使用 pop_df.index.str.startswith 但无法正常工作。如果您能提供任何帮助，我将不胜感激。提前致谢!

Answer 1

Geo 似乎是索引，所以你可以这样做:

result = df[df.index.str.match(rf'\b{"|".join(EuropeanUnion)}')]

输出 (虚拟)

           2014      2015      2016      2017      2018      2019
Geo                                                              
BE011  134332.0  131054.0  129056.0  125579.0  120978.0  118948.0
DE13   139774.0  140222.0  140752.0  141368.0  141994.0  142565.0

来自 str.match 上的文档:

Determine if each string starts with a match of a regular expression.

表达式 rf'\b{"|".join(EuropeanUnion)}' 构建了一个匹配任何国家代码的正则表达式模式。