python - Pandas 过滤 : Returning True/False vs the Actual Value

Question

我的数据框:

df_all_xml_mfiles_tgther

      file_names     searching_for                                 everything
0          a.txt             where              Dave Ran Away. Where is Dave?
1          a.txt             candy                                mmmm, candy
2          b.txt              time                We are looking for the book.
3          b.txt             where                   where the red fern grows

我的问题:

我正在尝试筛选包含在我的搜索条件中找到的词的记录。我需要一次检查 1 条记录并返回实际记录，而不仅仅是单词 true。

我尝试过的:

search_content_array = ['where', 'candy', 'time']
file_names_only = ['a.txt', 'b.txt']


for cc in range(0, len(file_names_only), 1):
     for bb in range(0, len(search_content_array), 1):

            stuff = `df_all_xml_mfiles_tgther[cc:cc+1].everything.str.contains(search_content_array[bb], flags=re.IGNORECASE, na=False, regex=True)`

            if not regex_stuff.empty:
                 regex_stuff_new = pd.DataFrame([regex_stuff.rename(None)])
                 regex_stuff_new.columns = ['everything']
                 regex_stuff_new['searched_for_found'] = search_content_array[bb]
                 regex_stuff_new['file_names'] = file_names_only[cc]

            regex_stuff_new = regex_stuff_new[['file_names', 'searched_for_found', 'everything']] ##This rearranges the columns

            df_regex_test =  df_regex_test.append(regex_stuff_new, ignore_index=True, sort=False)

我得到的结果是这样的:

    file_names  searched_for_found  everything
0        a.txt               where        True
1        a.txt               candy        True
2        b.txt               where        True

我想要的结果是这样的:

    file_names  searched_for_found                           everything
0        a.txt               where        Dave Ran Away. Where is Dave?
1        a.txt               candy                          mmmm, candy
3        b.txt               where             where the red fern grows

如何获取返回结果的实际值，而不仅仅是 true/false？

Answer 1

使用列表推导式按元素执行此操作。

df[[y.lower() in x.lower() for x, y in zip(df['everything'], df['searching_for'])]]

或者，

df[[y.lower() in x.lower() 
    for x, y in df[['everything', 'searching_for']].values.tolist()]]


  file_names searching_for                     everything
0      a.txt         where  Dave Ran Away. Where is Dave?
1      a.txt         candy                    mmmm, candy
3      b.txt         where       where the red fern grows