熊猫列的列表以分隔行

Question

I've got a dataframe that contains analysed news articles w/ each row referencing an article and columns w/ some information about that article (eg tone). One column of that df contains a list of FIPS country codes of the locations that were mentioned in that article.

I want to "extract" these country codes such that I get a dataframe in which each mentioned location has its own row, along with the other columns of the original row in which that location was referenced (there will be multiple rows with the same information, but different locations, as the same article may mention multiple locations).

I tried something like this, but iterrows() is notoriously slow, so is there any faster/more efficient way for me to do this? Thanks a lot.

• 'events' is the column that contains the locations
• 'event_cols' are the columns from the original df that I want to retain in the new df.
• 'df_events' is the new data frame

for i, row in df.iterrows():
  for location in df.events.loc[i]:
    try:
        df_storage = pd.DataFrame(row[event_cols]).T
        df_storage['loc'] = location 
        df_events = df_events.append(df_storage)
    except ValueError as e:
        continue

Answer 1

I would group the DataFrame with groupby() , explode the lists with a combination of apply and a lambda function, and then reset the index and drop the level column that is created to clean up the resulting DataFrame .

df_events = df.groupby(['event_col1', 'event_col2', 'event_col3'])['events']\
                 .apply(lambda x: pd.DataFrame(x.values[0]))\
                 .reset_index().drop('level_3', axis = 1)

In general, I always try to find a way to use apply() before most other methods, because it is often much faster than iterating over each row.