Expanding pandas data frame with date range in columns

Question

I have a pandas dataframe with dates and strings similar to this:

Start        End           Note    Item
2016-10-22   2016-11-05    Z       A
2017-02-11   2         


        

Answer 1

You can iterate over each row and create a new dataframe and then concatenate them together

pd.concat([pd.DataFrame({'Start': pd.date_range(row.Start, row.End, freq='W-SAT'),
               'Note': row.Note,
               'Item': row.Item}, columns=['Start', 'Note', 'Item']) 
           for i, row in df.iterrows()], ignore_index=True)

       Start Note Item
0 2016-10-22    Z    A
1 2016-10-29    Z    A
2 2016-11-05    Z    A
3 2017-02-11    W    B
4 2017-02-18    W    B
5 2017-02-25    W    B

Answer 2

You don't need iteration at all.

df_start_end = df.melt(id_vars=['Note','Item'],value_name='date')

df = df_start_end.groupby('Note').apply(lambda x: x.set_index('date').resample('W').pad()).drop(columns=['Note','variable']).reset_index()

Answer 3

If the number of unique values of df['End'] - df['Start'] is not too large, but the number of rows in your dataset is large, then the following function will be much faster than looping over your dataset:

def date_expander(dataframe: pd.DataFrame,
                  start_dt_colname: str,
                  end_dt_colname: str,
                  time_unit: str,
                  new_colname: str,
                  end_inclusive: bool) -> pd.DataFrame:
    td = pd.Timedelta(1, time_unit)

    # add a timediff column:
    dataframe['_dt_diff'] = dataframe[end_dt_colname] - dataframe[start_dt_colname]

    # get the maximum timediff:
    max_diff = int((dataframe['_dt_diff'] / td).max())

    # for each possible timediff, get the intermediate time-differences:
    df_diffs = pd.concat([pd.DataFrame({'_to_add': np.arange(0, dt_diff + end_inclusive) * td}).assign(_dt_diff=dt_diff * td)
                          for dt_diff in range(max_diff + 1)])

    # join to the original dataframe
    data_expanded = dataframe.merge(df_diffs, on='_dt_diff')

    # the new dt column is just start plus the intermediate diffs:
    data_expanded[new_colname] = data_expanded[start_dt_colname] + data_expanded['_to_add']

    # remove start-end cols, as well as temp cols used for calculations:
    to_drop = [start_dt_colname, end_dt_colname, '_to_add', '_dt_diff']
    if new_colname in to_drop:
        to_drop.remove(new_colname)
    data_expanded = data_expanded.drop(columns=to_drop)

    # don't modify dataframe in place:
    del dataframe['_dt_diff']

    return data_expanded