将df的日期范围扩展为每天一行 [英] Expand df with range of dates to one row per day

问题描述

我有一个df，每个项目包含一行，并带有日期范围，我需要将其扩展为每个项目每天包含一行.

I have a df that contains one line per item with a range of dates, and I need to expand it to contain one row per day per item.

它看起来像这样:

  from       to         id
1 25/02/2019 27/02/2019 A
2 15/07/2019 16/07/2019 B

我想要这个:

  date       id
1 25/02/2019 A
2 26/07/2019 A
3 27/07/2019 A
4 15/07/2019 B
5 16/07/2019 B

我设法编写了一个可以运行的代码，但是运行需要一个多小时，所以我想知道是否有更有效的方法来实现它.

I managed to write a code that works but it takes over one hour to run, so I am wondering if there is a more efficient way to do it.

我的代码:

df_dates = pd.DataFrame()

for i in range(len(df)):

    start = df.loc[i]['from']
    end = df.loc[i]['to'] + np.timedelta64(1,'D') #includes last day of the range
    dates = np.arange(start, end, dtype='datetime64[D]')

    temp = pd.DataFrame()
    temp = temp.append([df.loc[i]]*len(dates), ignore_index=True)
    temp['datadate'] = dates

    df_dates = df_dates.append(temp, ignore_index=True)

这花费了很长时间，因为真实范围大约有50年，包含1700多个项目，因此新的df非常庞大，但是也许您知道可以更快地完成相同操作的窍门:)

It takes long because the real ranges are of about 50 years with over 1700 items so the new df is massive, but maybe you know a trick to do the same faster :)

推荐答案

尝试:

df['from'] = pd.to_datetime(df['from'])
df['to'] = pd.to_datetime(df['to'])
pd.concat([pd.DataFrame({'date': pd.date_range(row['from'], row['to'], freq='D'), 'id': row['id']})
           for i, row in df.iterrows()], ignore_index=True)
        date id
0 2019-02-25  A
1 2019-02-26  A
2 2019-02-27  A
3 2019-07-15  B
4 2019-07-16  B

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