如何为第 "start_date" 列和 "end_date" 列的每一行创建 pandas.date_range()？

Question

我有一个像这样的 df:

id | start_date | end_date  | price
1  | 2020-10-01 | 2020-10-3 | 1
1  | 2020-10-03 | 2020-10-4 | 1
2  | 2020-10-04 | 2020-10-6 | 2
3  | 2020-10-05 | 2020-10-5 | 3

列“start_date”和“end_date”是 datetime64[ns]。

我想根据日期范围创建一个“日期”列。

最简单的方法是创建一个 pandas.date_range(start_date, end_date, freq="D")，然后使用 .explode()。

最终结果应如下所示：

id | start_date | end_date  | price | date
1  | 2020-10-01 | 2020-10-3 | 1     | 2020-10-01
1  | 2020-10-01 | 2020-10-3 | 1     | 2020-10-02
1  | 2020-10-01 | 2020-10-3 | 1     | 2020-10-03
1  | 2020-10-03 | 2020-10-4 | 1     | 2020-10-03
1  | 2020-10-03 | 2020-10-4 | 1     | 2020-10-04
2  | 2020-10-04 | 2020-10-6 | 2     | 2020-10-04
2  | 2020-10-04 | 2020-10-6 | 2     | 2020-10-05
2  | 2020-10-04 | 2020-10-6 | 2     | 2020-10-06
3  | 2020-10-05 | 2020-10-5 | 3     | 2020-10-05

到目前为止尝试过：

df["daterange"] = pd.date_range(df["start_date"], df["end_date"])

TypeError: Cannot convert input [0 2020-10-01 1 2020-10-01

for row in df.itertuples():
    df["daterange"] = pd.date_range(start=row.start_date, end=row.end_date)

ValueError: Length of values (3) does not match length of index (9)

Lambdas、apply、melt 等对于我的数据帧大小来说太慢了，无法使用！

/编辑

目前发现的Fastet方法：

https://github.com/Garve/scikit-bonus
skbonus.pandas.preprocessing.DateTimeExploder(
        "date",
        start_column="start_date",
        end_column="end_date",
        frequency="d",
        drop=False,
    )

Answer 1

使用DataFrame.apply:

df["daterange"] = df.apply(lambda x: pd.date_range(x.start_date, x.end_date), axis=1)
df = df.explode('daterange').reset_index(drop=True)
print (df)
   id  start_date   end_date  price  daterange
0   1  2020-10-01  2020-10-3      1 2020-10-01
1   1  2020-10-01  2020-10-3      1 2020-10-02
2   1  2020-10-01  2020-10-3      1 2020-10-03
3   1  2020-10-03  2020-10-4      1 2020-10-03
4   1  2020-10-03  2020-10-4      1 2020-10-04
5   2  2020-10-04  2020-10-6      2 2020-10-04
6   2  2020-10-04  2020-10-6      2 2020-10-05
7   2  2020-10-04  2020-10-6      2 2020-10-06
8   3  2020-10-05  2020-10-5      3 2020-10-05

选择：

s = pd.concat([pd.Series(r.Index,pd.date_range(r.start_date, r.end_date)) for r in df.itertuples()])
s = pd.Series(s.index, s)

df = df.join(s.rename('daterange')).reset_index(drop=True)
print (df)
   id  start_date   end_date  price  daterange
0   1  2020-10-01  2020-10-3      1 2020-10-01
1   1  2020-10-01  2020-10-3      1 2020-10-02
2   1  2020-10-01  2020-10-3      1 2020-10-03
3   1  2020-10-03  2020-10-4      1 2020-10-03
4   1  2020-10-03  2020-10-4      1 2020-10-04
5   2  2020-10-04  2020-10-6      2 2020-10-04
6   2  2020-10-04  2020-10-6      2 2020-10-05
7   2  2020-10-04  2020-10-6      2 2020-10-06
8   3  2020-10-05  2020-10-5      3 2020-10-05

Answer 2

目前发现的禁食方法：

https://github.com/Garve/scikit-bonus
from skbonus.pandas.preprocessing import DateTimeExploder
df = DateTimeExploder(
        "date",
        start_column="start_date",
        end_column="end_date",
        frequency="d",
        drop=False,
    )

如何为第 "start_date" 列和 "end_date" 列的每一行创建 pandas.date_range()？

How to create pandas.date_range() for each row from column "start_date" and column "end_date"?

pandas

date

date-range