Pandas - 将具有开始和结束日期的数据框转换为每日数据

2022-01-11 00:00:00 pandas time-series date

问题描述

每个 ID 有一个记录，包括开始日期和结束日期

I have one record per ID with start date and end date

id  age state   start_date  end_date
123 18  CA     2/17/2019    5/4/2019
223 24  AZ     1/17/2019    3/4/2019

我想为开始日和结束日之间的每一天创建一条记录，以便将每日活动数据加入其中.目标输出看起来像这样

I want to create a record for each day between the start and end day, so I can join daily activity data to it. The target output would look something like this

id  age state   start_date
123 18  CA      2/17/2019
123 18  CA      2/18/2019
123 18  CA      2/19/2019
123 18  CA      2/20/2019
123 18  CA      2/21/2019
            …
123 18  CA      5/2/2019
123 18  CA      5/3/2019
123 18  CA      5/4/2019

当然，对数据集中的所有 id 及其各自的开始日期执行此操作.非常感谢任何帮助 - 谢谢！

And of course do this for all ids and their respective start dates in the dataset. Any help is much appreciated - thanks!

解决方案

melt, GroupBy, resample &填充

首先我们melt (unpivot) 你的两个日期列合二为一.然后我们resample 按天计算:

melt, GroupBy, resample & ffill

First we melt (unpivot) your two date columns to one. Then we resample on day basis:

melt = df.melt(id_vars=['id', 'age', 'state'], value_name='date').drop('variable', axis=1)
melt['date'] = pd.to_datetime(melt['date'])

melt = melt.groupby('id').apply(lambda x: x.set_index('date').resample('d').first())
           .ffill()
           .reset_index(level=1)
           .reset_index(drop=True)

输出

          date     id   age state
0   2019-02-17  123.0  18.0    CA
1   2019-02-18  123.0  18.0    CA
2   2019-02-19  123.0  18.0    CA
3   2019-02-20  123.0  18.0    CA
4   2019-02-21  123.0  18.0    CA
..         ...    ...   ...   ...
119 2019-02-28  223.0  24.0    AZ
120 2019-03-01  223.0  24.0    AZ
121 2019-03-02  223.0  24.0    AZ
122 2019-03-03  223.0  24.0    AZ
123 2019-03-04  223.0  24.0    AZ

[124 rows x 4 columns]

编辑:

我不得不在一个项目中重新审视这个问题，看起来像使用 DataFrame.apply 和 pd.date_range 和 DataFrame.explode 是快了近 3 倍:

I had to revisit this problem in a project, and looks like using DataFrame.apply with pd.date_range and DataFrame.explode is almost 3x faster:

df["date"] = df.apply(
    lambda x: pd.date_range(x["start_date"], x["end_date"]), axis=1
)
df = (
    df.explode("date", ignore_index=True)
    .drop(columns=["start_date", "end_date"])
)

输出

      id  age state       date
0    123   18    CA 2019-02-17
1    123   18    CA 2019-02-18
2    123   18    CA 2019-02-19
3    123   18    CA 2019-02-20
4    123   18    CA 2019-02-21
..   ...  ...   ...        ...
119  223   24    AZ 2019-02-28
120  223   24    AZ 2019-03-01
121  223   24    AZ 2019-03-02
122  223   24    AZ 2019-03-03
123  223   24    AZ 2019-03-04

[124 rows x 4 columns]

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