从每日中提取每周数据，并使用 Pandas 将其从长格式重塑为宽格式

Question

给定如下示例数据，我希望每周提取一个数据条目，如果该周有多个条目，那么我将使用该周最大的工作日数据：

          date  variable      value
0    2020-11-4  quantity   564.0000
1   2020-11-11  quantity   565.0000
2   2020-11-18  quantity   566.0000
3   2020-11-25  quantity   566.0000
4    2020-11-2     price  1829.1039
5    2020-11-3     price  1789.5883
6    2020-11-4     price  1755.4307
7    2020-11-5     price  1750.0727
8    2020-11-6     price  1746.7239
9    2020-11-9     price  1756.1005
10  2020-11-10     price  1752.0820
11  2020-11-11     price  1814.3693
12  2020-11-12     price  1833.7922
13  2020-11-13     price  1833.7922
14  2020-11-16     price  1784.2302
15  2020-11-17     price  1764.1376
16  2020-11-18     price  1770.1654
17  2020-11-19     price  1757.4400
18  2020-11-20     price  1770.1654

要获取每个日期的周数，我使用df['week_number'] = pd.to_datetime(df['date']).dt.week。

          date  variable      value  week_number
0    2020-11-4  quantity   564.0000           45  --> to keep
1   2020-11-11  quantity   565.0000           46  --> to keep
2   2020-11-18  quantity   566.0000           47  --> to keep
3   2020-11-25  quantity   566.0000           48  --> to keep
4    2020-11-2     price  1829.1039           45
5    2020-11-3     price  1789.5883           45
6    2020-11-4     price  1755.4307           45
7    2020-11-5     price  1750.0727           45
8    2020-11-6     price  1746.7239           45  --> to keep, since it's the largest weekday for this week
9    2020-11-9     price  1756.1005           46
10  2020-11-10     price  1752.0820           46
11  2020-11-11     price  1814.3693           46
12  2020-11-12     price  1833.7922           46
13  2020-11-13     price  1833.7922           46  --> to keep, since it's the largest weekday for this week
14  2020-11-16     price  1784.2302           47
15  2020-11-17     price  1764.1376           47
16  2020-11-18     price  1770.1654           47
17  2020-11-19     price  1757.4400           47
18  2020-11-20     price  1770.1654           47  --> to keep, since it's the largest weekday for this week

最后，我将指示to_keep 的行重新整形为预期结果，如下所示：

   variable  the_45th_week  the_46th_week  the_47th_week  the_48th_week
0  quantity       564.0000       565.0000       566.0000          566.0
1     price      1756.1005      1833.7922      1770.1654            NaN

如何操作数据以获得预期结果？真诚的感谢。

编辑：

df = df.sort_values(by=['variable','date'], ascending=False)
df.drop_duplicates(['variable', 'week_number'], keep='last')

输出：

          date  variable      value  week_number
0    2020-11-4  quantity   564.0000           45
3   2020-11-25  quantity   566.0000           48
2   2020-11-18  quantity   566.0000           47
1   2020-11-11  quantity   565.0000           46
4    2020-11-2     price  1829.1039           45
14  2020-11-16     price  1784.2302           47
10  2020-11-10     price  1752.0820           46

Answer 1

在您的解决方案中可以添加pivot 和rename：

df['week_number'] = pd.to_datetime(df['date']).dt.week
df = df.sort_values(by=['variable','date'], ascending=False)
df = df.drop_duplicates(['variable', 'week_number'], keep='last')

f = lambda x: f'the_{x}th_week'
out = df.pivot('variable','week_number','value').rename(columns=f)
print(out)
week_number  the_45th_week  the_46th_week  the_47th_week  the_48th_week
variable                                                               
price            1829.1039       1752.082      1784.2302            NaN
quantity          564.0000        565.000       566.0000          566.0

或者删除DataFrame.drop_duplicates，所以可以将DataFrame.pivot_table与聚合函数last一起使用：

df['week_number'] = pd.to_datetime(df['date']).dt.week
df = df.sort_values(by=['variable','date'], ascending=False)

f = lambda x: f'the_{x}th_week'
out = df.pivot_table(index='variable',columns='week_number',values='value', aggfunc='last').rename(columns=f)

编辑：获得与预期完全相同的结果：

out.reset_index().rename_axis(None, axis=1)

输出：

   variable  the_45th_week  the_46th_week  the_47th_week  the_48th_week
0     price      1829.1039       1752.082      1784.2302            NaN
1  quantity       564.0000        565.000       566.0000          566.0