如何从日期时间索引中每月切出第 n 天？

Question

我有一个如下所示的索引：

DatetimeIndex(['2005-01-03', '2005-01-04', '2005-01-05', '2005-01-07',
               '2005-01-10', '2005-01-11', '2005-01-12', '2005-01-13',
               '2005-01-14', '2005-01-17',
               ...
               '2021-11-18', '2021-11-19', '2021-11-22', '2021-11-23',
               '2021-11-24', '2021-11-25', '2021-11-26', '2021-11-29',
               '2021-11-30', '2021-12-01'],

在索引中为每个月分割第 n 天的最佳方法是什么？如果第 n 天在一个月内不存在，则应包括索引中的下一天。

Answer 1

IIUC，计算每一天与第N天的差值后使用resample：

设置一个MRE:

dti = pd.date_range('2021', '2022', freq='6D')
df = pd.DataFrame({'A': np.random.random(len(dti))}, index=dti)
print(df)

# Output:
                   A
2021-01-01  0.771107
2021-01-07  0.513512
2021-01-13  0.613989
2021-01-19  0.842237
2021-01-25  0.056986
...              ...
2021-12-03  0.002022
2021-12-09  0.943162
2021-12-15  0.665152
2021-12-21  0.350673
2021-12-27  0.109381

[61 rows x 1 columns]

Select 右行

# N = 12
>>> df.loc[df.assign(diff=abs(N - df.index.day)).resample('M')['diff'].idxmin().values]

                   A
2021-01-13  0.613989  # nearest day
2021-02-12  0.847639  # day N
2021-03-14  0.356432  # nearest day
2021-04-13  0.279536  # nearest day
2021-05-13  0.628562  # nearest day
2021-06-12  0.859800  # day N
2021-07-12  0.112848  # day N
2021-08-11  0.933266  # nearest day
2021-09-10  0.484271  # nearest day
2021-10-10  0.289343  # nearest day
2021-11-09  0.963057  # nearest day
2021-12-09  0.943162  # nearest day

如何从日期时间索引中每月切出第 n 天？

How to slice day n each month from datetime index?

python

datetime

slice

pandas