在日期 +/- 2 个工作日内对 pandas 时间序列进行切片

问题描述

具有以下时间序列:

In [65]: p
Out[65]: 
Date
2008-06-02    125.20
2008-06-03    124.47
2008-06-04    124.40
2008-06-05    126.89
2008-06-06    122.84
2008-06-09    123.14
2008-06-10    122.53
2008-06-11    120.73
2008-06-12    121.19
Name: SPY

如何在特定日期 +/- 2 个相邻(工作日)进行切片，即如果 d = '2008-06-06':

how can I slice on a specfic date +/- 2 neighbouring (business) days, so ie if d = '2008-06-06':

 -2   2008-06-04    124.40
 -1   2008-06-05    126.89
  0   2008-06-06    122.84
  1   2008-06-09    123.14
  2   2008-06-10    122.53

推荐答案

你可以使用索引方法get_loc，然后切片:

You could use the index method get_loc, and then slice:

d = pd.to_datetime('2008-06-06')
loc = s.index.get_loc(d)

In [12]: loc
Out[12]: 4

In [13]: s[loc-2:loc+3]
Out[13]: 
2008-06-04    124.40
2008-06-05    126.89
2008-06-06    122.84
2008-06-09    123.14
2008-06-10    122.53
Name: SPY

.

如果你只是在两天内对这些感兴趣:

In [14]: dt = datetime.timedelta(1)

In [15]: s[d - 2*dt:d + 2*dt]
Out[15]: 
2008-06-04    124.40
2008-06-05    126.89
2008-06-06    122.84
Name: SPY

这篇关于在日期 +/- 2 个工作日内对 pandas 时间序列进行切片的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！