以天为单位的 Pandas Timedelta [英] Pandas Timedelta in Days

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本文介绍了以天为单位的 Pandas Timedelta的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我在 Pandas 中有一个名为munged_data"的数据框,其中包含两列entry_date"和dob",我已使用 pd.to_timestamp 将其转换为时间戳.我想弄清楚如何根据时间计算人的年龄'entry_date' 和 'dob' 之间的差异,为此我需要得到两列之间的天数差异(这样我就可以做一些像 round(days/365.25) 这样的事情.我似乎无法找到一种使用矢量化操作执行此操作的方法.当我执行 munged_data.entry_date-munged_data.dob 时,我得到以下信息:

I have a dataframe in pandas called 'munged_data' with two columns 'entry_date' and 'dob' which i have converted to Timestamps using pd.to_timestamp.I am trying to figure out how to calculate ages of people based on the time difference between 'entry_date' and 'dob' and to do this i need to get the difference in days between the two columns ( so that i can then do somehting like round(days/365.25). I do not seem to be able to find a way to do this using a vectorized operation. When I do munged_data.entry_date-munged_data.dob i get the following :

internal_quote_id
2                    15685977 days, 23:54:30.457856
3                    11651985 days, 23:49:15.359744
4                     9491988 days, 23:39:55.621376
7                     11907004 days, 0:10:30.196224
9                    15282164 days, 23:30:30.196224
15                  15282227 days, 23:50:40.261632  

但是我似乎无法将天数提取为整数,以便我可以继续计算.任何帮助表示赞赏.

However i do not seem to be able to extract the days as an integer so that i can continue with my calculation. Any help appreciated.

推荐答案

为此你需要 0.11(0.11rc1 已发布,下周最终测试)

You need 0.11 for this (0.11rc1 is out, final prob next week)

In [9]: df = DataFrame([ Timestamp('20010101'), Timestamp('20040601') ])

In [10]: df
Out[10]: 
                    0
0 2001-01-01 00:00:00
1 2004-06-01 00:00:00

In [11]: df = DataFrame([ Timestamp('20010101'), 
                          Timestamp('20040601') ],columns=['age'])

In [12]: df
Out[12]: 
                  age
0 2001-01-01 00:00:00
1 2004-06-01 00:00:00

In [13]: df['today'] = Timestamp('20130419')

In [14]: df['diff'] = df['today']-df['age']

In [16]: df['years'] = df['diff'].apply(lambda x: float(x.item().days)/365)

In [17]: df
Out[17]: 
                  age               today                diff      years
0 2001-01-01 00:00:00 2013-04-19 00:00:00 4491 days, 00:00:00  12.304110
1 2004-06-01 00:00:00 2013-04-19 00:00:00 3244 days, 00:00:00   8.887671

你最后需要这个奇怪的应用程序,因为还没有完全支持 timedelta64[ns] 标量(例如,我们现在如何将时间戳用于 datetime64[ns],在 0.12 中)

You need this odd apply at the end because not yet full support for timedelta64[ns] scalars (e.g. like how we use Timestamps now for datetime64[ns], coming in 0.12)

这篇关于以天为单位的 Pandas Timedelta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IT屋!

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