将时间序列数据分成一行时间间隔(PythonicWay)-每小时 [英] Split Time Series Data Into Time Intervals in one line (PythonicWay) - Hourly

问题描述

我有一个带有时间列的分钟数据.我想仅使用日期时间格式的hours创建一个新列，例如format ='%Y-%m-%d %H:%M:%S'.我知道在R中，我们可以使用类似

I have a minute data that has the time column. I want to create new column with just hours with date time format, for example format ='%Y-%m-%d %H:%M:%S'. I know in R, we can use something like,

value$hour<- cut(as.POSIXct(paste(value$time),
                          format="%Y-%m-%d %H:%M:%S"), breaks="hour")

执行此操作时，得到以下输出(我需要)

When I do this, I get the following output, (which i need)

time                 hour
2017-02-10 00:00:00  2017-02-10 00:00:00
2017-02-10 00:01:00  2017-02-10 00:00:00
2017-02-10 00:02:00  2017-02-10 00:00:00
2017-02-10 00:03:00  2017-02-10 00:00:00
....
2017-12-1 10:05:00   2017-12-01 10:00:00
2017-12-1 10:06:00   2017-12-01 10:00:00

我也意识到有很多关于dt.date，dt.hour等的讨论.我可以在python中像这样进行以下操作，

I am also aware that there are many threads that discusses about dt.date, dt.hour etc. I can do the following in python like this,

value['date'] = value['time'].dt.date
value['hour'] = value['time'].dt.hour

我是否可以在python中做任何与上面提到的R类似的方法? 任何想法将不胜感激.预先感谢！

Is there any way that I can do in python that is similar to R as mentioned above in one line? Any thoughts would be appreciated. Thanks in advance!

推荐答案

您需要如果需要转换为datetime列time，请添加 to_datetime :

If need convert to datetime column time add to_datetime:

df['hour'] = pd.to_datetime(df['time']).dt.floor('H')
print (df)
                  time                hour
0  2017-02-10 00:00:00 2017-02-10 00:00:00
1  2017-02-10 00:01:00 2017-02-10 00:00:00
2  2017-02-10 00:02:00 2017-02-10 00:00:00
3  2017-02-10 00:03:00 2017-02-10 00:00:00
4   2017-12-1 10:05:00 2017-12-01 10:00:00
5   2017-12-1 10:06:00 2017-12-01 10:00:00

这篇关于将时间序列数据分成一行时间间隔(PythonicWay)-每小时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！