将“TimeStamp”列截断为pandas`DataFrame`中的小时精度

Question

I have a pandas.DataFrame called df which has an automatically generated index, with a column dt :

df['dt'].dtype, df['dt'][0]
# (dtype('<M8[ns]'), Timestamp('2014-10-01 10:02:45'))

What I'd like to do is create a new column truncated to hour precision. I'm currently using:

df['dt2'] = df['dt'].apply(lambda L: datetime(L.year, L.month, L.day, L.hour))

This works, so that's fine. However, I've an inkling there's some nice way using pandas.tseries.offsets or creating a DatetimeIndex or similar.

So if possible, is there some pandas wizardry to do this?

Answer 1

In pandas 0.18.0 and later, there are datetime floor , ceil and round methods to round timestamps to a given fixed precision/frequency. To round down to hour precision, you can use:

>>> df['dt2'] = df['dt'].dt.floor('h')
>>> df
                      dt                     dt2
0    2014-10-01 10:02:45     2014-10-01 10:00:00
1    2014-10-01 13:08:17     2014-10-01 13:00:00
2    2014-10-01 17:39:24     2014-10-01 17:00:00

---

Here's another alternative to truncate the timestamps. Unlike floor , it supports truncating to a precision such as year or month.

You can temporarily adjust the precision unit of the underlying NumPy datetime64 datatype, changing it from [ns] to [h] :

df['dt'].values.astype('<M8[h]')

This truncates everything to hour precision. For example:

>>> df
                       dt
0     2014-10-01 10:02:45
1     2014-10-01 13:08:17
2     2014-10-01 17:39:24

>>> df['dt2'] = df['dt'].values.astype('<M8[h]')
>>> df
                      dt                     dt2
0    2014-10-01 10:02:45     2014-10-01 10:00:00
1    2014-10-01 13:08:17     2014-10-01 13:00:00
2    2014-10-01 17:39:24     2014-10-01 17:00:00

>>> df.dtypes
dt     datetime64[ns]
dt2    datetime64[ns]

The same method should work for any other unit: months 'M' , minutes 'm' , and so on:

• Keep up to year: '<M8[Y]'
• Keep up to month: '<M8[M]'
• Keep up to day: '<M8[D]'
• Keep up to minute: '<M8[m]'
• Keep up to second: '<M8[s]'

Answer 2

我过去用来实现这个目标的方法如下（与你已经在做的非常相似，但我想我还是把它扔出去了）：

df['dt2'] = df['dt'].apply(lambda x: x.replace(minute=0, second=0))