pandas 如何去除日期、小时和秒的熊猫日期时间
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How to strip a pandas datetime of date, hours and seconds
提问by Josh Kidd
How do I remove Date, Hours and Seconds from a pandas datetime, so that I'm left with only the minutes? I have a table of dates in the format:
如何从Pandas日期时间中删除日期、小时和秒，以便我只剩下分钟？我有一个格式的日期表：
Date
2015-04-18 23:33:58
2015-04-19 14:32:08
2015-04-20 18:42:44
2015-04-20 21:41:19
and I want:
而且我要：
Date
33
32
42
41
The code I'm trying to use is:
我尝试使用的代码是：
fiveMin['Date'] = fiveMin['Date'] - pd.Timedelta(fiveMin['Date'], unit='s')
to remove the seconds, however I'm getting the error:
删除秒，但是我收到错误：
Value must be Timedelta, string, integer, float, timedelta or convertible
回答by EdChum
If the dtype is already datetimeyou can use dtaccessor to return just the minute attribute:
如果 dtype 已经是datetime你可以使用dt访问器只返回分钟属性：
In [43]:
df['Date'].dt.minute
Out[43]:
0 33
1 32
2 42
3 41
Name: Date, dtype: int64
If needed convert using to_datetime:
如果需要转换使用to_datetime：
df['Date'] = pd.to_datetime(df['Date'])`
If the dates are strings and are well formed you could just split on :and extract the second to last split:
如果日期是字符串并且格式正确，您可以拆分:并提取倒数第二个拆分：
In [46]:
df['Date'].str.split(':').str[-2]
Out[46]:
0 33
1 32
2 42
3 41
Name: Date, dtype: object
This returns a string series however, if you want ints then you can cast to int using astype(int):
但是，这将返回一个字符串系列，如果您想要整数，则可以使用astype(int)以下方法转换为 int ：
In [47]:
(df['Date'].str.split(':').str[-2]).astype(int)
Out[47]:
0 33
1 32
2 42
3 41
Name: Date, dtype: int32
