python - 如何在分钟内转换(可能为负)Pandas TimeDelta( float )？

Question

我有这样一个数据框

df[['timestamp_utc','minute_ts','delta']].head()
Out[47]: 
            timestamp_utc           minute_ts                    delta
0 2015-05-21 14:06:33.414 2015-05-21 12:06:00 -1 days +21:59:26.586000
1 2015-05-21 14:06:33.414 2015-05-21 12:07:00 -1 days +22:00:26.586000
2 2015-05-21 14:06:33.414 2015-05-21 12:08:00 -1 days +22:01:26.586000
3 2015-05-21 14:06:33.414 2015-05-21 12:09:00 -1 days +22:02:26.586000
4 2015-05-21 14:06:33.414 2015-05-21 12:10:00 -1 days +22:03:26.586000

其中 df['delta']=df.minute_ts-df.timestamp_utc

timestamp_utc     datetime64[ns]
minute_ts         datetime64[ns]
delta            timedelta64[ns]

问题是，我想获取 timestamp_utc 和 minutes_ts 之间的分钟数(可能为负数)，忽略秒部分。

所以对于第一行，我想得到 -120。实际上，2015-05-21 12:06:00 比 2015-05-21 14:06:33.414 早 120 分钟。

最 Pandas 式的做法是什么？

非常感谢!

Answer 1

您可以使用:

df['a'] = df['delta'] / np.timedelta64(1, 'm')
print (df)
            timestamp_utc           minute_ts                    delta  \
0 2015-05-21 14:06:33.414 2015-05-21 12:06:00 -1 days +21:59:26.586000   
1 2015-05-21 14:06:33.414 2015-05-21 12:07:00 -1 days +22:00:26.586000   
2 2015-05-21 14:06:33.414 2015-05-21 12:08:00 -1 days +22:01:26.586000   
3 2015-05-21 14:06:33.414 2015-05-21 12:09:00 -1 days +22:02:26.586000   
4 2015-05-21 14:06:33.414 2015-05-21 12:10:00 -1 days +22:03:26.586000   

          a  
0 -120.5569  
1 -119.5569  
2 -118.5569  
3 -117.5569  
4 -116.5569  

然后将float转换为int:

df['a'] = (df['delta'] / np.timedelta64(1, 'm')).astype(int)
print (df)
            timestamp_utc           minute_ts                    delta    a
0 2015-05-21 14:06:33.414 2015-05-21 12:06:00 -1 days +21:59:26.586000 -120
1 2015-05-21 14:06:33.414 2015-05-21 12:07:00 -1 days +22:00:26.586000 -119
2 2015-05-21 14:06:33.414 2015-05-21 12:08:00 -1 days +22:01:26.586000 -118
3 2015-05-21 14:06:33.414 2015-05-21 12:09:00 -1 days +22:02:26.586000 -117
4 2015-05-21 14:06:33.414 2015-05-21 12:10:00 -1 days +22:03:26.586000 -116