我有一个数据框,有两列不同的字符串格式的时间,我想找到两列之间的差异,所以我使用下面的代码
operational_data_clean['Pick/pack start-time'] = pd.to_datetime(operational_data_clean['Pick/pack start-time'])
operational_data_clean['Flight launched-time'] = pd.to_datetime(operational_data_clean['Flight launched-time'])
operational_data_clean['time_to_launch'] = 0
operational_data_clean['time_to_launch'] = operational_data_clean['Flight launched-time'] - operational_data_clean['Pick/pack start-time']
但是当我第一次运行时,这段代码我得到了很好的结果,但是当我第二次运行时,它会在“Pick / pack start-time”和“Flight launch-time”值上添加今天的日期。
如何将此时间仅转换为数小时而不会使日期变得混乱我的数据。
我假设您使用jupyter笔记本运行代码。
执行代码时,变量operational_data_clean['Pick/pack start-time']
变为pd.to_datetime(operational_data_clean['Pick/pack start-time'])
。
因此,当您再次执行该块时,jupyter会记住您的变量,因此将再次执行相同的功能,基本上这样做:pd.to_datetime(pd.to_datetime(operational_data_clean['Pick/pack start-time']))
。
至于pd.to_datetime()
本身,我建议透过大熊猫docs。