Python 如何将字符串中的时间转换为日期时间格式,以便获得两列之间的差异
我有一个数据帧,它有两列,时间不同,是字符串格式,我想找出这两列之间的差异,所以我使用下面的代码Python 如何将字符串中的时间转换为日期时间格式,以便获得两列之间的差异,python,pandas,datetime,Python,Pandas,Datetime,我有一个数据帧,它有两列,时间不同,是字符串格式,我想找出这两列之间的差异,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['
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']
但这段代码在我第一次运行时得到了很好的结果,但在我第二次运行时,它在“拾取/打包开始时间”和“航班启动时间”值上添加了今天的日期
我如何才能将这段时间转换为小时,而不获取扰乱我数据的日期。我假设您正在使用jupyter笔记本运行代码 执行代码时,变量
operational\u data\u clean['Pick/pack start time']
变为pd.to\u datetime(operational\u data\u 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函数,基本上是这样做的:
pd.to_datetime(pd.to_datetime(操作数据清理['Pick/pack start time'))
至于pd.to_datetime()本身,我建议你仔细看看熊猫