首页 > 解决方案 > pandas str 到 datetime 转换错误(60 分钟字段) ParserError: minute must be in 0..59: 2015-04-24 12:60:46

问题描述

我正在尝试使用给定的str熊猫系列转换为日期时间pd.to_datetime()。但是,我有一些str其中 60 分钟字段。

['2015-04-24 12:60:46',
 '2015-04-24 11:60:20',
 '2015-03-14 12:60:02',
 '2015-05-11 12:60:53',
 '2015-04-26 11:60:44',
 '2015-05-31 15:60:59',
 '2015-04-02 07:60:10',
 '2015-04-23 12:60:59',
 '2015-05-07 18:60:11',
 '2015-04-27 12:60:39',
 '2015-04-10 09:60:26',
 '2015-04-03 18:60:05',
 '2015-05-20 08:60:37',
 '2015-05-08 12:60:17',
 '2015-04-16 12:60:50',
 '2015-03-26 09:60:51',
 '2015-03-20 08:60:29',
 '2015-03-21 13:60:19',
 '2015-03-07 01:60:16',
 '2015-05-31 14:60:56',
 '2015-03-06 18:60:01',
 '2015-05-17 14:60:46',
 '2015-03-10 04:60:18',
 '2015-05-23 12:60:30',
 '2015-04-17 09:60:53',
 '2015-04-23 17:60:34',
 '2015-03-31 12:60:50',
.....]

知道如何解决这个问题吗?

标签: pythonpandasdatetime

解决方案


按空格拆分值,Series.str.split然后将日期部分转换为日期时间,并添加转换为 timedeltas 的时间部分to_timedelta

d = ['2015-04-24 12:60:46', '2015-04-24 11:60:20', 
     '2015-03-14 12:60:02', '2015-05-11 12:60:53']
df = pd.DataFrame({'date':d})

s = df['date'].str.split()

df['date'] = pd.to_datetime(s.str[0]) + pd.to_timedelta(s.str[1])
print (df)
                 date
0 2015-04-24 13:00:46
1 2015-04-24 12:00:20
2 2015-03-14 13:00:02
3 2015-05-11 13:00:53

如果总是60存在所有数据的另一个想法,否则使用以前的解决方案:

df['date'] = (pd.to_datetime(df['date'].replace(":60:",":59:", regex=True)) +  
              pd.Timedelta(1, 'min'))
print (df)
                 date
0 2015-04-24 13:00:46
1 2015-04-24 12:00:20
2 2015-03-14 13:00:02
3 2015-05-11 13:00:53

推荐阅读