Question

我导入了一个txt文件，这是一个结果：

   Place  Bib    Athlete Name          City State Age Gender FinishTime  \
0      1  120     Runner 1            Bronx    NY  31      M    1:21:40   
1      2  910     Runner 2            Bronx    NY  38      M    1:23:16   
2      3  352     Runner 3         New York    NY  45      M    1:24:28   


  Unnamed: 8  
0        NaN  
1        NaN  
2        NaN

我想将FinishTime字符串对象转换为时间格式。运用 pd.to_datetime（race [＆＃39; FinishTime＆＃39;]）我得到ValueError：给定日期字符串不太可能是日期时间。有关如何做到这一点的任何建议？我想按时计算，比如Runner 1比Runner 2快x％。谢谢。

Answer 1

to_datetime可以使用format参数来指定时间戳的外观：

>>> pd.to_datetime('1:23:16', format='%H:%M:%S')
Timestamp('1900-01-01 01:23:16')

关于你的数据：

pd.to_datetime(race['FinishTime'], format='%H:%M:%S')

Answer 2

您可以使用to_timedelta将列转换为timedelta。

然后Athlete Name列set_index，然后按loc选择值。最后得到timedeltas的差异：

df.FinishTime = pd.to_timedelta(df.FinishTime)
df = df.set_index('Athlete Name')
runner1 = df.loc['Runner 1', 'FinishTime']
runner2 = df.loc['Runner 2', 'FinishTime']

print('Runner 1 is {} faster than runner 2.'.format(runner2 - runner1))
Runner 1 is 0 days 00:01:36 faster than runner 2.

Answer 3

严格地说，完成时间不是一个时间，而是一个timedelta对象，表示持续时间，完成比赛需要多长时间。

以下代码可用于将字符串转换为timedelta并进行计算：

from datetime import timedelta

finish_time_1 = '1:23:16'
finish_time_2 = '1:24:28'

hours, minutes, seconds = finish_time_1.split(':')
duration_1 = timedelta(hours=int(hours), minutes=int(minutes), seconds=int(seconds))

hours, minutes, seconds = finish_time_2.split(':')
duration_2 = timedelta(hours=int(hours), minutes=int(minutes), seconds=int(seconds))

print('Runner 1 is {} faster than runner 2'.format(duration_2 - duration_1))

结果是：＆＃34;跑步者1比跑步者2＆＃34更快0:01:12;

Answer 4

这是我最终编码的方式：

#not sure if I needed to do this
race['FinishTime']=race.FinishTime.str.rjust(8, fillchar='0')

#split out each component of the string
race['hour'] = race.FinishTime.str.split(':').str[0]
race['mins'] = race.FinishTime.str.split(':').str[1]
race['secs'] = race.FinishTime.str.split(':', n=2).str[-1]

#concat each component into a new column
race['FTime'] = race['hour'] + ':' + race['mins'] + ':' + race['secs']

#convert the string to timedelta
race.FTime = pd.to_timedelta(race.FTime)

如何转换＆＃39; 1：30：20＆＃39; pandas中的字符串到时间格式

4 个答案: