如果我使用这个功能pd.DatetimeIndex(dfTrain['datetime']).weekday
我得到当天的号码,但是我找不到任何给出de day名字的函数...所以我需要将0转换为星期一,1转换为Tuestday等等。
以下是我的数据框示例:
datetime season holiday workingday weather temp atemp humidity windspeed count
0 2011-01-01 00:00:00 1 0 0 1 9.84 14.395 81 0.0000 16
1 2011-01-01 01:00:00 1 0 0 1 9.02 13.635 80 0.0000 40
2 2011-01-01 02:00:00 1 0 0 1 9.02 13.635 80 0.0000 32
3 2011-01-01 03:00:00 1 0 0 1 9.84 14.395 75 0.0000 13
4 2011-01-01 04:00:00 1 0 0 1 9.84 14.395 75 0.0000 1
5 2011-01-01 05:00:00 1 0 0 2 9.84 12.880 75 6.0032 1
6 2011-01-01 06:00:00 1 0 0 1 9.02 13.635 80 0.0000 2
7 2011-01-01 07:00:00 1 0 0 1 8.20 12.880 86 0.0000 3
8 2011-01-01 08:00:00 1 0 0 1 9.84 14.395 75 0.0000 8
9 2011-01-01 09:00:00 1 0 0 1 13.12 17.425 76 0.0000 14
另一个问题是,pandas.DatetimeIndex.dayofweek
和pandas.DatetimeIndex.weekday
之间的区别是什么?
答案 0 :(得分:12)
一种方法,只要datetime已经是datetime列,就是应用datetime.strftime
来获取工作日的字符串:
In [105]:
df['weekday'] = df[['datetime']].apply(lambda x: dt.datetime.strftime(x['datetime'], '%A'), axis=1)
df
Out[105]:
datetime season holiday workingday weather temp atemp \
0 2011-01-01 00:00:00 1 0 0 1 9.84 14.395
1 2011-01-01 01:00:00 1 0 0 1 9.02 13.635
2 2011-01-01 02:00:00 1 0 0 1 9.02 13.635
3 2011-01-01 03:00:00 1 0 0 1 9.84 14.395
4 2011-01-01 04:00:00 1 0 0 1 9.84 14.395
5 2011-01-01 05:00:00 1 0 0 2 9.84 12.880
6 2011-01-01 06:00:00 1 0 0 1 9.02 13.635
7 2011-01-01 07:00:00 1 0 0 1 8.20 12.880
8 2011-01-01 08:00:00 1 0 0 1 9.84 14.395
9 2011-01-01 09:00:00 1 0 0 1 13.12 17.425
humidity windspeed count weekday
0 81 0.0000 16 Saturday
1 80 0.0000 40 Saturday
2 80 0.0000 32 Saturday
3 75 0.0000 13 Saturday
4 75 0.0000 1 Saturday
5 75 6.0032 1 Saturday
6 80 0.0000 2 Saturday
7 86 0.0000 3 Saturday
8 75 0.0000 8 Saturday
9 76 0.0000 14 Saturday
关于您的其他问题,dayofweek
和weekday
之间没有区别。
在工作日定义工作日地图到String等效和调用地图会更快:
dayOfWeek={0:'Monday', 1:'Tuesday', 2:'Wednesday', 3:'Thursday', 4:'Friday', 5:'Saturday', 6:'Sunday'}
df['weekday'] = df['datetime'].dt.dayofweek.map(dayOfWeek)
对于0.15.0
之前的版本,以下内容应该有效:
import datetime as dt
df['weekday'] = df['datetime'].apply(lambda x: dt.datetime.strftime(x, '%A'))
版本0.18.1及更新
现在有了一种新的便捷方法dt.weekday_name
来执行上述操作
版本0.23.0及更新
weekday_name现在已被删除,有利于dt.day_name
。
答案 1 :(得分:12)
在版本0.18.1
中,您可以使用新方法dt.weekday_name
:
df['weekday'] = df['datetime'].dt.weekday_name
print df
datetime season holiday workingday weather temp atemp \
0 2011-01-01 00:00:00 1 0 0 1 9.84 14.395
1 2011-01-01 01:00:00 1 0 0 1 9.02 13.635
2 2011-01-01 02:00:00 1 0 0 1 9.02 13.635
3 2011-01-01 03:00:00 1 0 0 1 9.84 14.395
4 2011-01-01 04:00:00 1 0 0 1 9.84 14.395
5 2011-01-01 05:00:00 1 0 0 2 9.84 12.880
6 2011-01-01 06:00:00 1 0 0 1 9.02 13.635
7 2011-01-01 07:00:00 1 0 0 1 8.20 12.880
8 2011-01-01 08:00:00 1 0 0 1 9.84 14.395
9 2011-01-01 09:00:00 1 0 0 1 13.12 17.425
humidity windspeed count weekday
0 81 0.0000 16 Saturday
1 80 0.0000 40 Saturday
2 80 0.0000 32 Saturday
3 75 0.0000 13 Saturday
4 75 0.0000 1 Saturday
5 75 6.0032 1 Saturday
6 80 0.0000 2 Saturday
7 86 0.0000 3 Saturday
8 75 0.0000 8 Saturday
9 76 0.0000 14 Saturday
答案 2 :(得分:5)
使用dt.weekday_name
是deprecated since pandas 0.23.0
,而是使用dt.day_name()
:
df.datetime.dt.day_name()
0 Saturday
1 Saturday
2 Saturday
3 Saturday
4 Saturday
5 Saturday
6 Saturday
7 Saturday
8 Saturday
9 Saturday
Name: datetime, dtype: object
答案 3 :(得分:0)
在@jezrael的上一个正确答案中,您可以使用以下方法:
import calendar
df['weekday'] = pd.Series(pd.Categorical(df['datetime'].dt.weekday_name, categories=list(calendar.day_name)))
还会根据this为您的新类别变量提供 order (在此示例中为“ Monday”,...,“ Sunday”)。这可能对下一步的分析很有帮助。