If I use this funtion pd.DatetimeIndex(dfTrain[\'datetime\']).weekday
I get number of the day, but I don\'t find any function which give the name of de day... So I
In version 0.18.1
you can use new method 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
One method, so long as datetime is already a datetime column is to apply datetime.strftime
to get the string for the weekday:
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
As to your other question, there is no difference between dayofweek
and weekday
.
It will be quicker to define a map of the weekday to String equivalent and call map on the weekday:
dayOfWeek={0:'Monday', 1:'Tuesday', 2:'Wednesday', 3:'Thursday', 4:'Friday', 5:'Saturday', 6:'Sunday'}
df['weekday'] = df['datetime'].dt.dayofweek.map(dayOfWeek)
For version prior to 0.15.0
the following should work:
import datetime as dt
df['weekday'] = df['datetime'].apply(lambda x: dt.datetime.strftime(x, '%A'))
Version 0.18.1 and newer
There is now a new convenience method dt.weekday_name to do the above
Version 0.23.0 and newer
weekday_name is now depricated in favour of dt.day_name.
Using dt.weekday_name
is deprecated since pandas 0.23.0, instead, use 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
Adding to the previous correct answer from @jezrael, you can use this:
import calendar
df['weekday'] = pd.Series(pd.Categorical(df['datetime'].dt.weekday_name, categories=list(calendar.day_name)))
which also provides your new categorical variable with order (in this example: 'Monday', ..., 'Sunday') according to this. This will possibly be helpful on next steps of your analysis.