KeyError: 'Dayofweek' - python-3.x

Question

Seems fastai library is not working on Python. However, I have tried to add feature using the following lines of code with an objective that should identify whether a given day is
Monday/Friday or Tuesday/Wednesday/Thursday.
The code is as follows
data['mon_fri'] = 0
for i in range(0,len(data)):
if (data['Dayofweek'][i] == 0 or data['Dayofweek'][i] == 4):
data['mon_fri'][i] = 1
else:
data['mon_fri'][i] = 0
when i run, getting the following error -
KeyError: 'Dayofweek'
Can anyone help me on this?

Answer 1

New answer showing the entire dataframe.
In [51]: df = pd.DataFrame({'col1':range(9)})
In [52]: df['d'] = pd.date_range('2016-12-31','2017-01-08',freq='D')
In [53]: df
Out[53]:
col1 d
0 0 2016-12-31
1 1 2017-01-01
2 2 2017-01-02
3 3 2017-01-03
4 4 2017-01-04
5 5 2017-01-05
6 6 2017-01-06
7 7 2017-01-07
8 8 2017-01-08
Now adding column for day of the week
In [54]: df['dow'] = df['d'].dt.dayofweek
In [55]: df
Out[55]:
col1 d dow
0 0 2016-12-31 5
1 1 2017-01-01 6
2 2 2017-01-02 0
3 3 2017-01-03 1
4 4 2017-01-04 2
5 5 2017-01-05 3
6 6 2017-01-06 4
7 7 2017-01-07 5
8 8 2017-01-08 6
Finally doing the calculation, 1 for M/Th 0 for other days
In [56]: df['feature'] = df['dow'].apply(lambda x: int((x==1) or (x==4)))
In [57]: df
Out[57]:
col1 d dow feature
0 0 2016-12-31 5 0
1 1 2017-01-01 6 0
2 2 2017-01-02 0 0
3 3 2017-01-03 1 1
4 4 2017-01-04 2 0
5 5 2017-01-05 3 0
6 6 2017-01-06 4 1
7 7 2017-01-07 5 0
8 8 2017-01-08 6 0

Answer 2

Assuming you are using pandas you can just use the built in dayofweek function
In [32]: d = pd.date_range('2016-12-31', '2017-01-08', freq='D').to_series()
In [33]: d
Out[33]:
2016-12-31 2016-12-31
2017-01-01 2017-01-01
2017-01-02 2017-01-02
2017-01-03 2017-01-03
2017-01-04 2017-01-04
2017-01-05 2017-01-05
2017-01-06 2017-01-06
2017-01-07 2017-01-07
2017-01-08 2017-01-08
Freq: D, dtype: datetime64[ns]
In [34]: s = (d.dt.dayofweek==1) |(d.dt.dayofweek==4)
In [35]: s
Out[35]:
2016-12-31 False
2017-01-01 False
2017-01-02 False
2017-01-03 True
2017-01-04 False
2017-01-05 False
2017-01-06 True
2017-01-07 False
2017-01-08 False
Freq: D, dtype: bool
Then convert to 1/0 simply by
In [39]: t = s.apply(lambda x: int(x==True))
In [40]: t
Out[40]:
2016-12-31 0
2017-01-01 0
2017-01-02 0
2017-01-03 1
2017-01-04 0
2017-01-05 0
2017-01-06 1
2017-01-07 0
2017-01-08 0
Freq: D, dtype: int64