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Pandas Dataframe，为一周中的某一天指定值 . 特征提取

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我正在提取功能以寻找工作日 . 到目前为止我所拥有的是：

days = {0:'Mon', 1: 'Tues', 2:'Wed', 3:'Thurs', 4:'Fri', 5:'Sat', 6:'Sun'}
data['day_of_week'] = data['day_of_week'].apply(lambda x: days[x])
data['if_Weekday'] = np.where( (data['day_of_week'] == 'Mon') | (data['day_of_week'] == 'Tues') | (data['day_of_week'] == 'Wed') | (data['day_of_week'] == 'Thurs') | (data['day_of_week'] == 'Friday'), '1', '0')

此代码将Mon-Fri指定为 1 ，将Sat-Sun指定为 0 . 但是，我想为工作日分配不同的值 . 例如，Mon = 1，Tues = 2，Wed = 3，Thurs = 4，Fri = 5，Sat和Sun都应该等于0 .

任何帮助将非常感激 .

python pandas numpy multiple-columns dayofweek

2 回答

  • 0

    我想你可以使用mask：

    data = pd.DataFrame({'day_of_week':[0,1,2,3,4,5,6]})

#original column to new, add 1
data['if_Weekday'] = data['day_of_week'] + 1

#map days if necessary
days = {0:'Mon', 1: 'Tues', 2:'Wed', 3:'Thurs', 4:'Fri', 5:'Sat', 6:'Sun'}
data['day_of_week'] = data['day_of_week'].map(days)

#correct weekend days
data['if_Weekday'] = data['if_Weekday'].mask(data['if_Weekday'] >= 6, 0)

print (data)
  day_of_week  if_Weekday
0         Mon           1
1        Tues           2
2         Wed           3
3       Thurs           4
4         Fri           5
5         Sat           0
6         Sun           0
    回复于
  • 0

    好吧我认为最简单的方法是使用 np.where 来测试工作日是否大于或等于5，如果是，则分配 0 ，否则返回工作日值并添加 1 ：

    In [21]:
df['is_weekday'] = np.where(df['weekday'] >= 5, 0, df['weekday'] + 1)
df
Out[21]:
       dates  weekday  is_weekday
0 2016-01-01        4           5
1 2016-01-02        5           0
2 2016-01-03        6           0
3 2016-01-04        0           1
4 2016-01-05        1           2
5 2016-01-06        2           3
6 2016-01-07        3           4
7 2016-01-08        4           5
8 2016-01-09        5           0
9 2016-01-10        6           0
    回复于

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