我有以下数据框(不包括其余列):
| customer_id | department |
| ----------- | ----------------------------- |
| 11 | ['nail', 'men_skincare'] |
| 23 | ['nail', 'fragrance'] |
| 25 | [] |
| 45 | ['skincare', 'men_fragrance'] |
我正在预处理我的数据以适合模型。我想将部门变量转换为每个独特部门类别的虚拟变量(无论可能有多少独特的部门,不仅限于这里的内容)。
想要得到这个结果:
| customer_id | department | nail | men_skincare | fragrance | skincare | men_fragrance |
| ----------- | ---------- | ---- | ------------ | --------- | -------- | ------------- |
| 11 | ['nail', 'men_skincare'] | 1 | 1 | 0 | 0 | 0 |
| 23 | ['nail', 'fragrance'] | 1 | 0 | 1 | 0 | 0 |
| 25 | [] | 0 | 0 | 0 | 0 | 0 |
| 45 | ['skincare', 'men_fragrance'] | 0 | 0 | 0 | 1 | 1 |
我试过这个链接,但是当我拼接它时,它把它当作一个字符串来对待,并且只为字符串中的每个字符创建一个列;我用的是什么:
df['1st'] = df['department'].str[0]
df['2nd'] = df['department'].str[1]
df['3rd'] = df['department'].str[2]
df['4th'] = df['department'].str[3]
df['5th'] = df['department'].str[4]
df['6th'] = df['department'].str[5]
df['7th'] = df['department'].str[6]
df['8th'] = df['department'].str[7]
df['9th'] = df['department'].str[8]
df['10th'] = df['department'].str[9]
然后我尝试使用以下方法拆分字符串并变成一个列表:
df['new_column'] = df['department'].apply(lambda x: x.split(","))
然后再次尝试,仍然只为每个字符创建列。
有什么建议么?
编辑:我使用 anky 发送过来的链接找到了答案,特别是我使用了这个:https ://stackoverflow.com/a/29036042
什么对我有用:
df['department'] = df['department'].str.replace("'",'').str.replace("]",'').str.replace("[",'').str.replace(' ','')
df['department'] = df['department'].apply(lambda x: x.split(","))
s = df['department']
df1 = pd.get_dummies(s.apply(pd.Series).stack()).sum(level=0)
df = pd.merge(df, df1, right_index=True, left_index=True, how = 'left')