Transform pandas Data Frame to use for MultiLabelBinarizer

耗尽温柔 提交于 2019-12-24 04:41:16

问题


My question is: How can I transform a Data Frame like this to eventually use it in scikit's MulitLabelBinarizer:

d1 = {'ID':[1,2,3,4], 'km':[80,90,90,100], 'weight':[10,20,20,30], 'label':['A','B','C','D','E']}
df1 = pd.DataFrame(data=d1)
df1

    ID  km  weight label
0   1   80      10     A
1   2   90      20     B
2   2   90      20     C
3   4  100      30     D

It should tourn ot like this:

d2 ={'km':[80,90,100], 'weight':[10,20,30], 'label':['A',('B','C'),'D']}
df2 = pd.DataFrame(data=d2)
df2

    km  weight   label
0   80      10       A
1   90      20  (B, C)
2  100      30       D

So I can juse the data properly in the MultiLabelBinarizer:

from sklearn.preprocessing import MultiLabelBinarizer

mlb = MultiLabelBinarizer()
mlb.fit(df2['label'])
mlb.transform(df2['label'])

array([[1, 0, 0, 0],
       [0, 1, 1, 0],
       [0, 0, 0, 1]])

Note: the raw data has more than 1 million rows.


回答1:


I think you need this:

d1 = {'ID':[1,2,3,4], 'km':[80,90,90,100], 'weight':[10,20,20,30], 'label':['A','B','C','D']}
df1 = pd.DataFrame(data=d1)
#Groupby and get tuple, like you need 
df2 = pd.DataFrame(df1.groupby(['km','weight'])['label'].apply(lambda x: tuple(x.values)))
df2.reset_index(inplace=True)

from sklearn.preprocessing import MultiLabelBinarizer
mlb = MultiLabelBinarizer()
mlb.fit(df2['label'])
mlb.transform(df2['label'])


来源:https://stackoverflow.com/questions/53494873/transform-pandas-data-frame-to-use-for-multilabelbinarizer

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!