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python - sklearn MinMaxScaler() 与 groupby pandas

转载 作者:行者123 更新时间:2023-12-03 08:15:50 24 4
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我有两个特征排名评级,针对不同类别下的不同产品ID,这些特征是在不同日期从电子商务网站上抓取的。

此处提供示例数据框:

import pandas as pd
import numpy as np
import warnings; warnings.simplefilter('ignore')
from sklearn.preprocessing import MinMaxScaler
from sklearn.preprocessing import StandardScaler
from sklearn.preprocessing import RobustScaler

df=pd.read_csv('https://raw.githubusercontent.com/amanaroratc/hello-world/master/testdf.csv')
df.head()

category bid date rank ratings
0 Aftershave ASCDBNYZ4JMSH42B 2021-10-01 61.0 462.0
1 Aftershave ASCDBNYZ4JMSH42B 2021-10-02 69.0 462.0
2 Aftershave ASCDBNYZ4JMSH42B 2021-10-05 89.0 463.0
3 Aftershave ASCE3DZK2TD7G4DN 2021-10-01 309.0 3.0
4 Aftershave ASCE3DZK2TD7G4DN 2021-10-02 319.0 3.0

我想使用 sklearn 中的 MinMaxScaler() 标准化 rank ratings

我试过了

cols=['rank','ratings']
features=df[cols]
scaler1=MinMaxScaler()
df_norm[['rank_norm_mm', 'ratings_norm_mm']] = scaler1.fit_transform(features)

这对整个数据集进行标准化。我想使用 groupby 对每个特定日期的每个类别执行此操作。

最佳答案

使用GroupBy.apply :

file = 'https://raw.githubusercontent.com/amanaroratc/hello-world/master/testdf.csv'
df=pd.read_csv(file)

from sklearn.preprocessing import MinMaxScaler

cols=['rank','ratings']

def f(x):
scaler1=MinMaxScaler()
x[['rank_norm_mm', 'ratings_norm_mm']] = scaler1.fit_transform(x[cols])
return x

df = df.groupby(['category', 'date']).apply(f)

另一个解决方案:

file = 'https://raw.githubusercontent.com/amanaroratc/hello-world/master/testdf.csv'
df=pd.read_csv(file)

from sklearn.preprocessing import MinMaxScaler

scaler1=MinMaxScaler()
cols=['rank','ratings']

df= df.join(df.groupby(['category', 'date'])[cols]
.apply(lambda x: pd.DataFrame(scaler1.fit_transform(x), index=x.index))
.add_prefix('_norm_mm'))

关于python - sklearn MinMaxScaler() 与 groupby pandas,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/69476352/

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