将 StandardScaler 应用于部分数据集

Question

我想使用sklearn 的StandardScaler。是否可以将其应用于某些特征列但不能应用于其他列？

例如，假设我的data 是：

data = pd.DataFrame({'Name' : [3, 4,6], 'Age' : [18, 92,98], 'Weight' : [68, 59,49]})

   Age  Name  Weight
0   18     3      68
1   92     4      59
2   98     6      49


col_names = ['Name', 'Age', 'Weight']
features = data[col_names]

我适应并改造了data

scaler = StandardScaler().fit(features.values)
features = scaler.transform(features.values)
scaled_features = pd.DataFrame(features, columns = col_names)

       Name       Age    Weight
0 -1.069045 -1.411004  1.202703
1 -0.267261  0.623041  0.042954
2  1.336306  0.787964 -1.245657

当然，名称不是真正的整数而是字符串，我不想将它们标准化。如何仅在 Age 和 Weight 列上应用 fit 和 transform 方法？

Answer 1

v0.20 中引入了 ColumnTransformer，它将转换器应用于数组或 pandas DataFrame 的一组指定列。

import pandas as pd
data = pd.DataFrame({'Name' : [3, 4,6], 'Age' : [18, 92,98], 'Weight' : [68, 59,49]})

col_names = ['Name', 'Age', 'Weight']
features = data[col_names]

from sklearn.compose import ColumnTransformer
from sklearn.preprocessing import StandardScaler

ct = ColumnTransformer([
        ('somename', StandardScaler(), ['Age', 'Weight'])
    ], remainder='passthrough')

ct.fit_transform(features)

注意：与 Pipeline 一样，它也有一个速记版本 make_column_transformer，不需要命名转换器

输出

-1.41100443,  1.20270298,  3.       
 0.62304092,  0.04295368,  4.       
 0.78796352, -1.24565666,  6.       

Answer 2

更新：

目前处理此问题的最佳方法是使用 ColumnTransformer，如 here 所述。

---

首先创建数据框的副本：

scaled_features = data.copy()

不要在转换中包含名称列：

col_names = ['Age', 'Weight']
features = scaled_features[col_names]
scaler = StandardScaler().fit(features.values)
features = scaler.transform(features.values)

现在，不要创建新数据框，而是将结果分配给这两列：

scaled_features[col_names] = features
print(scaled_features)


        Age  Name    Weight
0 -1.411004     3  1.202703
1  0.623041     4  0.042954
2  0.787964     6 -1.245657

Answer 3

聚会迟到了，但这是我的首选解决方案：

#load data
data = pd.DataFrame({'Name' : [3, 4,6], 'Age' : [18, 92,98], 'Weight' : [68, 59,49]})

#list for cols to scale
cols_to_scale = ['Age','Weight']

#create and fit scaler
scaler = StandardScaler()
scaler.fit(data[cols_to_scale])

#scale selected data
data[cols_to_scale] = scaler.transform(data[cols_to_scale])

Answer 4

另一种选择是在缩放之前删除名称列，然后将其重新合并在一起：

data = pd.DataFrame({'Name' : [3, 4,6], 'Age' : [18, 92,98], 'Weight' : [68, 59,49]})
from sklearn.preprocessing import StandardScaler

# Save the variable you don't want to scale
name_var = data['Name']

# Fit scaler to your data
scaler.fit(data.drop('Name', axis = 1))

# Calculate scaled values and store them in a separate object
scaled_values = scaler.transform(data.drop('Name', axis = 1))

data = pd.DataFrame(scaled_values, index = data.index, columns = data.drop('ID', axis = 1).columns)
data['Name'] = name_var

print(data)

Answer 5

我找到的最简单的方法是：

from sklearn.preprocessing import StandardScaler
# I'm selecting only numericals to scale
numerical = temp.select_dtypes(include='float64').columns
# This will transform the selected columns and merge to the original data frame
temp.loc[:,numerical] = StandardScaler().fit_transform(temp.loc[:,numerical])

输出

         Age  Name    Weight
0 -1.411004     3  1.202703
1  0.623041     4  0.042954
2  0.787964     6 -1.245657

Answer 6

一种更 Pythonic 的方式来做到这一点 -

from sklearn.preprocessing import StandardScaler
data[['Age','Weight']] = data[['Age','Weight']].apply(
                           lambda x: StandardScaler().fit_transform(x))
data 

输出 -

         Age  Name    Weight
0 -1.411004     3  1.202703
1  0.623041     4  0.042954
2  0.787964     6 -1.245657