在使用Robust Scaler将模型拟合到缩放数据之后，是否可以对LASSO回归中的截距和系数进行逆变换？
我正在使用LASSO回归来预测未经标准化且在LASSO上效果不佳的数据的值，除非事先进行缩放。缩放数据并拟合LASSO模型后，理想情况下，我希望能够看到模型截距和系数是多少，但以原始单位（而非缩放版本）为准。我问了类似的问题here，但看来这是不可能的。如果没有，为什么？谁可以给我解释一下这个？我正在尝试加深对LASSO和Robust Scaler的工作方式的了解。
下面是我正在使用的代码。在这里，我尝试使用transformer_x对系数进行逆变换，并使用transformer_y对截距进行逆变换。但是，听起来这是不正确的。
import pandas as pd
from sklearn.preprocessing import RobustScaler
from sklearn.linear_model import Lasso
df = pd.DataFrame({'Y':[5, -10, 10, .5, 2.5, 15], 'X1':[1., -2., 2., .1, .5, 3], 'X2':[1, 1, 2, 1, 1, 1],
'X3':[6, 6, 6, 5, 6, 4], 'X4':[6, 5, 4, 3, 2, 1]})
X = df[['X1','X2', 'X3' ,'X4']]
y = df[['Y']]
#Scaling
transformer_x = RobustScaler().fit(X)
transformer_y = RobustScaler().fit(y)
X_scal = transformer_x.transform(X)
y_scal = transformer_y.transform(y)
#LASSO
lasso = Lasso()
lasso = lasso.fit(X_scal, y_scal)
def pred_val(X1,X2,X3,X4):
print('X1 entered: ', X1)
#Scale X value that user entered - by hand
med_X = X.median()
Q1_X = X.quantile(0.25)
Q3_X = X.quantile(0.75)
IQR_X = Q3_X - Q1_X
X_scaled = (X1 - med_X)/IQR_X
print('X1 scaled by hand: ', X_scaled[0].round(2))
#Scale X value that user entered - by function
X_scaled2 = transformer_x.transform(np.array([[X1,X2]]))
print('X1 scaled by function: ', X_scaled2[0][0].round(2))
#Intercept by hand
med_y = y.median()
Q1_y = y.quantile(0.25)
Q3_y = y.quantile(0.75)
IQR_y = Q3_y - Q1_y
inv_int = med_y + IQR_y*lasso.intercept_[0]
#Intercept by function
inv_int2 = transformer_y.inverse_transform(lasso.intercept_.reshape(-1, 1))[0][0]
#Coefficient by hand
inv_coef = lasso.coef_[0]*IQR_y
#Coefficient by function
inv_coef2 = transformer_x.inverse_transform(reg.coef_.reshape(1,-1))[0]
#Prediction by hand
preds = inv_int + inv_coef*X_scaled[0]
#Prediction by function
preds_inner = lasso.predict(X_scaled2)
preds_f = transformer_y.inverse_transform(preds_inner.reshape(-1, 1))[0][0]
print('\nIntercept by hand: ', inv_int[0].round(2))
print('Intercept by function: ', inv_int2.round(2))
print('\nCoefficients by hand: ', inv_coef[0].round(2))
print('Coefficients by function: ', inv_coef2[0].round(2))
print('\nYour predicted value by hand is: ', preds[0].round(2))
print('Your predicted value by function is: ', preds_f.round(2))
print('Perfect Prediction would be 80')
pred_val(10,1,1,1)
更新：我已经更新了代码，以显示我要创建的预测函数的类型。我只是在尝试创建一个功能，其功能与.predict完全相同，但也以其未缩放单位显示截距和系数。
当前输出：
Out[1]:
X1 entered: 10
X1 scaled by hand: 5.97
X1 scaled by function: 5.97
Intercept by hand: 34.19
Intercept by function: 34.19
Coefficients by hand: 7.6
Coefficients by function: 8.5
Your predicted value by hand is: 79.54
Your predicted value by function is: 79.54
Perfect Prediction would be 80
理想的输出：
Out[1]:
X1 entered: 10
X1 scaled by hand: 5.97
X1 scaled by function: 5.97
Intercept by hand: 34.19
Intercept by function: 34.19
Coefficients by hand: 7.6
Coefficients by function: 7.6
Your predicted value by hand is: 79.54
Your predicted value by function is: 79.54
Perfect Prediction would be 80
答案 0 :(得分：3)
基于链接的SO线程，您要做的就是获取未缩放的预测值。那正确吗？
如果是，那么您要做的就是：
# Scale the test dataset
X_test_scaled = transformer_x.transform(X_test)
# Predict with the trained model
prediction = lasso.predict(X_test_scaled)
# Inverse transform the prediction
prediction_in_dollars = transformer_y.inverse_transform(prediction)
更新：
假设火车数据仅包含一个名为X的要素。这是RobustScaler的工作：
X_scaled = (X - median(X))/IQR(X)
y_scaled = (y - median(y))/IQR(y)
然后，套索回归将给出如下预测：
a * X_scaled + b = y_scaled
您必须计算方程式，以查看未缩放数据上的模型系数：
# Substituting X_scaled and y_scaled from the 1st equation
# In this equation `median(X), IQR(X), median(y) and IQR(y) are plain numbers you already know from the training phase
a * (X - median(X))/IQR(X) + b = (y - median(y))/IQR(y)
如果您尝试以此来制作类似a_new * x + b_new = y的方程式，您将得到：
a_new = (a * (X - median(X)) / (X * IQR(X))) * IQR(y)
b_new = b * IQR(y) + median(y)
a_new * X + b_new = y
您会看到未缩放系数（a_new）取决于X。因此，您可以使用未缩放的X直接进行预测，但在两者之间可以间接应用转换。
更新2
我已经修改了您的代码，现在它说明了如何获取原始比例的系数。该脚本只是我上面显示的公式的实现。
import pandas as pd
import numpy as np
from sklearn.preprocessing import RobustScaler
from sklearn.linear_model import Lasso
df = pd.DataFrame({'Y':[5, -10, 10, .5, 2.5, 15], 'X1':[1., -2., 2., .1, .5, 3], 'X2':[1, 1, 2, 1, 1, 1],
'X3':[6, 6, 6, 5, 6, 4], 'X4':[6, 5, 4, 3, 2, 1]})
X = df[['X1','X2','X3','X4']]
y = df[['Y']]
#Scaling
transformer_x = RobustScaler().fit(X)
transformer_y = RobustScaler().fit(y)
X_scal = transformer_x.transform(X)
y_scal = transformer_y.transform(y)
#LASSO
lasso = Lasso()
lasso = lasso.fit(X_scal, y_scal)
def pred_val(X_test):
print('X entered: ',)
print (X_test.values[0])
#Scale X value that user entered - by hand
med_X = X.median()
Q1_X = X.quantile(0.25)
Q3_X = X.quantile(0.75)
IQR_X = Q3_X - Q1_X
X_scaled = ((X_test - med_X)/IQR_X).fillna(0).values
print('X_test scaled by hand: ',)
print (X_scaled[0])
#Scale X value that user entered - by function
X_scaled2 = transformer_x.transform(X_test)
print('X_test scaled by function: ',)
print (X_scaled2[0])
#Intercept by hand
med_y = y.median()
Q1_y = y.quantile(0.25)
Q3_y = y.quantile(0.75)
IQR_y = Q3_y - Q1_y
a = lasso.coef_
coef_new = ((a * (X_test - med_X).values) / (X_test * IQR_X).values) * float(IQR_y)
coef_new = np.nan_to_num(coef_new)[0]
b = lasso.intercept_[0]
intercept_new = b * float(IQR_y) + float(med_y)
custom_pred = sum((coef_new * X_test.values)[0]) + intercept_new
pred = lasso.predict(X_scaled2)
final_pred = transformer_y.inverse_transform(pred.reshape(-1, 1))[0][0]
print('Original intercept: ', lasso.intercept_[0].round(2))
print('New intercept: ', intercept_new.round(2))
print('Original coefficients: ', lasso.coef_.round(2))
print('New coefficients: ', coef_new.round(2))
print('Your predicted value by function is: ', final_pred.round(2))
print('Your predicted value by hand is: ', custom_pred.round(2))
X_test = pd.DataFrame([10,1,1,1]).T
X_test.columns = ['X1', 'X2', 'X3', 'X4']
pred_val(X_test)
您可以看到自定义预测使用原始值（X_test.values）。
结果：
X entered:
[10 1 1 1]
X_test scaled by hand:
[ 5.96774194 0. -6.66666667 -1. ]
X_test scaled by function:
[ 5.96774194 0. -6.66666667 -1. ]
Original intercept: 0.01
New intercept: 3.83
Original coefficients: [ 0.02 0. -0. -0. ]
New coefficients: [0.1 0. 0. 0. ]
Your predicted value by function is: 4.83
Your predicted value by hand is: 4.83
如上所述，新系数取决于X_test。这意味着您不能将其当前值与另一个测试样本一起使用。对于不同的输入，它们的值将不同。