支持向量回归
Support vector regression
执行此代码后,y_pred 太高了
我试过我的代码
import numpy as py
import matplotlib.pyplot as plt
import pandas as pd
dataset = pd.read_csv('Position_Salaries.csv')
X = dataset.iloc[:,1:2].values
y= dataset.iloc[:, 2].values
from sklearn.preprocessing import StandardScaler
sc_X = StandardScaler()
sc_y = StandardScaler()
X = sc_X.fit_transform(X)
y= sc_y.fit_transform(y.reshape(-1,1))
# Fitting SVR to the dataset
from sklearn.svm import SVR
regressor = SVR(kernel = 'rbf')
regressor.fit(X, y)
# Predicting a new result
y_pred=regressor.predict([[6.5]])
y_pred = sc_y.inverse_transform(y_pred)
为什么y_pred的价值这么高?我的代码有什么错误吗
我找到了解决方案:
我需要使用
而不是第 31 行和第 32 行
y_pred = sc_y.inverse_transform(regressor.predict(sc_X.transform(np.array([[6.5]))))
执行此代码后,y_pred 太高了
我试过我的代码
import numpy as py
import matplotlib.pyplot as plt
import pandas as pd
dataset = pd.read_csv('Position_Salaries.csv')
X = dataset.iloc[:,1:2].values
y= dataset.iloc[:, 2].values
from sklearn.preprocessing import StandardScaler
sc_X = StandardScaler()
sc_y = StandardScaler()
X = sc_X.fit_transform(X)
y= sc_y.fit_transform(y.reshape(-1,1))
# Fitting SVR to the dataset
from sklearn.svm import SVR
regressor = SVR(kernel = 'rbf')
regressor.fit(X, y)
# Predicting a new result
y_pred=regressor.predict([[6.5]])
y_pred = sc_y.inverse_transform(y_pred)
为什么y_pred的价值这么高?我的代码有什么错误吗
我找到了解决方案:
我需要使用
而不是第 31 行和第 32 行y_pred = sc_y.inverse_transform(regressor.predict(sc_X.transform(np.array([[6.5]))))