Question

我在.csv文件中有一个数据集（dataTrain.csv＆amp; dataTest.csv），格式为：

Temperature(K),Pressure(ATM),CompressibilityFactor(Z)
273.1,24.675,0.806677258
313.1,24.675,0.888394713
...,...,...

能够使用此代码构建回归模型和预测：

import pandas as pd
from sklearn import linear_model

dataTrain = pd.read_csv("dataTrain.csv")
dataTest = pd.read_csv("dataTest.csv")
# print df.head()

x_train = dataTrain['Temperature(K)'].reshape(-1,1)
y_train = dataTrain['CompressibilityFactor(Z)']

x_test = dataTest['Temperature(K)'].reshape(-1,1)
y_test = dataTest['CompressibilityFactor(Z)']

ols = linear_model.LinearRegression()
model = ols.fit(x_train, y_train)

print model.predict(x_test)[0:5]

但是，我想要做的是多元回归。因此，模型将为CompressibilityFactor(Z) = intercept + coef*Temperature(K) + coef*Pressure(ATM)

如何在scikit-learn中做到这一点？

Answer 1

如果您的上述代码适用于单变量，请尝试使用

import pandas as pd
from sklearn import linear_model

dataTrain = pd.read_csv("dataTrain.csv")
dataTest = pd.read_csv("dataTest.csv")
# print df.head()

x_train = dataTrain[['Temperature(K)', 'Pressure(ATM)']].reshape(-1,2)
y_train = dataTrain['CompressibilityFactor(Z)']

x_test = dataTest[['Temperature(K)', 'Pressure(ATM)']].reshape(-1,2)
y_test = dataTest['CompressibilityFactor(Z)']

ols = linear_model.LinearRegression()
model = ols.fit(x_train, y_train)

print model.predict(x_test)[0:5]

Answer 2

那是正确的，您需要使用.values.reshape（-1,2）

此外，如果您想知道表达式的系数和截距：

CompressibilityFactor（Z）=截距+ coef 温度（K）+ coef 压力（ATM）

您可以通过以下方式获得它们：

系数= model.coef_
拦截= model.intercept _

Scikit中的多变量/多元线性回归学习？

2 个答案: