具有强制非零 y 截距的线性回归

Question

我想在 y 截距强制为 0.115 的情况下运行线性回归。这是我尝试过的代码。我设置为 fit_intercept=True 以获得非零 y 截距，但我可以将其设置为一个值吗？

另外，我怎样才能绘制出最适合的线而不是连接每个点的线？

提前致谢。

import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from sklearn.metrics import r2_score
from sklearn.linear_model import LinearRegression
x=np.array([0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100]).reshape(-1,1)
y=np.array([0.113, 0.116, 0.130, 0.150, 0.150, 0.160, 0.180, 0.210, 0.220, 0.260, 0.280])
regression=LinearRegression(fit_intercept=True).fit(x,y)
r_sq=round(regression.score(x,y),4)
m=round(regression.coef_[0],4)
b=round(regression.intercept_,4)
print("r_sq:", r_sq,"m:",m,"b:",b)
plt.figure()
plt.scatter(x,y)
plt.title('A')
plt.ylabel('X')
plt.xlabel('Y')
plt.plot(x,y,'r--',label='measured')
plt.legend(loc='best')

Answer 1

从数据中减去要修正的 y 截距并设置 fit_intercept=False。

例如

import matplotlib.pyplot as plt
import numpy as np
from sklearn.linear_model import LinearRegression

x = np.array([0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100]).reshape(-1, 1)
y = np.array([0.113, 0.116, 0.130, 0.150, 0.150, 0.160, 0.180, 0.210, 0.220, 0.260, 0.280])

fig, ax = plt.subplots()

for fit, y_intercept in zip((True, False), (0.0, 0.115)):
    regression = LinearRegression(fit_intercept=fit)
    regression.fit(x, y - y_intercept)

    r_sq = regression.score(x, y - y_intercept)
    m = regression.coef_[0]
    b = regression.intercept_ + y_intercept

    print(f"Fit intercept: {regression.fit_intercept}")
    print(f"r_sq: {r_sq:0.4f}\nm: {m:0.4f}\nb: {b:0.4f}")

    ax.plot(x, y, "bo")
    ax.plot(
        x,
        regression.predict(x) + y_intercept,
        "r" + "--" * fit,
        label=f"Fit Intercept: {regression.fit_intercept}",
    )

ax.set_title("A")
ax.set_ylabel("X")
ax.set_xlabel("Y")

ax.legend(loc="best")

plt.show()

哪些打印：

Fit intercept: True
r_sq: 0.9473
m: 0.0017
b: -0.0192
Fit intercept: False
r_sq: 0.9112
m: 0.0014
b: 0.0000

Answer 2

fit_intercept 上的帖子 https://stackoverflow.com/questions/46779605

import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from sklearn.metrics import r2_score
from sklearn.linear_model import LinearRegression

x=np.array([0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100]).reshape(-1,1)
y=np.array([0.113, 0.116, 0.130, 0.150, 0.150, 0.160, 0.180, 0.210, 0.220, 0.260, 0.280])

lr_fi_true = LinearRegression(fit_intercept=True)
lr_fi_false = LinearRegression(fit_intercept=False)

lr_fi_true.fit(x, y)
lr_fi_false.fit(x, y)

print('Intercept when fit_intercept=True : {:.5f}'.format(lr_fi_true.intercept_))
print('Intercept when fit_intercept=False : {:.5f}'.format(lr_fi_false.intercept_))

lr_fi_true_yhat = np.dot(x, lr_fi_true.coef_) + lr_fi_true.intercept_
lr_fi_false_yhat = np.dot(x, lr_fi_false.coef_) + lr_fi_false.intercept_

plt.scatter(x, y, label='Actual points')
plt.plot(x, lr_fi_true_yhat, 'r--', label='fit_intercept=True')
plt.plot(x, lr_fi_false_yhat, 'r-', label='fit_intercept=False')
plt.legend()

plt.vlines(0, 0, y.max())
plt.hlines(0, x.min(), x.max())

plt.show()

印刷：

Intercept when fit_intercept=True : 0.09577
Intercept when fit_intercept=False : 0.00000

Answer 3

我找到了一个通用解决方案，它给出了相同的答案，但也允许我通过简单地修改函数来拟合非线性方程。

import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
import numpy as np

#set y-intercept
b=0.115

#Fitting function
def func(x, m):
    return (x*m)+b

#Experimental x and y data points    
x_A1 = np.array([0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100])
y_A1 = np.array([0.113, 0.116, 0.130, 0.150, 0.150, 0.160, 0.180, 0.210, 0.220, 0.260, 0.280])

#Plot experimental data points
plt.plot(x_A1, y_A1, 'bo', label='experimental')

#Perform the curve-fit
popt, pcov = curve_fit(func, x_A1, y_A1) #, initialGuess)
#print(popt)

#x values for the fitted function
x_A1_Fit = np.arange(x_A1[0], x_A1[-1], 0.1)

residuals = y_A1- func(x_A1, *popt)
ss_res = np.sum(residuals**2)
ss_tot = np.sum((y_A1-np.mean(y_A1))**2)
r_sq = 1 - (ss_res / ss_tot)

#Plot the fitted function
plt.plot(x_A1_Fit, func(x_A1_Fit, *popt), 'r--', label='fitted: m=%5.4f' % tuple(popt))

plt.xlabel('x')
plt.ylabel('y')
plt.legend()
plt.show()
print ('r_sq=', "%.4f"%r_sq, 'm=', "%.4f"%popt, "b=", "%.4f"%b)