在Python上具有两个变量的线性回归

Question

我正在开发代码来分析两个变量之间的关系。我正在使用 DataFrame 将变量保存在两列中，如下所示：

column A = 132.54672, 201.3845717, 323.2654551  
column B = 51.54671995,  96.38457166, 131.2654551

我尝试使用 statsmodels ，但是它表示我没有足够的样本。

有人可以帮助我吗？我需要定义系数和截距才能计算其他变量。

y = coefficient * x + intercept

Answer 1

好的，这是使用DataFrame的解决方案。我跳过了导入命令，只显示了相关部分。如果您想知道它们是什么，请给我评论。

我正在使用NumPy的polyfit进行1阶线性回归。您可以打印拟合（fit）以获得斜率和截距。 fit[0]是截距，fit[1]是斜率（或您所称的系数）

column_A= [132.54672, 201.3845717, 323.2654551]
column_B= [51.54671995, 96.38457166, 131.2654551]
df = pd.DataFrame({'A': column_A, 'B': column_B})

fit = np.poly1d(np.polyfit(df['A'], df['B'], 1))

A_mesh = np.linspace(min(df['A']), max(df['A']), 100)

plt.plot(df['A'], df['B'], 'bx', label='Data', ms=10)
plt.plot(A_mesh, fit(A_mesh), '-b', label='Linear fit')

print (fit)
# 0.4028 x + 4.833

Answer 2

您可以使用curve_fit进行此操作：

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

x = np.array([132.54672, 201.3845717, 323.2654551])
y = np.array([51.54671995, 96.38457166, 131.2654551])

linear = lambda x, a, b: a * x + b

popt, pcov = curve_fit(linear, x, y, p0=[1, 1])
plt.plot(x, y, "rx")
plt.plot(x, linear(x, *popt), "b-")
plt.title("f(x)=a*x+b, a={:.2f}, b={:.2f}".format(*popt))
plt.show()

图：

Answer 3

除了以前的出色答案外，这是一个图形钳工，具有3D散点图，3D表面图和轮廓图。

import numpy, scipy, scipy.optimize
import matplotlib
from mpl_toolkits.mplot3d import  Axes3D
from matplotlib import cm # to colormap 3D surfaces from blue to red
import matplotlib.pyplot as plt

graphWidth = 800 # units are pixels
graphHeight = 600 # units are pixels

# 3D contour plot lines
numberOfContourLines = 16


def SurfacePlot(func, data, fittedParameters):
    f = plt.figure(figsize=(graphWidth/100.0, graphHeight/100.0), dpi=100)

    matplotlib.pyplot.grid(True)
    axes = Axes3D(f)

    x_data = data[0]
    y_data = data[1]
    z_data = data[2]

    xModel = numpy.linspace(min(x_data), max(x_data), 20)
    yModel = numpy.linspace(min(y_data), max(y_data), 20)
    X, Y = numpy.meshgrid(xModel, yModel)

    Z = func(numpy.array([X, Y]), *fittedParameters)

    axes.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap=cm.coolwarm, linewidth=1, antialiased=True)

    axes.scatter(x_data, y_data, z_data) # show data along with plotted surface

    axes.set_title('Surface Plot (click-drag with mouse)') # add a title for surface plot
    axes.set_xlabel('X Data') # X axis data label
    axes.set_ylabel('Y Data') # Y axis data label
    axes.set_zlabel('Z Data') # Z axis data label

    plt.show()
    plt.close('all') # clean up after using pyplot or else thaere can be memory and process problems


def ContourPlot(func, data, fittedParameters):
    f = plt.figure(figsize=(graphWidth/100.0, graphHeight/100.0), dpi=100)
    axes = f.add_subplot(111)

    x_data = data[0]
    y_data = data[1]
    z_data = data[2]

    xModel = numpy.linspace(min(x_data), max(x_data), 20)
    yModel = numpy.linspace(min(y_data), max(y_data), 20)
    X, Y = numpy.meshgrid(xModel, yModel)

    Z = func(numpy.array([X, Y]), *fittedParameters)

    axes.plot(x_data, y_data, 'o')

    axes.set_title('Contour Plot') # add a title for contour plot
    axes.set_xlabel('X Data') # X axis data label
    axes.set_ylabel('Y Data') # Y axis data label

    CS = matplotlib.pyplot.contour(X, Y, Z, numberOfContourLines, colors='k')
    matplotlib.pyplot.clabel(CS, inline=1, fontsize=10) # labels for contours

    plt.show()
    plt.close('all') # clean up after using pyplot or else thaere can be memory and process problems


def ScatterPlot(data):
    f = plt.figure(figsize=(graphWidth/100.0, graphHeight/100.0), dpi=100)

    matplotlib.pyplot.grid(True)
    axes = Axes3D(f)
    x_data = data[0]
    y_data = data[1]
    z_data = data[2]

    axes.scatter(x_data, y_data, z_data)

    axes.set_title('Scatter Plot (click-drag with mouse)')
    axes.set_xlabel('X Data')
    axes.set_ylabel('Y Data')
    axes.set_zlabel('Z Data')

    plt.show()
    plt.close('all') # clean up after using pyplot or else thaere can be memory and process problems


def func(data, a, alpha, beta):
    t = data[0]
    p_p = data[1]
    return a * (t**alpha) * (p_p**beta)


if __name__ == "__main__":
    xData = numpy.array([1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0])
    yData = numpy.array([11.0, 12.1, 13.0, 14.1, 15.0, 16.1, 17.0, 18.1, 90.0])
    zData = numpy.array([1.1, 2.2, 3.3, 4.4, 5.5, 6.6, 7.7, 8.0, 9.9])

    data = [xData, yData, zData]

    initialParameters = [1.0, 1.0, 1.0] # these are the same as scipy default values in this example

    # here a non-linear surface fit is made with scipy's curve_fit()
    fittedParameters, pcov = scipy.optimize.curve_fit(func, [xData, yData], zData, p0 = initialParameters)

    ScatterPlot(data)
    SurfacePlot(func, data, fittedParameters)
    ContourPlot(func, data, fittedParameters)

    print('fitted prameters', fittedParameters)

    modelPredictions = func(data, *fittedParameters) 

    absError = modelPredictions - zData

    SE = numpy.square(absError) # squared errors
    MSE = numpy.mean(SE) # mean squared errors
    RMSE = numpy.sqrt(MSE) # Root Mean Squared Error, RMSE
    Rsquared = 1.0 - (numpy.var(absError) / numpy.var(zData))
    print('RMSE:', RMSE)
    print('R-squared:', Rsquared)

Answer 4

使用scipy.stats：

import pandas as pd
from scipy import stats
import matplotlib.pyplot as plt


column_A= [132.54672, 201.3845717, 323.2654551]
column_B= [51.54671995, 96.38457166, 131.2654551]
df = pd.DataFrame({'A': column_A, 'B': column_B})

reg = stats.linregress(df.A, df.B)

plt.plot(df.A, df.B, 'bo', label='Data')
plt.plot(df.A, reg.intercept + reg.slope * df.A, 'k-', label='Linear Regression')
plt.xlabel('A')
plt.ylabel('B')
plt.legend()
plt.show()

您还可以从dir(reg)中找到有用的方法，其中包括

.intercept .pvalue .rvalue .slope .stderr

请参见here。