] 2
我想绘制回归线并计算回归系数。随之而来的是 R 平方/拟合优度。
我已经确定了曲线上的回归线并计算了该回归线的系数,但无法计算出它们的拟合优度。
请帮我解决这个问题。
代码计算回归系数并在图上绘制回归线。但我无法计算 R 平方/拟合优度。
还编写了拟合优度代码,但给出错误“TypeError:无法将类型'ndarray'转换为分子/分母”
该错误与“def确定系数部分”有关
from PIL import Image
import numpy as np
import matplotlib.pyplot as plt
from statistics import mean
from matplotlib import style
style.use('ggplot')
#loading of imageries
im1 = Image.open('D:\\code\\test\\subset\\image1.tif')
im2 = Image.open('D:\\code\\test\\subset\\image2.tif')
#im.show()
#Conversion into array
img1 = np.array(im1)
img2 = np.array(im2)
def estimate_coef(x, y):
# number of observations/points
n = np.size(x)
# mean of x and y vector
m_x, m_y = np.mean(x), np.mean(y)
# calculating cross-deviation and deviation about x
SS_xy = np.sum(y*x) - n*m_y*m_x
SS_xx = np.sum(x*x) - n*m_x*m_x
# calculating regression coefficients
b_1 = SS_xy / SS_xx
b_0 = m_y - b_1*m_x
return(b_0, b_1)
def squared_error(ys_orig,ys_line):
return sum((ys_line - ys_orig) * (ys_line - ys_orig))
def coefficient_of_determination(ys_orig,ys_line):
y_mean_line = [mean(ys_orig) for y in ys_orig]
squared_error_regr = squared_error(ys_orig, ys_line)
squared_error_y_mean = squared_error(ys_orig, ys_mean_line)
return 1 - (squared_error_regr/squared_error_y_mean)
def plot_regression_line(x, y, b):
# plotting the actual points as scatter plot
plt.scatter(x, y, color = "m",
marker = "o", s = 30)
# predicted response vector
y_pred = b[0] + b[1]*x
# plotting the regression line
plt.plot(x, y_pred, color = "g")
# putting labels
plt.xlabel('x')
plt.ylabel('y')
# function to show plot
plt.show()
b_0, b_1 = estimate_coef(img1, img2)
regression_line = [(b_0*x)+b_1 for x in img1]
r_squared = coefficient_of_determination(img2,regression_line)
print(r_squared)
def main():
# observations
x = img1
y = img2
# estimating coefficients
b = estimate_coef(x, y)
print("Estimated coefficients:\nb_0 = {} \ \nb_1 = {}".format(b[0], b[1]))
# plotting regression line
plot_regression_line(x, y, b)
if __name__ == "__main__":
main()