回归曲线近似

Question

我正在尝试找到曲线的切向量。曲线的方程是一个问题，我有不同的点，基于这些点我正在寻找函数的近似值，它描述曲线并拟合点。

当我绘制数据时，它看起来像这样：

应用多项式回归后（基于本文：https://www.statology.org/curve-fitting-in-r/）我得到以下结果：

fit <- lm(cl2[,3] ~ poly(cl2[,2], 3))
    summary(fit)

Call:
lm(formula = cl2[, 3] ~ poly(cl2[, 2], 3))

Residuals:
     Min       1Q   Median       3Q      Max 
-0.31834 -0.10187  0.02132  0.09577  0.27393 

Coefficients:
                     Estimate Std. Error   t value Pr(>|t|)    
(Intercept)        -109.89121    0.03789 -2900.217  < 2e-16 ***
poly(cl2[, 2], 3)1    7.33365    0.16516    44.403  < 2e-16 ***
poly(cl2[, 2], 3)2   -4.43572    0.16516   -26.857 4.25e-14 ***
poly(cl2[, 2], 3)3    1.14772    0.16516     6.949 4.66e-06 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.1652 on 15 degrees of freedom
Multiple R-squared:  0.9946,    Adjusted R-squared:  0.9935 
F-statistic: 913.7 on 3 and 15 DF,  p-value: < 2.2e-16

当我拟合曲线时，结果看起来不错：

lines(cl2[,2], predict(fit, data.frame(cl2[,2:3])))

根据系数 s，我假设曲线方程为：

1.14x**3-4.43x**2+7.33*x-109

当我计算 y 估计值时，我得到了非常奇怪的数字：

y_实际：

[1] -108.4569 -108.1504 -108.0895 -108.0728 -108.0461 -108.1777 -108.2751 -108.4619 -108.6918 [10] -108.9750 -109.3552 -109.7625 -110.3328 -110.9580 -111.4312 -112.0062 -112.7337 -113.5880 [19]-114.3681

y_预测：

[1] -8935267 -8980331 -9044297 -9115821 -9166614 -9270340 -9355643 -9456574 -9533497 -9602089 [11]-9631113 -9670175 -9715100 -9754453 -9798813 -9851816 -9880888 -9926067 -9940310

这里出了什么问题？

我尝试将 poly 函数的原始变量设置为 TRUE，得到了不同的系数，但问题仍然存在。

---

编辑

数据采用

dput

格式

y_actual <-
c(-108.4569, -108.1504, -108.0895, -108.0728, -108.0461, -108.1777, 
-108.2751, -108.4619, -108.6918, -108.975, -109.3552, -109.7625, 
-110.3328, -110.958, -111.4312, -112.0062, -112.7337, -113.588, 
-114.3681)

y_predicted <-
c(-8935267, -8980331, -9044297, -9115821, -9166614, -9270340, 
-9355643, -9456574, -9533497, -9602089, -9631113, -9670175, -9715100, 
-9754453, -9798813, -9851816, -9880888, -9926067, -9940310)

Answer 1

这是您的代码示例。因为没有提供 x 值，所以我估算了 x 值。我上面的评论是：“使用

raw=TRUE

表示非正交多项式”。

蓝线是预测函数的预测值。红线使用的是拟合系数。请注意，绿线使用相同的系数，但四舍五入为 5 位有效数字，并产生略有不同的曲线。
这是高阶方程的问题，微小的差异乘以就会变成很大的差异。一个混乱的系统。

y_actual <-
   c(-108.4569, -108.1504, -108.0895, -108.0728, -108.0461, -108.1777, 
     -108.2751, -108.4619, -108.6918, -108.975, -109.3552, -109.7625, 
     -110.3328, -110.958, -111.4312, -112.0062, -112.7337, -113.588, 
     -114.3681)

#approximating the x values
x<- rev(seq(-204.5, -197, length.out=19))

fit <- lm(y_actual ~ poly(x, 3, raw=TRUE))
summary(fit)

# Call:
#    lm(formula = y_actual ~ poly(x, 3, raw = TRUE))
# 
# Residuals:
#    Min        1Q    Median        3Q       Max 
# -0.123971 -0.030200 -0.000881  0.033129  0.075555 
# 
# Coefficients:
#    Estimate Std. Error t value Pr(>|t|)  
# (Intercept)             -1.924e+04  1.086e+04  -1.771   0.0968 .
# poly(x, 3, raw = TRUE)1 -2.537e+02  1.623e+02  -1.563   0.1390  
# poly(x, 3, raw = TRUE)2 -1.099e+00  8.087e-01  -1.359   0.1942  
# poly(x, 3, raw = TRUE)3 -1.546e-03  1.343e-03  -1.151   0.2678  
# ---
#    Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
# 
# Residual standard error: 0.05382 on 15 degrees of freedom
# Multiple R-squared:  0.9994,  Adjusted R-squared:  0.9993 
# F-statistic:  8647 on 3 and 15 DF,  p-value: < 2.2e-16

y_predicted <- predict(fit, data.frame(x))

#plotting the prediction
plot(x, y_actual)
lines(x, y_predicted, col="blue", lwd=3)

#plotting the prediction with the coefficients 
y_pred_2 <- fit$coefficients[4]*x**3 + fit$coefficients[3]*x**2 + fit$coefficients[2]*x + fit$coefficients[1]
lines(x, y_pred_2+0.001, col="red")

#round the coefficents
y_pred_3 <- -1.5455E-3*x**3 - 1.09927*x**2 - 253.68*x - 19238
lines(x, y_pred_3, col="green")