Question

的代码

x <- c(32,64,96,118,126,144,152.5,158)
y <- c(99.5,104.8,108.5,100,86,64,35.3,15)
plot(x,y,pch=19)

fit <- lm(y~poly(x,4,raw=TRUE))
summary(fit)

Call:lm(formula = y ~ poly(x, 4, raw = TRUE))

Residuals:
      1       2       3       4       5       6       7       8 
 0.1242 -0.6912  1.6355  1.4491 -5.1240  4.0360 -0.4692 -0.9604 

Coefficients:
                      Estimate Std. Error t value Pr(>|t|)
(Intercept)              7.474e+01  5.473e+01   1.366    0.265
poly(x, 4, raw = TRUE)1  1.426e+00  3.095e+00   0.461    0.676
poly(x, 4, raw = TRUE)2 -2.854e-02  5.729e-02  -0.498    0.653
poly(x, 4, raw = TRUE)3  2.878e-04  4.278e-04   0.673    0.549
poly(x, 4, raw = TRUE)4 -1.134e-06  1.113e-06  -1.018    0.384

Residual standard error: 4.04 on 3 degrees of freedom
Multiple R-squared:  0.9943,    Adjusted R-squared:  0.9868 
F-statistic: 131.5 on 4 and 3 DF,  p-value: 0.001064

鉴于我们认为这非常合适，我想知道拟合的精确多项式方程是什么。有没有办法实现这个目标？

[edit]

另外一个问题，我看到p值都倾向于显示自变量不够显着，但我们看到一个很好的拟合，有人可以解释

Answer 1

您可以使用包polynomial中的函数polynom来编写等式：

library(polynom)

x <- c(32,64,96,118,126,144,152.5,158)
y <- c(99.5,104.8,108.5,100,86,64,35.3,15)
plot(x,y,pch=19)

fit <- lm(y~poly(x,4,raw=TRUE))

p0 <- polynomial(coef(fit))
p0
# 74.73766 + 1.425813*x - 0.0285437*x^2 + 0.0002877714*x^3 - 1.133744e-06*x^4

使用signif来舍入系数：

p0 <- polynomial(signif(coef(fit), 3))
p0
# 74.7 + 1.43*x - 0.0285*x^2 + 0.000288*x^3 - 1.13e-06*x^4

您可以使用p0：

进行一些游戏

f0 <- as.function(p0)
f0(x)
# [1]  99.37580 105.49117 106.86449  98.55089  91.12402  59.96402  35.76922
# [8]  15.96039


predict(fit)
#         1         2         3         4         5         6         7         8 
#  99.37580 105.49117 106.86449  98.55089  91.12402  59.96402  35.76922  15.96039 


plot(x, y)
lines(x, f0(x), col = "grey", lwd = 2)        # bold grey line
lines(x, predict(fit), col = "red", lty = 2)  # dashed red line

Answer 2

摘要中有系数：

   f <- function(x) { 
         return(7.473766e+01 + 1.425813e+00*x -2.854370e-02*x^2 + 2.877714e-04*x^3 - 1.133744e-06*x^4 ) 
   }  
   plot(x, y)
   lines(x, f(x), col="red")

确定R中非线性回归的拟合方程

2 个答案: