R 求解最佳拟合多项式并绘制下拉线

我在Windows10上使用R3.3.1（64位）。我有一个x-y数据集，我用一个二阶多项式拟合。我想求解y=4时x的最佳拟合多项式，并绘制从y=4到x轴的下拉线

这将在dataframe v1中生成数据：

v1 <- structure(list(x = c(-5.2549, -3.4893, -3.5909, -2.5546, -3.7247, 
-5.1733, -3.3451, -2.8993, -2.6835, -3.9495, -4.9649, -2.8438, 
-4.6926, -3.4768, -3.1221, -4.8175, -4.5641, -3.549, -3.08, -2.4153, 
-2.9882, -3.4045, -4.6394, -3.3404, -2.6728, -3.3517, -2.6098, 
-3.7733, -4.051, -2.9385, -4.5024, -4.59, -4.5617, -4.0658, -2.4986, 
-3.7559, -4.245, -4.8045, -4.6615, -4.0696, -4.6638, -4.6505, 
-3.7978, -4.5649, -5.7669, -4.519, -3.8561, -3.779, -3.0549, 
-3.1241, -2.1423, -3.2759, -4.224, -4.028, -3.3412, -2.8832, 
-3.3866, -0.1852, -3.3763, -4.317, -5.3607, -3.3398, -1.9087, 
-4.431, -3.7535, -3.2545, -0.806, -3.1419, -3.7269, -3.4853, 
-4.3129, -2.8891, -3.0572, -5.3309, -2.5837, -4.1128, -4.6631, 
-3.4695, -4.1045, -7.064, -5.1681, -6.4866, -2.7522, -4.6305, 
-4.2957, -3.7552, -4.9482, -5.6452, -6.0302, -5.3244, -3.9819, 
-3.8123, -5.3085, -5.6096, -6.4557), y = c(0.99, 0.56, 0.43, 
2.31, 0.31, 0.59, 0.62, 1.65, 2.12, 0.1, 0.24, 1.68, 0.09, 0.59, 
1.23, 0.4, 0.36, 0.49, 1.41, 3.29, 1.22, 0.56, 0.1, 0.67, 2.38, 
0.43, 1.56, 0.07, 0.08, 1.53, -0.01, 0.12, 0.1, 0.04, 3.42, 0.23, 
0, 0.34, 0.15, 0.03, 0.19, 0.17, 0.2, 0.09, 2.3, 0.07, 0.15, 
0.18, 1.07, 1.21, 3.4, 0.8, -0.04, 0.02, 0.74, 1.59, 0.71, 10.64, 
0.64, -0.01, 1.06, 0.81, 4.58, 0.01, 0.14, 0.59, 7.35, 0.63, 
0.17, 0.38, -0.08, 1.1, 0.89, 0.94, 1.52, 0.01, 0.1, 0.38, 0.02, 
7.76, 0.72, 4.1, 1.36, 0.13, -0.02, 0.13, 0.42, 1.49, 2.64, 1.01, 
0.08, 0.22, 1.01, 1.53, 4.39)), .Names = c("x", "y"), class = "data.frame", row.names = c(NA, 
-95L))

---

v1 4，因此我想使用下拉行突出显示规格零件中将产生的x值范围。
您有一个二次方程

0.73198 * x^2 + 6.08073 * x + 12.75558 = 4
OR
0.73198 * x^2 + 6.08073 * x + 8.75558 = 0

你可以用二次公式来解析解这个问题。R给出了两个根：

(-6.08073 + sqrt(6.08073^2 -4*0.73198 * 8.75558)) / (2 * 0.73198)
[1] -1.853392
(-6.08073 - sqrt(6.08073^2 -4*0.73198 * 8.75558)) / (2 * 0.73198)
[1] -6.453843

abline（v=c（-1.853392，-6.453843））

您可以使用二次公式计算值：

betas <- coef(fit2)    # get coefficients
betas[1] <- betas[1] - 4    # adjust intercept to look for values where y = 4

# note degree increases, so betas[1] is c, etc.
betas
##             (Intercept) poly(x, 2, raw = TRUE)1 poly(x, 2, raw = TRUE)2 
##               8.7555833               6.0807302               0.7319848 

solns <- c((-betas[2] + sqrt(betas[2]^2 - 4 * betas[3] * betas[1])) / (2 * betas[3]), 
           (-betas[2] - sqrt(betas[2]^2 - 4 * betas[3] * betas[1])) / (2 * betas[3]))

solns
## poly(x, 2, raw = TRUE)1 poly(x, 2, raw = TRUE)1 
##               -1.853398               -6.453783 

segments(solns, -1, solns, 4, col = 'green')    # add segments to graph

因为它返回一个复数向量，如果你想把它传递给
段
，我完全理解这个简单的二次多项式有一个解析解。我给你们展示数值解的原因是你们在回归设置中问这个问题。当你有更复杂的回归曲线时，数值解通常是你的解

下面我将使用
uniroot
函数。如果您不熟悉，请先阅读以下简短答案：



这是用代码生成的绘图。你就快到了。这是一个根查找问题，您可以在数字上使用
uniroot
。让我们定义一个函数：

f <- function (x) {
  ## subtract 4
  predict(fit2, newdata = data.frame(x = x)) - 4
  }

现在，您可以将垂直线从这些点向下放置到x轴：

y1 <- f(x1) + 4  ## add 4 back
y2 <- f(x2) + 4  

abline(h = 0, col = 4)  ## x-axis
segments(x1, 0, x1, y1, lty = 2)
segments(x2, 0, x2, y2, lty = 2)

y1这里还有一个基于

已经提出了许多解决方案，下面是另一个

显然，我们有兴趣找到满足多项式（二次）方程
a_0+a_1.x+a_2.x^2=4
的
x
值，其中
a_0，a_1，a_2
是拟合多项式的系数。我们可以将方程改写为标准二次方程
ax^2+bx+c=0
，并使用拟合多项式的系数和多项式回归，使用
Sridhar的
公式求根，如下所示：



f <- function (x) {
  ## subtract 4
  predict(fit2, newdata = data.frame(x = x)) - 4
  }

x1 <- uniroot(f, c(-7, -6))$root
#[1] -6.453769

x2 <- uniroot(f, c(-3, -1))$root
#[1] -1.853406

y1 <- f(x1) + 4  ## add 4 back
y2 <- f(x2) + 4  

abline(h = 0, col = 4)  ## x-axis
segments(x1, 0, x1, y1, lty = 2)
segments(x2, 0, x2, y2, lty = 2)

attach(v1)
fit2 = lm(y~poly(x,2,raw=TRUE))
xx = seq(-8,0, length=50)

vector1 = predict(fit2, data.frame(x=xx)) 
vector2= replicate(length(vector1),4)

# Find points where vector1 is above vector2.
above = vector1 > vector2

# Points always intersect when above=TRUE, then FALSE or reverse
intersect.points = which(diff(above)!=0)    

# Find the slopes for each line segment.
vector1.slopes = vector1[intersect.points+1] - vector1[intersect.points]
vector2.slopes = vector2[intersect.points+1] - vector2[intersect.points]

# Find the intersection for each segment.
x.points = intersect.points + ((vector2[intersect.points] - vector1[intersect.points]) / (vector1.slopes-vector2.slopes))
y.points = vector1[intersect.points] + (vector1.slopes*(x.points-intersect.points))

#Scale x.points to the axis value of xx
x.points = xx[1] + ((x.points - 1)/(49))*(xx[50]-xx[1])

plot(xx, y = vector1, type= "l", col = "blue")
points(x,y,pch = 20)
lines(x = c(x.points[1],x.points[1]), y = c(0,y.points[1]), col='red')
lines(x = c(x.points[2],x.points[2]), y = c(0,y.points[2]), col='red')

a <- fit2$coefficients[3]
b <- fit2$coefficients[2]
c <- fit2$coefficients[1] - 4

as.numeric((-b + sqrt(b^2-4*a*c)) / (2*a))
#[1] -1.853398
as.numeric((-b-+ sqrt(b^2-4*a*c)) / (2*a))
#[1] -6.453783

a <- fit2$coefficients  # fitted quadratic polynomial coefficients

f <- function(x) {
  as.numeric(a[1] + a[2]*x + a[3]*x^2-4)
}

df <- function(x) {
  as.numeric(a[2] + 2*a[3]*x)
} 

Newton.Raphson <- function(x0) {
  eps <- 1e-6
  x <- x0
  while(TRUE) {
    x <- x0 - f(x0) / df(x0)
    if (abs(x - x0) < eps) {
      return(x0)
    }
    x0 <- x
  }
}

t1 <- Sys.time()
x1 <- Newton.Raphson(-10)
x2 <- Newton.Raphson(10)
x1
#[1] -6.453783
x2
#[1] -1.853398
s2
print(paste('time taken to compute the roots:' ,Sys.time() - t1))
#[1] "time taken to compute the roots: 0.0160109996795654"
points(x1, 4, pch=19, col='green')
points(x2, 4, pch=19, col='green')
abline(v=x1, col='green')
abline(v=x2, col='green')