放大一条曲线，使其可以在R图中沿着另一条曲线显示

问题描述

我有一个R图，我想在其中显示 IF 红色 曲线（现在位于底部）图未正确显示）乘以常数，它可以 匹配 蓝色 曲线当前正在显示。

I have an R plot in which I want to show that IF the "red" curve (which is now at the bottom of plot not showing correctly) be multiplied by a constant, it can match the "blue" curve currently showing.

我想知道如何最好地 扩大规模 红色曲线，以使其与蓝色曲线完全 匹配 ？

I'm wondering how best I can scale up the "red" curve so that it exactly matches the "blue" curve?

（我的R代码在图片下方。）

(My R code is provided below the picture.)

这是我的R代码：

SIGMA = 2                  # Population SIGMA known
observations = seq(1, 30)  # observations drawn
n = length(observations)   # number of observations
x_bar = mean(observations) # mean of observations
SE = SIGMA / sqrt(n)       # 'S'tandard 'E'rror of the mean
x.min = x_bar - 4*SE
x.max = x_bar + 4*SE

Like = function(x) sapply(lapply(x, dnorm, x = observations, SIGMA), prod) # multiplication of densities to obtain Likelihood values

curve(dnorm(x, x_bar, SE), from = x.min, to = x.max, col = 'blue', lwd = 3, lty = 2 ) # Sampling Distribution of x_bar
curve(Like, from = x.min, to = x.max, col = 'red', lwd = 3, add = T) # Likelihood function of MU

推荐答案

基本上，我们需要缩放 cc2 $ y 按比例缩放，以使 cc2 $ y 的缩放值具有与<$ c $相同的范围（最小和最大） c> cc1 $ y 。我已经使用 scales 包的 rescale 函数来做到这一点

Basically, we need to scale values of cc2$y proportionally such that the scaled values of cc2$y have the same range (minimum and maximum) as cc1$y. I have used rescale function of scales package to do that

# Sampling Distribution of x_bar
cc1 = curve(dnorm(x, x_bar, SE), from = x.min, to = x.max, col = 'blue', lwd = 3, lty = 2 )

# Likelihood function of MU
cc2 = curve(Like, from = x.min, to = x.max, col = 'red', lwd = 3, add = T)

library(scales)
scale_factor = mean(rescale(cc2$y, range(cc1$y)) / cc2$y) #APPROXIMATE

plot(cc1, type = "l")
lines(cc2$x, cc2$y * scale_factor, col = "red")

这里是 rescale2 从 rescale 修改而来的缩放库的功能，如果您想在不加载库的情况下使用它

Here is the rescale2 modified from rescale function of scales library if you want to use it without loading the library

rescale2 = function (x, to = c(0, 1))
{
    (x - min(x))/diff(range(x)) * diff(to) + to[1]
}

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