问题描述
虽然我认为这是一个基本问题，但我似乎无法找到如何在 R 中计算:
Although I think this is a basic question, I can't seem to find out how to calculate this in R:
2 个或多个正态分布(拟合在直方图上)的交点(我需要 x 值)，例如具有以下参数:
the point of intersection (I need the x-value) of 2 or more normal distributions (fitted on a histogram) which have for example the following parameters:
d=data.frame(mod=c(1,2),mean=c(14,16),sd=c(0.9,0.6),prop=c(0.6,0.4))
用我的 2 条曲线的均值和标准差，并支撑每个 mod 对分布的贡献比例.
With the mean and standard deviation of my 2 curves, and prop the proportions of contribution of each mod to the distribution.
推荐答案
你可以使用uniroot:
f <- function(x) dnorm(x, m=14, sd=0.9) * .6 - dnorm(x, m=16, sd=0.6) * .4
uniroot(f, interval=c(12, 16))
$root
[1] 15.19999
$f.root
[1] 2.557858e-06
$iter
[1] 5
$estim.prec
[1] 6.103516e-05
ETA 一些说明:
ETA some exposition:
uniroot 是一个单变量根查找器，即给定一个变量 x 的函数 f，它找到 x 的值 求解方程 f(x) = 0.
uniroot is a univariate root finder, ie given a function f of one variable x, it finds the value of x that solves the equation f(x) = 0.
要使用它，您需要提供函数 f，以及假定解值所在的区间.在这种情况下，f 只是两个密度的差值；它们相交的点将是 f 为零的地方.在这个例子中，我通过绘制一个图并看到它们在 x=15 附近相交，得到了区间 (12, 16).
To use it, you supply the function f, along with an interval within which the solution value is assumed to lie. In this case, f is just the difference between the two densities; the point where they intersect will be where f is zero. I got the interval (12, 16) in this example by making a plot and seeing that they intersected around x=15.
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