如何找到在R中改变一个常数的函数的最大值? [英] How to find the maximum value of a function changing one constant value in R?

问题描述

我有以下线性化的图:

a和b是带有数据的向量，c是常数.任务是找到一个c值，该值使R^2最大化以进行线性回归

a and b are vectors with data, c is a constant. The task is find a value of c that maximizes R^2 for a linear regression

a <- c(56.60, 37.56, 15.80, 27.65, 9.20, 5.05, 3.54)
b <- c(23.18, 13.49, 10.45, 7.24, 5.44, 4.19, 3.38)
c <- 1

x <- log(a)
y <- log((c*(a/b))-1)

rsq <- function(x, y) summary(lm(y~x))$r.squared
rsq(x, y)

optimise(rsq, maximum = TRUE)

推荐答案

这有效:

a <- c(56.60, 37.56, 15.80, 27.65, 9.20, 5.05, 3.54)
b <- c(23.18, 13.49, 10.45, 7.24, 5.44, 4.19, 3.38)

rsq <- function(c) {
  x <- log(a)
  y <- log((c*(a/b))-1)
  stopifnot(all(is.finite(y)))  
  summary(lm(y ~ x))$r.squared
}
optimise(rsq, maximum = TRUE, interval=c(0.8, 3))

.

# > optimise(rsq, maximum = TRUE, interval=c(0.8, 3))
# $maximum
# [1] 1.082352
# 
# $objective
# [1] 0.8093781

您也可以有一个不错的情节:

You can also have a nice plot:

plot(Vectorize(rsq), .8, 3)
grid()

要为观察设置条件

rsq <- function(c) {
  xy <- data.frame(a=a, y=(c*(a/b))-1)
  summary(lm(log(y) ~ log(a), data=subset(xy, y>0)))$r.squared
}
optimise(rsq, maximum = TRUE, interval=c(0.1, 3))

...和有趣的情节:

... and the interesting plot:

plot(Vectorize(rsq), .3, 1.5)
grid()

rsq(0.4) 

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