Question

在下面我的R代码的最后一行中，我使用optimize()查找使df2函数最小化的ncp_diff。

但是，我想知道是否可以uniroot()代替optimize()来实现这种最小化？

alpha = c(.025, .975); df1 = 3; peta = .3   # The input

f <- function(alpha, q, df1, df2, ncp){     # Notice `ncp` is the unknown
  alpha - suppressWarnings(pf(q = (peta / df1) / ((1 - peta)/df2), df1, df2, ncp, lower = FALSE))
}

ncp <- function(df2){      # Root finding: finds 2 `ncp` for a given `df2`

 b <- sapply(c(alpha[1], alpha[2]),
       function(x) uniroot(f, c(0, 1e7), alpha = x, q = peta, df1 = df1, df2 = df2)[[1]])

 b / (b + (df2 + 4))
}

ncp_diff <- function(df2, target = 0.15){
 the_ncp <- ncp(df2)
  return(abs(abs(the_ncp[2] - the_ncp[1]) - target))
 }

optimize(ncp_diff, c(0, 1000)) ## HERE can I use `uniroot()` instead of `optimize()`

Answer 1

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编辑：

为了使用相同的间隔（0,1000），我们可以寻找一种方法来解决以下情况：下限值和上限值都在数字线的同一侧产生结果。由于这是r中的错误，我们可以通过alpha = c(.025, .975); df1 = 3; peta = .3 # The input f <- function(alpha, q, df1, df2, ncp){ # Notice `ncp` is the unknown alpha - suppressWarnings(pf(q = (peta / df1) / ((1 - peta)/df2), df1, df2, ncp, lower = FALSE)) } ncp <- function(df2){ # Root finding: finds 2 `ncp` for a given `df2` b <- sapply(c(alpha[1], alpha[2]), function(x) uniroot(f, c(0, 1e7), alpha = x, q = peta, df1 = df1, df2 = df2)[[1]]) b / (b + (df2 + 4)) } ncp_diff <- function(df2, target = 0.15){ the_ncp <- ncp(df2) return((the_ncp[2] - the_ncp[1]) - target) } uniroot(ncp_diff, c(100, 1000)) # $root [1] 336.3956 $f.root [1] 3.74663e-09 $iter [1] 7 $init.it [1] NA $estim.prec [1] 6.103516e-05

tryCatch

我们可以使用uniroot（）而不是optimize（）来实现R中的这种最小化吗？

1 个答案: