The rows of data frame "pars" hold the two parameters defining logistical curves:

library(ggplot2)
library(purrr)

pars <- data.frame(
  diff = c(-1.5, 2.5),
  disc = c(1.2, 2.5)
)

These two curves can be plotted with map() and ggplot() like this.

icc <- function(x) map(
  1:nrow(pars),
  ~ stat_function(fun = function(x)
    (exp(pars$disc[.x]*(x - pars$diff[.x])))/(1   exp(pars$disc[.x]*(x - pars$diff[.x]))))
)
ggplot(data.frame(x = -5 : 5))  
  aes(x)  
  icc()

The corresponding derivations can be plotted like this:

disc1 <- 1.2
disc2 <- 2.5
diff1 <- -1.5
diff2 <- 2.5

icc1 <- function(x) (exp(disc1*(x - diff1)))/(1   exp(disc1*(x - diff1)))
icc2 <- function(x) (exp(disc2*(x - diff2)))/(1   exp(disc2*(x - diff2)))

info1 <- Deriv(icc1, "x")
info2 <- Deriv(icc2, "x")

ggplot(data.frame(x = -5 : 5))  
  aes(x)  
  stat_function(fun = info1)  
  stat_function(fun = info2)

However, I'd like to use a more generic approach with preferably purrr() for the derivations as well since I'll need a function for a varying number of curves. Maybe there's a solution with pmap() that could iterate through a data frame with parameters and apply function and derivation to each row. Unfortunately, I was unlucky so far. I am extremely grateful for any helpful answers.

CodePudding user response：

One option may look like so:

1. I have put the parameters for your curves in a data.frame
2. Making use of a function factory and pmap to loop over the params df to create a list of your icc functions.

The rest is pretty straighforward.

3. Loop over the list of functions to get the derivatives.

4. Use map to add the stat_function layers.

library(ggplot2)
library(Deriv)
#> Warning: package 'Deriv' was built under R version 4.1.2
library(purrr)

df <- data.frame(
  disc = c(1.2, 2.5),
  diff = c(-1.5, 2.5)
)

icc <- function(disc, diff) {
  function(x) (exp(disc*(x - diff)))/(1   exp(disc*(x - diff)))
}

icc_list <- pmap(df, function(disc, diff) icc(disc, diff))
info_list <- map(icc_list, Deriv, "x")

ggplot(data.frame(x = -5 : 5))  
  aes(x)  
  map(info_list, ~ stat_function(fun = .x))

CodePudding user response：

You could also do it in a very similar way to the icc plotting:

deriv_icc <- function(x) map(
  1:nrow(pars),
  ~ stat_function(fun = function(x){
    exb <- exp(pars$disc[.x]*(x-pars$diff[.x]))
    pars$disc[.x]*(exb/(1 exb) - exb^2/(1 exb)^2)
    }))

ggplot(data.frame(x = -5 : 5))  
  aes(x)  
  deriv_icc()

This simply recognizes the fact that the derivative of the logistic CDF for this problem is the discrimination parameter times the logistic PDF: