在mutate dplyr r内耦合一个函数：计算二阶导数

Question

I would like to calculate the following formula on all my data grouped_by(id)

$$\frac{y''}{((1+(y')^2)^{3/2}}$$

I'm getting stuck on passing a function through mutate because it gives me unequal column sizes.

Sample data is:

require(dplyr)

df<-data.frame(Time=seq(65),
               SkinTemp=rnorm(65,37,0.5),
               id=rep(1:10,c(5,4,10,6,7,8,9,8,4,4)))

First derivative (OK):

df <-df %>% group_by(id) %>% filter(n() > 1) %>%
  mutate(first_d = SkinTemp - lag(SkinTemp))
df<-as.data.frame(df)  #Back to data frame

Second derivative (stuck):

drv <- function(x, y) diff(y) / diff(x) #Defined this to pass to mutate

middle_pts <- function(x) x[-1] - diff(x) / 2 

second_d <-df %>% group_by(id)  %>%
  mutate(second_d = drv(middle_pts(Time), drv(Time, SkinTemp)))

Answer 1

The diff returns a vector of length one less than the original vector . The function mutate requires the column to return the same length. Therefore, we need to do some appending with NA or value of choice

drv <- function(x, y) c(NA, (diff(y) /diff(x))) 
middle_pts <- function(x) c(NA, (x[-1] - diff(x) /2))

Now, the OP's code should work

df %>%
   group_by(id)  %>%
   mutate(second_d = drv(middle_pts(Time), drv(Time, SkinTemp)))