Find max number from row then divide and subtract previous number-CodePudding

For example:

v1 <- c(10, 11, 20, 30, 40, 50, 30, 25, 20, 10)

In v1 50 is a max number, I need the below result

result <- c(0, 0, 0, 0, 0, (50/40)-1, (30/50)-1, (25/30)-1, (20/25)-1, (10/20)-1)

The final array will look like this

c(0, 0, 0, 0, 0, 0.25, -0.4, -0.16, -0.2, -0.5)

CodePudding user response：

Here is an approach that leverages data.table::shift()

f <- function(v) {
  mx = which.max(v); 
  c(rep(0,mx-1), (v/shift(v)-1)[mx:length(v)])
}

Usage:

f(v1)

Output:

 [1]  0.0000000  0.0000000  0.0000000  0.0000000  0.0000000  0.2500000 -0.4000000 -0.1666667 -0.2000000 -0.5000000

CodePudding user response：

in Base R we will do:

idx <- which.max(v1)
index <- seq_along(v1) >= idx - 1
c(numeric(idx - 1), tail(v1[index],-1)/head(v1[index], -1) - 1)

[1]  0.0000000  0.0000000  0.0000000  0.0000000  0.0000000  0.2500000 -0.4000000
 [8] -0.1666667 -0.2000000 -0.5000000

CodePudding user response：

We may also do

library(dplyr)
replace((v1/lag(v1))-1, seq_along(v1) < which.max(v1), 0)

-output

[1]  0.0000000  0.0000000  0.0000000  0.0000000  0.0000000  0.2500000 -0.4000000 -0.1666667 -0.2000000 -0.5000000

CodePudding user response：

You can do this in the following, somewhat clunky, way using a combination of base R's which.max() and dplyr's lag() function and some indexing:

c(rep(0,which.max(v1) - 1), v1[which.max(v1):length(v1)] / (dplyr::lag(v1[(which.max(v1)-1):length(v1)]))[-1]-1)

Output:

#[1]  0.0000000  0.0000000  0.0000000  0.0000000  0.0000000  0.2500000 -0.4000000 -0.1666667 -0.2000000
#[10] -0.5000000

To break this down to more palatable code:

numerator <- v1[which.max(v1):length(v1)]
denominator <- lag(v1[(which.max(v1) - 1):length(v1)])[-1]
leading_zeros <- rep(0, which.max(v1) - 1)

c(leading_zeros, numerator/denominator - 1)

#[1]  0.0000000  0.0000000  0.0000000  0.0000000  0.0000000  0.2500000 -0.4000000 -0.1666667 -0.2000000
#[10] -0.5000000