Examine if a value is in an interval using R-CodePudding

Having the following vector:

t <- c(2, 6, 8, 20, 22, 30, 40, 45, 60)

I would like to find the values that fall between the following intervals:

g <- list(c(1,20), c(20, 40))

The desired output is:

1, 20 c(2, 6, 8)
20, 40 c(20, 22, 30)

Using the dplyr library, I do the following:

library(dplyr)
for(i in t){
    for(h in g){
        if(between(i, h[[1]], h[[2]])==TRUE){print(c(i, h[[1]], h[[2]]))}
}}

Is there a better way of doing this in R?

CodePudding user response：

We can loop over the list 'g' and extract the 't' elements based on the first and second values by creating a logical vector with >/< and extract the elements of 't'

lapply(g, function(x) t[t >= x[1] & t  < x[2]])

-output

[[1]]
[1] 2 6 8

[[2]]
[1] 20 22 30

CodePudding user response：

library(purrr)
library(dplyr)

map(g,~keep(t,between(t,.[1],.[2])))

[[1]]
[1]  2  6  8 20

[[2]]
[1] 20 22 30 40

CodePudding user response：

You may find findInterval() from base R useful:

g <- c(1, 20, 40)
t <- c(2, 6, 8, 20, 22, 30, 40, 45, 60)
findInterval(t, g)
#> [1] 1 1 1 2 2 2 3 3 3

So t[1], t[2] and t[3] are in the first interval, t[4], t[5] and t[6] in the second, and t[7], t[8] and t[9] the third (meaning that these values are bigger than the right end point of the second interval.) If you had values lower than one they would be labelled by 0:

t2 <- c(-1, 0, 2, 6, 8, 20, 22, 30, 40, 45, 60)
findInterval(t2, g)
#>  [1] 0 0 1 1 1 2 2 2 3 3 3

You can save the result of findInterval() as e.g. y and use which(y==1) to find which entries correspond to the first interval.

CodePudding user response：

We can try cut is.na like below

lapply(
  g,
  function(x) {
    t[!is.na(cut(t, x, include.lowest = TRUE))]
  }
)

which gives

[[1]]
[1]  2  6  8 20

[[2]]
[1] 20 22 30 40