I am trying to identify Outiers in a Dataset, where each sample has a different amount of Data points. Meaning the Matrix length differs from sample to sample. Since I am working on a big data set it is really Time consuming to change matrix length manually. I hope to find a solution for that problem here:) Also suggestions for R are welcome!

Here is an example data set.

df1 <-
  structure(list(
    Time = c(5, 17, 18, 24, 4, 8, 9, 10, 16, 15, 4, 13, 9, 17), 
    ID = c(1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3)), 
    row.names = c(NA, -14L), 
    class = c("tbl_df", "tbl", "data.frame" ))

I used the Function:

=IF(OR(B2>QUARTILE($B$2:$B$5;3)+1,5*(QUARTILE($B$2:$B$5;3)-QUARTILE($B$2:$B$5;1));F2<QUARTILE($B$2:$B$51;1)-1,5*(QUARTILE($B$2:$B$5;3)-QUARTILE($B$2:$B$5;1)));1;0)  

But as described above, with that approach I need to change the matrix manually from sample to sample.
The outlier is then identified with "1".

Here are two ways, with quantile/IQR and with the statistics boxplot uses. The latter is fastest.

outlier <- function(x, na.rm = FALSE) {
  qq <- quantile(x, c(1/4, 3/4), na.rm = na.rm)
  iqr <- IQR(x, na.rm = na.rm)
  x < qq[1L] - 1.5*iqr | x > qq[2L] + 1.5*iqr
}
outlier_bp <- function(x, na.rm = FALSE) {
  x %in% boxplot.stats(x)$out
}

sapply(iris[1:4], outlier) |> head()
#>      Sepal.Length Sepal.Width Petal.Length Petal.Width
#> [1,]        FALSE       FALSE        FALSE       FALSE
#> [2,]        FALSE       FALSE        FALSE       FALSE
#> [3,]        FALSE       FALSE        FALSE       FALSE
#> [4,]        FALSE       FALSE        FALSE       FALSE
#> [5,]        FALSE       FALSE        FALSE       FALSE
#> [6,]        FALSE       FALSE        FALSE       FALSE
sapply(iris[1:4], outlier_bp) |> head()
#>      Sepal.Length Sepal.Width Petal.Length Petal.Width
#> [1,]        FALSE       FALSE        FALSE       FALSE
#> [2,]        FALSE       FALSE        FALSE       FALSE
#> [3,]        FALSE       FALSE        FALSE       FALSE
#> [4,]        FALSE       FALSE        FALSE       FALSE
#> [5,]        FALSE       FALSE        FALSE       FALSE
#> [6,]        FALSE       FALSE        FALSE       FALSE

library(microbenchmark)

mb <- microbenchmark(
  qnt = sapply(iris[1:4], outlier),
  bxp = sapply(iris[1:4], outlier_bp)
)
print(mb, order = "median")
#> Unit: microseconds
#>  expr   min     lq    mean median     uq    max neval cld
#>   bxp 233.3 321.15 388.300 365.55 454.15  738.9   100   b
#>   qnt 551.2 742.65 875.958 817.95 949.30 3593.9   100  a

^{Created on 2023-09-25 with reprex v2.0.2}

Edit

After the OP comments, here is a better use case for the function outlier above.
Split Time by ID and apply the function to each sub-vector. This is done with tapply.

i_out <- with(df1, tapply(Time, ID, FUN = outlier))
i_out
#> $`1`
#> [1]  TRUE FALSE FALSE FALSE
#> 
#> $`2`
#> [1] FALSE FALSE FALSE
#> 
#> $`3`
#> [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE

out <- with(df1, tapply(Time, ID, FUN = \(x) x[outlier(x)]))
out
#> $`1`
#> [1] 5
#> 
#> $`2`
#> numeric(0)
#> 
#> $`3`
#> numeric(0)

^{Created on 2023-09-30 with reprex v2.0.2}

Or with a function that returns the outliers, not an index to them,

outlier2 <- function(x, na.rm = FALSE) {
  qq <- quantile(x, c(1/4, 3/4), na.rm = na.rm)
  iqr <- IQR(x, na.rm = na.rm)
  i <- x < qq[1L] - 1.5*iqr | x > qq[2L] + 1.5*iqr
  x[i]
}

with(df1, tapply(Time, ID, FUN = outlier2))
#> $`1`
#> [1] 5
#> 
#> $`2`
#> numeric(0)
#> 
#> $`3`
#> numeric(0)

^{Created on 2023-09-30 with reprex v2.0.2}