Function in R to Identify Out of Specification Values

Question

I have the following dataframe called data:

I am trying to create a function which determines out of specification values of "Assayperc" as follows:

is_oos <- function(x){
  return(x < (data$Assayperc < 0.90 && data$Assayperc > 1.10))
}

In other words out of specifiation are Assayperc values less than 90% or greater than 110%.

I then want to use the "is_oos" function to add a out of specificatio column to the data frame as follows:

data <- data %>%
  group_by(Manufacturer) %>%
  mutate(OutOfSpec = ifelse(is_oos(Assayperc), BatchNumber, ""))

The output is as follows which is not as expected as all values are being labelled as out of specification:

Kindly assist with the function.

Regards Chris

if you can use this `dput(df)` that would be helpful for other images are hard to replicate — PesKchan, Sep 16 '22 at 11:04

diomedesdata · Accepted Answer · 2022-09-16T12:10:50.007

0

In other words out of specifiation are Assayperc values less than 90% or greater than 110%.

As pointed out by Robert, you need | rather than &&.

data <- data %>%
  group_by(Manufacturer) %>%
  mutate(OutOfSpec = ifelse(Assayperc < 0.90 | Assayperc > 1.10, BatchNumber, ""))

Please update the question in line with the comments, it is very difficult to read as is.

Just to add an alternative, you could do this with data.table as follows, which would update by reference. I use fifelse below (fast ifelse).

library(data.table)
setDT(data)

data[, OutOfSpec := fifelse(Assayperc < 0.90 | Assayperc > 1.10, BatchNumber, ""),
 by = .(Manufacturer)]

edited Sep 16 '22 at 12:10

answered Sep 16 '22 at 11:58

diomedesdata

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I have some doubts `Assayperc < 0.90 && Assayperc > 1.10` can ever be true. I guess there should be `|` instead of `&&` but that's to much guessing for me to try to answer. – Robert Hacken Sep 16 '22 at 12:03
@RobertHacken Indeed, I just realised the same thing. On some test data that I assume is similar to what OP intended, this seems to work. – diomedesdata Sep 16 '22 at 12:04
@user5684073 no worries, and going off what is in the question I agree with Konrad that the grouping will make no difference, so unless there is something going on that isn't included in the question, feel free to remove that step. – diomedesdata Sep 17 '22 at 02:54

score 0 · Answer 2 · answered Sep 16 '22 at 12:15

Your is_oos function does something weird: you first compute a logical value based on the vector data$Assayperc, and then you compare that logical value to the argument x.

What you probably intended was to compute a logical value based on the argument x instead:

is_oos <- function (x) {
  x < 0.90 | x > 1.10
}

Note that I’ve also replaced && with | as noted in the other answer (and it needs to be | rather than || since we need the computation to be vectorised). I also removed the return() call, since it’s misleading clutter.

Your usage of this function is essentially correct, but note that group_by has no effect here and is therefore unnecessary:

data <- data %>%
  mutate(OutOfSpec = ifelse(is_oos(Assayperc), BatchNumber, ""))

Function in R to Identify Out of Specification Values

2 Answers2