R coding: creating a name variable to use with select

Question

I am wondering if this is correct:

VAR <- c("City")
data %>% 
 select(salary, age, VAR)
 melt(id.vars = c("Salary2", "Age2","City")) -> finalData

I get the message:

Note: Using an external vector in selections is ambiguous.
i Use `all_of(VAR)` instead of `VAR` to silence this message.
i See <https://tidyselect.r-lib.org/reference/faq-external-vector.html>.

The resulting finalData seems correct, but I wonder if this is the way to go. Thanks!

Answer 1

Answer

Like the error says, use all_of (imported by tidyverse from tidyselect):

var <- "Sepal.Length"
iris %>% 
 select(tidyselect::all_of(var)) %>% 
 str
# 'data.frame': 150 obs. of 1 variable:
# $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...

Rationale

Let's say we want to select a column:

iris[, "Sepal.Length"]

In base R, we can easily select columns with variables:

var <- "Sepal.Length"
iris[, var]

---

However, you are using tidyverse functions. For example, select is from the dplyr package. The tidyverse packages rely on a special type of non-standard evaluation. As such, the correct way of selecting the "Sepal.Length" column is:

select(iris, Sepal.Length) 
# more typically written as:
iris %>% select(Sepal.Length)

As you can see, we are no longer using quotation marks. This is similar to iris$Sepal.Length in base R. A more extensive explanation of design choices in dplyr can be found here. The core reasons for designing tidyverse this way are that it is more intuitive to use, and that is often faster to write.

---

Let's consider your case:

VAR <- c("City")
iris %>% select(VAR)

What it is doing, is looking for "VAR" inside iris. Since it cannot find it, it will then evaluate VAR, which yield "City". It will then look for "City", which in this case it won't find. It does work if we specify a column that is present (like in your example):

VAR <- c("Sepal.Length")
iris %>% select(VAR)

---

Where it goes wrong

So, what happens if we have a column called VAR, and supply a variable called VAR?

iris$VAR <- 1
VAR <- c("Sepal.Length")
iris %>% select(VAR) %>% str
#'data.frame': 150 obs. of 1 variable:
# $ VAR: num 1 1 1 1 1 1 1 1 1 1 ...

Here, it will return the column with ones, NOT what we intended. Using all_of, we can explicitly indicate that select should consider an EXTERNAL variable. This is to ensure that you get the results that you expect:

iris$VAR <- 1
VAR <- c("Sepal.Length")
iris %>% select(all_of(VAR)) %>% str
#'data.frame': 150 obs. of 1 variable:
# $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...

• Elina Bochkova: Thanks, however, I got confused as to which package I am using. I use the select function, from dplyr, not tidyverse. So It should work in my case. Also, in my case VAR is not the name of any column, its value is.
• slamballais: dplyr is part of tidyverse. If you type ?select, you'll find that dplyr actually imports select from tidyselect :)
• slamballais: And I realise that VAR is not one of your columns. The point I was making is that it is dangerous not to use all_of( ) because one of your columns may have the same name. Using all_of( ) ensures that your code behaves as you expect it to. It is safer to include it, although in your specific example it is not needed.
• Elina Bochkova: Thanks, all clear now :)

Answer 2

Here the VAR created another instance that is not part of data. Remember in R, everything is a vector.

• Elina Bochkova: Thanks. How to create a variable that can replace tens of others one so I don't have to replace manually?
• slamballais: This answer is not correct; running select(VAR) doesn't create anything. It just happens to evaluate VAR (since the column "VAR" is not present) to then run select(City), which is a column in the original dataset. If it had not been a column, it would give an error.
• Balakumar Natarajan: Thanks! now I too have got it cleared.
• J. Mini: Not everything in R is a vector. For example, functions aren't.