Using Tidy Eval with dplyr::filter

· by Nic Crane · Read in about 5 min · (872 words) ·

I previously blogged about using tidy eval with dplyr::mutate, and found that post handy to refer back to. I still haven’t got round to having an in-depth look at the principles of tidy eval, so instead I’m continuing to explore problems as and when they come up. In this post, I’ll be taking a look at using tidy eval with dplyr::filter. Once again, I’ll be using the iris dataset to create examples that should be simple to follow.

I’ve updated this blog post to reflect a number of conversations I’ve had via Twitter since publishing it. The one thing I’d say is that if, as a newcomer to tidy eval, this is hard to get your head around the theory, hopefully the code examples should at least be useful!

Typical Usage

Let’s start off with the typical way of using filter. If I wanted to return just the rows from the iris dataset that contain the value “versicolor” in the column called “Species”, I might run the following line of code:

filter(iris, Species == "versicolor") %>%
  head()
##   Sepal.Length Sepal.Width Petal.Length Petal.Width    Species
## 1          7.0         3.2          4.7         1.4 versicolor
## 2          6.4         3.2          4.5         1.5 versicolor
## 3          6.9         3.1          4.9         1.5 versicolor
## 4          5.5         2.3          4.0         1.3 versicolor
## 5          6.5         2.8          4.6         1.5 versicolor
## 6          5.7         2.8          4.5         1.3 versicolor

The input to the filter function, Species == "versicolor" is an expression which is evaluated in the context of the relevant data frame.

As a recap of how dplyr works (thanks to @_lionelhenry for this): “dplyr operates on name-expression pairs. The names are purely quoted while the expressions are quoted and then evaluated in the data frame. Unquoting provides an escape hatch (programmability) for both names and expressions.”

Note that there is a terminology difference here between base R and dplyr (thanks again @lionelhenry for the clarification):

“…base R treats names and symbols as synonyms (e.g as.name ()) but we prefer to keep the term names for the strings returned by names()”

Therefore, instead of providing a string directly to filter, we could also use a symbol mySpecies, which R will evaluate when it evaluates the expression, as in the example below.

mySpecies <- "versicolor"
filter(iris, Species == mySpecies) %>%
  head()
##   Sepal.Length Sepal.Width Petal.Length Petal.Width    Species
## 1          7.0         3.2          4.7         1.4 versicolor
## 2          6.4         3.2          4.5         1.5 versicolor
## 3          6.9         3.1          4.9         1.5 versicolor
## 4          5.5         2.3          4.0         1.3 versicolor
## 5          6.5         2.8          4.6         1.5 versicolor
## 6          5.7         2.8          4.5         1.3 versicolor

We get exactly the same result as before. The bigger challenge comes when we want to use a variable in the place of the name, and that’s what I’ll be looking at in the next section.

Advanced Usage

I’m going to start off with what doesn’t work.

I might naively run the two lines of code below, and expect to get identical results to those in the previous section.

a <- "Species"
filter(iris, a == "versicolor") %>%
  head()
## [1] Sepal.Length Sepal.Width  Petal.Length Petal.Width  Species     
## <0 rows> (or 0-length row.names)

However, this hasn’t worked and I get an empty data frame returned. This is because dplyr has tried to filter the iris dataset by rows which have the value “versicolor” in a column called “a”, and so we need to take a different approach.

The next thing I might try is unquoting a. The !! (bang bang) operator can be used to unquote things, so let’s give it a try.

Bang Bang

a <- "Species"
filter(iris, !!a == "versicolor") %>%
  head()
## [1] Sepal.Length Sepal.Width  Petal.Length Petal.Width  Species     
## <0 rows> (or 0-length row.names)

This is close, but still not perfect. Let’s take a look at part of the help file for !!:

“The !! operator unquotes its argument. It gets evaluated immediately in the surrounding context.”

The example above doesn’t work because we want a to be evaluated to Species to mirror the syntax in the first, typical usage example I gave at the start of this post. However, a is being evaluated as "Species" which is why it isn’t working.

Create a Symbol

And so, the final thing we need to do is turn the character string "Species" into Species, a symbol. We can do this by first calling the sym function from rlang on the variable a, and then using the !! operator to evaluate its results.

a <- "Species"
filter(iris, !!rlang::sym(a) == "versicolor") %>%
  head()
##   Sepal.Length Sepal.Width Petal.Length Petal.Width    Species
## 1          7.0         3.2          4.7         1.4 versicolor
## 2          6.4         3.2          4.5         1.5 versicolor
## 3          6.9         3.1          4.9         1.5 versicolor
## 4          5.5         2.3          4.0         1.3 versicolor
## 5          6.5         2.8          4.6         1.5 versicolor
## 6          5.7         2.8          4.5         1.3 versicolor

Hoorah, it worked! I hope this helped further explain tidy eval - I certainly learned a lot by using trial and error to find the correct syntax and then working backwards to figure out why it worked (and then some fantastic people on Twitter kindly pointing out where I’d gone terribly wrong and patiently explaining the correct answer!). I’d love to hear your feedback; I’m @nic_crane on Twitter. Massive thanks to @_lionelhenry and @romain_francois for your input on this! :)