Matifou - 2 months ago 5x
R Question

# Use pipe without feeding first argument

Is the

%>%
pipe operator always feeding the left-hand side (LHS) to the first argument of the right-hand side (RHS)? Even if the first argument is specified again in the RHS call?

Say I want to specify which variable to use in
cor()
:

library(magrittr)
iris %>%
cor(x=.\$Sepal.Length, y=.\$Sepal.Width)

But this fails, it looks like it call something like
cor(., x=.\$Sepal.Length, y=.\$Sepal.Width)
?

I know I could use instead

iris %\$%
cor(x=Sepal.Length, y=Sepal.Width)

But wanted to find a solution with
%>%
...

Is the %>% pipe operator always feeding the left-hand side (LHS) to the first argument of the right-hand side (RHS)? Even if the first argument is specified again in the RHS call?

No. You’ve noticed the exception yourself: if the right-hand side uses ., the first argument of the left-hand side is not fed in. You need to pass it manually.

However, this is not happening in your case because you’re not using . by itself, you’re using it inside an expression. To avoid the left-hand side being fed as the first argument, you additionally need to use braces:

iris %>% {cor(x = .\$Sepal.Length, y = .\$Sepal.Width)}

Or:

iris %\$% cor(x = Sepal.Length, y = Sepal.Width)

— after all, that’s what %\$% is there for, as opposed to %>%.

But compare:

iris %>% lm(Sepal.Width ~ Sepal.Length, data = .)

Here, we’re passing the left-hand side expression explicitly as the data argument to lm. By doing so, we prevent it being passed as the first argument to lm.

Source (Stackoverflow)