CODEWITHSUNDEEP
r
Koinos
Koinos

Reputation: 301

Quotations marks and expression

I've started today studying the R language.

After defining two vectors x,y i applied the outer product between them, applying what the documentation says.

f= outer (x, y, FUN = "*")

And all went great. Then i decide to try an anonymous function, so i wrote

f= outer (x, y, FUN = "function(x,y) cos(y) / (1+x^2)")

But i got an error. However if i write:

`f= outer (x, y, FUN = function(x,y) cos(y) / (1+x^2))`

It works.

Why the quotations marks work with * but not with this anonymous function ? Thanks for your time

Upvotes: 0

Views: 47

Answers (1)

MichaelChirico
MichaelChirico

Reputation: 34763

I'm not sure what you mean by this code "working":

`f= outer (x, y, FUN = function(x,y) cos(y) / (1+x^2))`

What this will do is attempt to access a variable named f= outer (x, y, FUN = function(x,y) cos(y) / (1+x^2)), e.g. I could write:

`f= outer (x, y, FUN = function(x,y) cos(y) / (1+x^2))` = 3
`f= outer (x, y, FUN = function(x,y) cos(y) / (1+x^2))`
# [1] 3

This is an insanely confusing thing to do, but surrounding objects in backticks (`) allows you to declare arbitrarily-named objects.

Again, don't do this unless you absolutely know what you're doing (i.e. well past your first day of R programming).

Anyway, regarding why * needs quotation marks, it's because * is an operator. Everything in R is a function (almost), but operators are special functions in that they let you write them from left to right like:

variable1 operator variable2

instead of the normal

function(variable1, variable2)

That said, all operators also have a functional form; but, to distinguish the functional form from the operator form, the R parser needs them to be set apart with quotes (or with backticks); the following are the same:

3 * 4
'*'(3, 4)
"*"(3, 4) #single and double quotes in R are basically the same
`*`(3, 4)

the FUN argument of outer requires functions, not operators, so we have to serve * in its functional form, e.g. "*".

One more bit of detail -- you can observe R internally changing the operator form into the functional form as part of the code-parsing process (i.e., before evaluation, R will already convert the operator form into functional form):

as.character(substitute(3 * 4))
# [1] "*" "3" "4"

# alternatively
pryr::ast(3 * 4)
# \- ()
#   \- `*
#   \-  3
#   \-  4

Upvotes: 1

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