combining mathematical notation and value substitution
On Sun, 22 Jun 2003, Faheem Mitha wrote:
If I'm doing this correctly, R does not seem to think it is a call.
is.call("Monotonic Multigamma run (" * n == len * ", " * theta == t1
* ").")
Error in "Monotonic Multigamma run (" * n :
non-numeric argument to binary operator
R is trying to *evaluate*
"Monotonic Multigamma run ("* n==len etc
which doesn't work. Remember, is.call(), like any normal function, will
be passed the *value* of its arguments.
You could try
is.call(quote("Monotonic Multigamma run("*n==len))
which is TRUE.
It considers it a valid R expression though.
(mode(expression("Monotonic Multigamma run (" * n == len * ", " * theta
== t1 * ")."))) [1] "expression"
That's because expression() returns an expression.
The clearest description I have seen of a call is in S Poetry, where it says "Mode call represents objects that are calls to a function. The first component of a call is the name (mode name) of the function being called. The rest of the call is the arguments given." This certainly is how calls are constructed using call(...), but I'm not sure how it fits in with an expression like the one above. What is the function being called in that case, for example?
Well, we can find out. It must be either * or ==, but it isn't immediately obvious which one ends up at the top level
thing <- quote("Monotonic Multigamma Run ("*n==len* ", " * theta
==t1*").")
mode(thing)
[1] "call"
length(thing)
[1] 3
thing[[1]]
==
thing[[2]]
"Monotonic Multigamma Run (" * n == len * ", " * theta
thing[[3]]
t1 * ")."
mode(thing[[2]])
[1] "call"
mode(thing[[3]])
[1] "call"
thing[[2]][[1]]
==
thing[[3]][[1]]
* So it is a call to ==, with two arguments, each itself a call. The first arguemetn is also a call to == and the second is a call to *. And so on in a tree structure.
Maybe I'd understand this stuff better if I knew Lisp. I'm very ignorant about programming language paradigms. I've looked at S Poetry, and S Programming. While they were helpful, I don't feel I've got a clear picture. Further recommended reading?
I don't know about further reading. I think this is probably easier to learn when you have a specific problem to solve. I learned mostly by porting the survival package. -thomas