Say I have a function:
MyFunction=function(x,y,z,thelist){
a = x+3
b = y+4
c = z+2
mylist=thelist
myfunc1=function(form=a) a+2
myfunc2=function(form=b) b+2
myfunc3=function(form=c) c+2
function(...){
.x = as.list(substitute(list(...)))[-1]
CombinedList = lapply(seq_along(mylist), function (y,.x) {
numbercols = length(eval(.x[[1]])(form=eval(formals(eval(.x[[1]]))$form))[[1]])
numbercolsplusone = numbercols+1
elementz = matrix(NA,nrow=length(.x),ncol=numbercolsplusone)
for (i in seq_along(.x)) {
elementz[i,1:numbercols] = eval(.x[[i]])(form=eval(formals(eval(.x[[i]]))$form))[[y]]
elementz[i,numbercolsplusone] = substr(as.character(.x[[i]]),1,nchar(as.character(.x[[i]]))-4)
}
....
Some more calculations....
....
},.x)
return(CombinedList)
}
}
I then choose, say myfunc1 and myfunc2 from the parent function, and call my toyfunction:
try = Myfunction(x,y,z,thelist)(myfunc1,myfunc2)
This is a toyfunction, and my real function is much longer and takes a lot of time to calculate. I guess the problem lies in the for-loop inside the child function - probably my use of eval.
This function takes longer time to calculate then the the other similar function which only differs in that I call all the functions in my parent function before proceeding to the child function, function(...){. And not choose functions like in this toy-example.
Is there some other, quicker way?
Related
I'm trying to run a loop over different functions with different number of arguments. The variables are created at runtime inside the loop, and I want to use eval at each iteration to instantiate a Struct using the variable :symbol. However, I can't do this since eval only works in the global scope. This is the MWE for the case that works:
function f1(x); return x; end
function f2(x1,x2); return x1+x2; end
handles = [f1,f2]
args =[:(x1),:(x1,x2)]
x1 = 1; x2 = 1;
for (i,f) in enumerate(handles)
params = eval(args[i])
#show f(params...)
end
f(params...) = 1
f(params...) = 2
However, if I move the variable definitions inside the loop, which is what I actually want, it doesn't work after restarting Julia to clear the workspace.
function f1(x); return x; end
function f2(x1,x2); return x1+x2; end
handles = [f1,f2]
args =[:(x1),:(x1,x2)]
for (i,f) in enumerate(handles)
x1 = 1; x2 = 1;
params = eval(args[i])
#show f(params...)
end
ERROR: UndefVarError: x1 not defined
I've tried several of the answers, such as this one, but I can't seem to make it work. I could write a custom dispatch function that takes[x1,x2] and calls f1 or f2 with the correct arguments. But still, is there any way to do this with eval or with an alternative elegant solution?
EDIT: here are more details as to what I'm trying to do in my code. I have a config struct for each algorithm, and in this I want to define beforehand the arguments it takes
KMF_config = AlgConfig(
name = "KMF",
constructor = KMC.KMF,
parameters = :(mu,N,L,p),
fit = KMC.fit!)
MF_config = AlgConfig(
name = "MF",
constructor = KMC.MF,
parameters = :(mu,N,L),
fit = KMC.fit!)
alg_config_list = [KMF_config, MF_config]
for (i,alg_config) in enumerate(alg_config_list)
mu,N,L,p,A,B,C,D,data = gen_vars() #this returns a bunch of variables that are used in different algorithms
method = alg_config.constructor(eval(method.parameters)...)
method.fit(data)
end
One possible solution is to have a function take all the variables and method, and return a tuple with a subset of variables according to method.name. But I'm not sure if it's the best way to do it.
Here's an approach using multiple dispatch rather than eval:
run_a(x, y) = x + 10*y
run_b(x, y, z) = x + 10*y + 100*z
extract(p, ::typeof(run_a)) = (p.x, p.y)
extract(p, ::typeof(run_b)) = (p.x, p.y, p.z)
genvars() = (x=1, y=2, z=3)
function doall()
todo = [
run_a,
run_b,
]
for runalg in todo
v = genvars()
p = extract(v, runalg)
#show runalg(p...)
end
end
In your example you would replace run_a and run_b with KMC.KMF and KMC.MF.
Edit: Cleaned up example to avoid structs that don't exist in your example.
How can I use function call in a for loop where one variable is the current iteration of the loop? My current code (very simplified) looks like this:
funCall <- call('FUN', arg1 = 10, arg2 = get('i'))
for(i in 1:x){
ans[i] <- eval(funCall)
}
Given a FUN like arg1 * arg2 the evaluated function should return 10 for i = 1, 20 for i = 2 etc.
Using get('i') I get the error that object 'i' is not found and I don't understand
1) whyget is evaluated right away inside call
2) how I can implement this callto get the correct i
edit
The solution to the problem is to use another call() inside the call like
call('FUN', arg1 = 10, arg2 = call('get','i')) so the get()function is only evaluated inside the loop.
In the function shown below, there is no return. However, after executing it, I can confirm that the value entered d normally.
There is no return. Any suggestions in this regard will be appreciated.
Code
#installed plotly, dplyr
accumulate_by <- function(dat, var) {
var <- lazyeval::f_eval(var, dat)
lvls <- plotly:::getLevels(var)
dats <- lapply(seq_along(lvls), function(x) {
cbind(dat[var %in% lvls[seq(1, x)], ], frame = lvls[[x]])
})
dplyr::bind_rows(dats)
}
d <- txhousing %>%
filter(year > 2005, city %in% c("Abilene", "Bay Area")) %>%
accumulate_by(~date)
In the function, the last assignment is creating 'dats' which is returned with bind_rows(dats) We don't need an explicit return statement. Suppose, if there are two objects to be returned, we can place it in a list
In some languages like python, for memory efficiency, generators are used which will yield instead of creating the whole output in memory i.e. Consider two functions in python
def get_square(n):
result = []
for x in range(n):
result.append(x**2)
return result
When we run it
get_square(4)
#[0, 1, 4, 9]
The same function can be written as a generator. Instead of returning anything,
def get_square(n):
for x in range(n):
yield(x**2)
Running the function
get_square(4)
#<generator object get_square at 0x0000015240C2F9E8>
By casting with list, we get the same output
list(get_square(4))
#[0, 1, 4, 9]
There is always a return :) You just don't have to be explicit about it.
All R expressions return something. Including control structures and user-defined functions. (Control-structures are just functions, by the way, so you can just remember that everything is a value or a function call, and everything evaluates to a value).
For functions, the return value is the last expression evaluated in the execution of the function. So, for
f <- function(x) 2 + x
when you call f(3) you will invoke the function + with two parameters, 2 and x. These evaluate to 2 and 3, respectively, so `+`(2, 3) evaluates to 5, and that is the result of f(3).
When you call the return function -- and remember, this is a function -- you just leave the control-flow of a function early. So,
f <- function(x) {
if (x < 0) return(0)
x + 2
}
works as follows: When you call f, it will call the if function to figure out what to do in the first statement. The if function will evaluate x < 0 (which means calling the function < with parameters x and 0). If x < 0 is true, if will evaluate return(0). If it is false, it will evaluate its else part (which, because if has a special syntax when it comes to functions, isn't shown, but is NULL). If x < 0 is not true, f will evaluate x + 2 and return that. If x < 0 is true, however, the if function will evaluate return(0). This is a call to the function return, with parameter 0, and that call will terminate the execution of f and make the result 0.
Be careful with return. It is a function so
f <- function(x) {
if (x < 0) return;
x + 2
}
is perfectly valid R code, but it will not return when x < 0. The if call will just evaluate to the function return but not call it.
The return function is also a little special in that it can return from the parent call of control structures. Strictly speaking, return isn't evaluated in the frame of f in the examples above, but from inside the if calls. It just handles this special so it can return from f.
With non-standard evaluation this isn't always the case.
With this function
f <- function(df) {
with(df, if (any(x < 0)) return("foo") else return("bar"))
"baz"
}
you might think that
f(data.frame(x = rnorm(10)))
should return either "foo" or "bar". After all, we return in either case in the if statement. However, the if statement is evaluated inside with and it doesn't work that way. The function will return baz.
For non-local returns like that, you need to use callCC, and then it gets more technical (as if this wasn't technical enough).
If you can, try to avoid return completely and rely on functions returning the last expression they evaluate.
Update
Just to follow up on the comment below about loops. When you call a loop, you will most likely call one of the built-in primitive functions. And, yes, they return NULL. But you can write your own, and they will follow the rule that they return the last expression they evaluate. You can, for example, implement for in terms of while like this:
`for` <- function(itr_var, seq, body) {
itr_var <- as.character(substitute(itr_var))
body <- substitute(body)
e <- parent.frame()
j <- 1
while (j < length(seq)) {
assign(x = itr_var, value = seq[[j]], envir = e)
eval(body, envir = e)
j <- j + 1
}
"foo"
}
This function, will definitely return "foo", so this
for(i in 1:5) { print(i) }
evalutes to "foo". If you want it to return NULL, you have to be explicit about it (or just let the return value be the result of the while loop -- if that is the primitive while it returns NULL).
The point I want to make is that functions return the last expression they evaluate has to do with how the functions are defined, not how you call them. The loops use non-standard evaluation, so the last expression in the loop body you provide them might be the last value they evaluate and might not. For the primitive loops, it is not.
Except for their special syntax, there is nothing magical about loops. They follow the rules all functions follow. With non-standard evaluation it can get a bit tricky to work out from a function call what the last expression they will evaluate might be, because the function body looks like it is what the function evaluates. It is, to a degree, if the function is sensible, but the loop body is not the function body. It is a parameter. If it wasn't for the special syntax, and you had to provide loop bodies as normal parameters, there might be less confusion.
There is one thing I really don't like about R reference class: the order you write the methods matters. Suppose your class goes like this:
myclass = setRefClass("myclass",
fields = list(
x = "numeric",
y = "numeric"
))
myclass$methods(
afunc = function(i) {
message("In afunc, I just call bfunc...")
bfunc(i)
}
)
myclass$methods(
bfunc = function(i) {
message("In bfunc, I just call cfunc...")
cfunc(i)
}
)
myclass$methods(
cfunc = function(i) {
message("In cfunc, I print out the sum of i, x and y...")
message(paste("i + x + y = ", i+x+y))
}
)
myclass$methods(
initialize = function(x, y) {
x <<- x
y <<- y
}
)
And then you start an instance, and call a method:
x = myclass(5, 6)
x$afunc(1)
You will get an error:
Error in x$afunc(1) : could not find function "bfunc"
I am interested in two things:
Is there a way to work around this nuisance?
Does this mean I can never split a really long class file into multiple files? (e.g. one file for each method.)
Calling bfunc(i) isn't going to invoke the method since it doesn't know what object it is operating on!
In your method definitions, .self is the object being methodded on (?). So change your code to:
myclass$methods(
afunc = function(i) {
message("In afunc, I just call bfunc...")
.self$bfunc(i)
}
)
(and similarly for bfunc). Are you coming from C++ or some language where functions within methods are automatically invoked within the object's context?
Some languages make this more explicit, for example in Python a method with one argument like yours actually has two arguments when defined, and would be:
def afunc(self, i):
[code]
but called like:
x.afunc(1)
then within the afunc there is the self variable which referes to x (although calling it self is a universal convention, it could be called anything).
In R, the .self is a little bit of magic sprinkled over reference classes. I don't think you could change it to .this even if you wanted.
Consider a code snippet as follow:
f = function(y) function() y()
f(version)()
Error in f(version)() : could not find function "y"
P.s. It seems that the closure mechanism is quite different from C# Lambda. (?)
Q: How can I capture a function in the closure?
--EDIT--
Scenario: Actually, I would like to write a function factory, and I don't want to add parameter to the nested function.
Like this:
theme_factory = function(theme_fun)
{
function(device)
{
if (!is.onMac()) # Not Mac
{
(device == "RStudioGD") %?% theme_fun(): theme_fun(base_family="Heiti")
}
else
{
theme_fun(base_family="STHeiti")
}
}
}
And I defined two customized theme function for ggplot
theme_bw_rmd = theme_factory(theme_bw)
theme_grey_rmd = theme_factory(theme_grey)
Then I use them like:
function(device)
ggplot(data) + geom_point() something like that + theme_bw_rmd(device)
Thanks.
So the problem is with passing parameter? What about something like this:
alwaysaddone <- function(f) function(...) f(...)+1
biggersum <- alwaysaddone(sum)
sum(1:3)
# 6
biggersum(1:3)
# 7
You can use ... to "pass-through" any parameters you like.
Use eval(func, envir = list(... captured parameters)) or substitute(func, envir) to eval the captured function in a specific environment.