Get right output from for loop when in function

Question

I know there may be more elegant to do this - but right now I just want to grok this logic... My problem is that when I do a typical Y[t]=Y[t-1]+i[t] thing for simulation it does not work when put in a function. I want to have a data frame out with the correct variable value on each row for time or t 1-10. Now the cat inside the loop reveals I am going wrong.

Here is my code:

iter <- 10; i <- rep(0.1,iter);  
i <- c(0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1.0)
Y0 <- 0.25 ; O0 <- 4.16 
Y <- rep(0,iter); O <- rep(0,iter) 
Tot <- rep(0,iter)
t <- seq(1,iter) #start time and fill vector
Y[1]=Y0 #First iter
O[1]=O0
Tot[1]= Y0+O0
time<-rep(1,iter)#runtime unit

#This is a simplified version of the ICBM function, to test the logic and output
Isim <- function(i, h, Y0, O0,iter,time) {
  for (t in 2:iter) {
    time<- time+1
    Y[t]=(Y[t-1]+i[t-1]); O[t]=(O[t-1]+i[t-1]); Tot[t]=Y[t]+O[t];
    simout <- data.frame(i,Y0,O0,Y,O,Tot,time)
    cat(time)
  } 
  return(simout)
}

result <- Isim(i, h, Y0, O0,iter, time)

What is time doing there, despite being increase by 1? You really shouldn't use a for loop for this. It's very easy to use a vectorized approach here and it will be faster by some orders of magnitude. — Roland
– Roland, Commented Oct 13, 2013 at 9:19

flodel · Accepted Answer · 2013-10-13 10:38:51Z

1

(Let's make abstraction of the fact your code cannot run.) You have a scoping issue at the Y[t]=(Y[t-1]+i[t-1]) step. Let's look at a small reproducible example:

increment_a <- function() { a <- a + 1; print(a); }
a <- 0; print(a); increment_a(); print(a);
# [1] 0 # ok
# [1] 1 # ok
# [1] 0 # not what you expected

Inside increment_a, when R computes a + 1, it first looks for a in the function's environment. That environment being empty, it then looks for a in the environment from which the function was called: your global environment. It finds it there, with a value of 0. It adds 1 to that 0 and then goes to assign it (the a <- part of the statement) in the local environment, i.e. the function's environment.

How can you solve this? You could tell R to assign the result of a + 1 to the a found in the global environment: use <<- instead of <-:

increment_a <- function() { a <<- a + 1; print(a); }
a <- 0; print(a); increment_a(); print(a);
# [1] 0
# [1] 1
# [1] 1

BUT this is not recommended. It is dangerous and will cause you troubles down the line. Instead, you need to fully embrace the functional programming style. I suggest you read the first paragraph of http://en.wikipedia.org/wiki/Functional_programming, especially the part about functions not having side effects. This is how things work:

make your function only use objects that are passed to it as arguments.
make use of the function's output(s) via the return statement.
Do not make your function have side-effect, in particular, do not use <<-.
if your function needs to modify objects: pass them as arguments, make your function modify them and return them. Then re-assign the result as in:

increment <- function(x) { x <- x + 1; print(x); return(x); }
a <- 0; print(a); a <- increment(a); print(a);
# [1] 0
# [1] 1
# [1] 1

edited Oct 13, 2013 at 10:38

answered Oct 13, 2013 at 10:13

flodel

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2 Comments

Olle Andrén Over a year ago

Wow, two quick replies. Thanks lots. I will now pour a cup of coffee, put on a tinfoil hat and try to improve and correct. R is fun - but it takes some effort to forget old ways...

flodel Over a year ago

Don't forget to accept an answer if one of us did answer your question.

alexis_laz · Accepted Answer · 2013-10-13 10:33:57Z

0

In your loop, instead of time<- time+1, try to use time[t] <- t..

When you increment time by 1 in each iteration, you end with the vector:

    time
    [1] 10 10 10 10 10 10 10 10 10 10

which is the final input in simout.

simout$time is altered in each iteration as follows:

    2 2 2 2 ... 
    3 3 3 3 ...
    ...
    10 10 10 ...

so you should change only a specific element of time in each iteration.

Now (you can see it from cat) simout$time changes as:

    1 1 1 1 ...
    1 2 1 1 ...
    1 2 3 1 ...
    ...
    1 2 3 4 5 6 7 8 9 10 #this is the final input in "simout"

answered Oct 13, 2013 at 10:33

alexis_laz

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