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I can't seem to make apply function access/modify a variable that is declared outside... what gives?

 x = data.frame(age=c(11,12,13), weight=c(100,105,110)) x testme <- function(df) { i <- 0 apply(df, 1, function(x) { age <- x[1] weight <- x[2] cat(sprintf("age=%d, weight=%d\n", age, weight)) i <- i+1 #this could not access the i variable in outer scope z <- z+1 #this could not access the global variable }) cat(sprintf("i=%d\n", i)) i } z <- 0 y <- testme(x) cat(sprintf("y=%d, z=%d\n", y, z))

Results:

 age=11, weight=100 age=12, weight=105 age=13, weight=110 i=0 y=0, z=0
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  • 1
    You need to pass the variables to testme, and then to apply: testme <- function(x, z) { and apply(df, 1, function(x, i, z) {}, i, z) Commented Nov 30, 2012 at 6:59
  • @bdemarest: that won't work as the value of i will be reset at iteration of apply (ie, for every row of df). I think the OP wants to track which row they are on Commented Nov 30, 2012 at 7:32
  • @RicardoSaporta, you are quite right. Probably the OP would be better off not using apply, but instead a standard for loop: for (i in 1:nrow(df)) {...}. Currently, we can only guess at the underlying problem he/she is trying to solve. Commented Nov 30, 2012 at 7:42
  • 1
    this was only a test snip to demonstrate the problem i had :-) It turns out that I should return the results back to the caller i.e. assign the result of the apply call to another variable. That's a better functional style. Commented Nov 30, 2012 at 8:12

2 Answers 2

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45

Using the <<- operator you can write to variables in outer scopes: 

x = data.frame(age=c(11,12,13), weight=c(100,105,110)) x testme <- function(df) { i <- 0 apply(df, 1, function(x) { age <- x[1] weight <- x[2] cat(sprintf("age=%d, weight=%d\n", age, weight)) i <<- i+1 #this could not access the i variable in outer scope z <<- z+1 #this could not access the global variable }) cat(sprintf("i=%d\n", i)) i } z <- 0 y <- testme(x) cat(sprintf("y=%d, z=%d\n", y, z))

The result here:

age=11, weight=100 age=12, weight=105 age=13, weight=110 i=3 y=3, z=3

Note that the usage of <<- is dangerous, as you break up scoping. Do this only if really necessary and if you do, document that behavior clearly (at least in bigger scripts)

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Comments

8

try the following inside your apply. Experiment with the value of n. I believe that for i it should be one less than for z. 

 assign("i", i+1, envir=parent.frame(n=2)) assign("z", z+1, envir=parent.frame(n=3)) 


testme <- function(df) { i <- 0 apply(df, 1, function(x) { age <- x[1] weight <- x[2] cat(sprintf("age=%d, weight=%d\n", age, weight)) ## ADDED THESE LINES assign("i", i+1, envir=parent.frame(2)) assign("z", z+1, envir=parent.frame(3)) }) cat(sprintf("i=%d\n", i)) i }

OUTPUT

> z <- 0 > y <- testme(x) age=11, weight=100 age=12, weight=105 age=13, weight=110 i=3 > cat(sprintf("y=%d, z=%d\n", y, z)) y=3, z=3
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5 Comments

I would use assign over eval(parse(...)).
@RomanLuštrik, I had originally put assign in the answer and then edited it to eval(parse((.)). Why your preference?
I guess religious reasons. See the entry on fortune. stackoverflow.com/questions/11025031/… I wonder if this question sparked the question made by @CarlWitthoft a few minutes ago: stackoverflow.com/questions/13647046/…
@RomanLuštrik, there are certainly quite a few questions today on it. I added one myself, and @DWin made a pretty good argument. Changed the above back to assign()
@RicardoSaporta eval(parse(...)) allows addressing elements of a list as well, while assign doesn't support that, so I actually prefer the former option

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