CODEWITHSUNDEEP
r
Tyler Rinker
Tyler Rinker

Reputation: 109864

Count occurrences of condition in lapply

I am running a simulation that I need to keep track of number of occurrences in a function call of a particular condition. I attempted to accomplish this with an assignment to a global object. It works if you run the function but if you try to lapply the function as I'm doing then you get a single count of all the times the condition happened rather than a count for every time it happened for each element in the list fed to lapply.

Here's a dummy situation where the occurrence is evenness of a number:

FUN <- function(x){
    lapply(1:length(x), function(i) {
        y <- x[i]
        if (y %% 2 == 0){
            assign("count.occurrences", count.occurrences + 1, env=.GlobalEnv)   
        }
        print("do something")
    })
    list(guy="x", count=count.occurrences)
}

#works as expected
count.occurrences <- 0
FUN(1:10)


count.occurrences <- 0  
lapply(list(1:10, 1:3, 11:16, 9), FUN) 

#gives me...
#> count.occurrences
#[1] 9

#I want...
#> count.occurrences
#[1] 5  1  3  0

It's in a simulation so speed is an issue. I want this to be as fast as possible so I'm not married to the global assignment idea.

Upvotes: 4

Views: 3851

Answers (3)

Brian Diggs
Brian Diggs

Reputation: 58825

Rather than assign to the global environment, why not just assign to inside FUN's environment?

FUN <- function(x){
    count.occurances <- 0
    lapply(1:length(x), function(i) {
        y <- x[i]
        if (y %% 2 == 0){
            count.occurances <<- count.occurances + 1
        }
        print("do something")
    })
    list(guy="x", count=count.occurances)
}

Z <- lapply(list(1:10, 1:3, 11:16, 9), FUN)

Then you can just pull the counts out.

> sapply(Z, `[[`, "count")
[1] 5 1 3 0

Upvotes: 8

A5C1D2H2I1M1N2O1R2T1
A5C1D2H2I1M1N2O1R2T1

Reputation: 193517

I haven't done any benchmarking on this, but have you tried just using a for loop? I know that loops aren't generally encouraged in R, but they're also not always slower.

FUN <- function(x) {
  count.occurrences = 0
  for (i in 1:length(x)) {
    y = x[i]
    if (y %% 2 == 0) {
      count.occurrences = count.occurrences + 1
    }
    print("do something")
  }
  list(guy="x", count=count.occurrences)
}

lapply(list(1:10, 1:3, 11:16, 9), FUN)

Upvotes: 2

Tyler Rinker
Tyler Rinker

Reputation: 109864

I can get it like this:

count.occurances <- 0  
Z <-lapply(list(1:10, 1:3, 11:16, 9), FUN) 
diff(c(0, sapply(1:length(Z), function(x) Z[[x]]$count)))

I'm open to better ideas (faster).

Upvotes: 0

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