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I must admit to complete lunacy when trying to understand how functions within functions are defined and passed in R. The examples always presume you understand every nuance and don't provide descriptions of the process. I have yet to come across a plain English, idiots guide break down of the process. So the first question is do you know of one?

Now my physical problem.
I have a list of data.frames: fileData.
I want to use the rollapply() function on specific columns in each data.frame. I then want all the results(lists) combined. So starting with one of the data.frames using the built in mtcars dataframes as an example:

Of course I need to tell rollapply() to use the function PPI() along with the associated parameters which are the columns.

PPI <- function(a, b){  
    value = (a + b)  
    PPI = sum(value)  
    return(PPI)  
}

I tried this:

f <- function(x) PPI(x$mpg, x$disp)
fileData<- list(mtcars, mtcars, mtcars)
df <- fileData[[1]]

and got stopped at

rollapply(df, 20, f)
Error in x$mpg : $ operator is invalid for atomic vectors

I think this is related to Zoo using matrices but other numerous attempts couldn't resolve the rollapply issue. So moving onto what I believe is next:

lapply(fileData, function(x) rollapply ......

Seems a mile away. Some guidance and solutions would be very welcome.
Thanks.

Look Left
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    Just want to let you know your not the only person who thinks R has a (too) high learning curve, a lot of things are borderline impossible to understand if you don't have a lot of experience programming, or someone with tons of experience to help you out. – Thorst Jun 04 '14 at 05:59
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    +1 to @Soccerman's comment. R is hard to learn. That said, it would be helpful if you could provide a *reproducible* example, see here: http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example – Stephan Kolassa Jun 04 '14 at 06:32
  • @StephanKolassa I normally try and add an example which I've done now. Assistance with describing the process would also be helpful, not just an answer. Cheers. – Look Left Jun 04 '14 at 06:45

3 Answers3

2

I will Try to help you and show how you can debug the problem. One trick that is very helpful in R is to learn how to debug. Gnerelly I am using browser function.

problem :

Here I am changing you function f by adding one line :

f <- function(x) {
  browser()
  PPI(x$changeFactor_A, x$changeFactor_B)
}

Now when you run : 

rollapply(df, 1, f)

The debugger stops and you can inspect the value of the argument x:

Browse[1]> x
 [1,] 
1e+05

as you see is a scalar value , so you can't apply the $ operator on it, hence you get the error:

Error in x$changeFactor_A : $ operator is invalid for atomic vectors

general guides

Now I will explain how you should do this.

  • Either you change your PPI function, to have a single parameter excees: so you do the subtraction outside of it (easier)
  • Or you use mapply to get a generalized solution. (Harder but more general and very useful)
  • Avoid using $ within functions. Personally, I use it only on the R console.

complete solution:

I assume that you data.frames(zoo objects) have changeFactor_A and changeFactor_B columns.

sapply(fileData,function(dat){
  dat <- transform(dat,excess= changeFactor_A-changeFactor_B)
  rollapply(dat[,'excess'],2,sum)
}

Or More generally : 

sapply(fileData,function(dat){
  excess <- get_excess(dat,'changeFactor_A','changeFactor_B')
  rollapply(excess,2,sum)
}

Where 

   get_excess <- 
     function(data,colA,colB){
          ### do whatever you want here
          ### return a vector
          excess
     }
agstudy
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  • That looks like the right direction. I have a full IDE debugger through Revolution R but debugging could certainly be improved. Thanks for the browse tip. Your last item may answer my second part but I can't change the PPI function (it's more complicated than just a subtraction), so my first question still stands; how to call rollapply() with a function that requires paramaters. You will notice that I changed the question to provide a better example, which is why the dataframe now has different names.) – Look Left Jun 04 '14 at 09:18
  • Thanks @agstudy. I have used your examples to start and mash together something that works without disassembling the existing framework too much. A question, if you don't use $ in a high level function, how do you tell it what elements to perform the task on? – Look Left Jun 04 '14 at 23:40
  • @LookLeft by element you mean column? I'am not sure what do you mean by, but `dat$col1` is equivalent to `dat[,"col1"]` ? – agstudy Jun 05 '14 at 06:36
  • Yes I meant column. Ah, I see your nuance with not using $. Use the stricter form of column referencing. This helped in my final answer which I'm about to post. Cheers. – Look Left Jun 06 '14 at 02:19
1

Look at the "Usage" section of the help page to ?rollapply. I'll admit that R help pages are not easy to parse, and I see how you got confused.

The problem is that rollapply can deal with ts, zoo or general numeric vectors, but only a single series. You are feeding it a function that takes two arguments, asset and benchmark. Granted, your f and PPI can trivially be vectorized, but rollapply simply isn't made for that.

Solution: calculate your excess outside rollapply (excess is easily vectorially calculated, and it does not involve any rolling calculations), and only then rollapply your function to it:

> mtcars$excess <- mtcars$mpg-mtcars$disp
> rollapply(mtcars$excess, 3, sum)
 [1]  -363.2  -460.8  -663.1  -784.8  -893.9 ...

You may possibly be interested in mapply, which vectorizes a function for multiple arguments, similarly to apply and friends, which work on single arguments. However, I know of no analogue of mapply with rolling windows.

Stephan Kolassa
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  • Thanks @StephanKolassa but this only helps to explain the simple use of rollapply() which I can get to work using the examples. How do I use it in the problem I have outlined? That is to 'be called' by a function and 'to call' a function; all with parameters being passed. The actual PPI function operation is of not relevant to this question. It is just a simple example function to demonstrate the problem. – Look Left Jun 04 '14 at 09:15
  • Note that `rollapply` can handle multivariate series by using the `by.column=FALSE` argument. – G. Grothendieck Jun 04 '14 at 11:19
  • @LookLeft: it may be helpful to check out the coverage of "anonymous functions" in a few R textbooks, e.g. section 7.13 in *The Art of R Programming*. *The R Inferno* has only two mentions, but its section 5.1 may be useful (though not easy reading). – Stephan Kolassa Jun 04 '14 at 14:35
1

I sweated away and took some time to slowly understand how to break down the process and protocol of calling a function with arguments from another function. A great site that helped was Advanced R from the one and only Hadley Wickham, again! The pictures showing the process breakdown are near ideal. Although I still needed my thinking cap on for a few details.

Here is a complete example with notes. Hopefully someone else finds it useful.

library(zoo)

#Create a list of dataframes for the example.
listOfDataFrames<- list(mtcars, mtcars, mtcars)
#Give each element a name.
names(listOfDataFrames) <- c("A", "B", "C")

#This is a simple function just for the example!
#I want to perform this function on column 'col' of matrix 'm'.
#Of course to make the whole task worthwhile, this function is usually something more complex.
fApplyFunction <- function(m,col){
    mean(m[,col])
}

#This function is called from lapply() and does 'something' to the dataframe that is passed.
#I created this function to keep lapply() very simply.
#The something is to apply the function fApplyFunction(), wich requires an argument 'thisCol'. 
fOnEachElement <- function(thisDF, thisCol){
    #Convert to matrix for zoo library.
    thisMatrix <- as.matrix(thisDF)
    rollapply(thisMatrix, 5, fApplyFunction, thisCol, partial = FALSE, by.column = FALSE)
}

#This is where the program really starts!
#
#Apply a function to each element of list.
#The list is 'fileData', with each element being a dataframe.
#The function to apply to each element is 'fOnEachElement'
#The additional argument for 'fOnEachElement' is "vs", which is the name of the column I want the function performed on.
#lapply() returns each result as an element of a list.
listResults <- lapply(listOfDataFrames, fOnEachElement, "vs")


#Combine all elements of the list into one dataframe.
combinedResults <- do.call(cbind, listResults)

#Now that I understand the argument passing, I could call rollapply() directly from lapply()...
#Note that ONLY the additional arguments of rollapply() are passed. The primary argurment is passed automatically by lapply().
listResults2 <- lapply(listOfDataFrames, rollapply, 5, fApplyFunction, "vs", partial = FALSE, by.column = FALSE)

Results:

> combinedResults
        A   B   C
 [1,] 0.4 0.4 0.4
 [2,] 0.6 0.6 0.6
 [3,] 0.6 0.6 0.6
 [4,] 0.6 0.6 0.6
 [5,] 0.6 0.6 0.6
 [6,] 0.8 0.8 0.8
 [7,] 0.8 0.8 0.8
 [8,] 0.8 0.8 0.8
 [9,] 0.6 0.6 0.6
[10,] 0.4 0.4 0.4
[11,] 0.2 0.2 0.2
[12,] 0.0 0.0 0.0
[13,] 0.0 0.0 0.0
[14,] 0.2 0.2 0.2
[15,] 0.4 0.4 0.4
[16,] 0.6 0.6 0.6
[17,] 0.8 0.8 0.8
[18,] 0.8 0.8 0.8
[19,] 0.6 0.6 0.6
[20,] 0.4 0.4 0.4
[21,] 0.2 0.2 0.2
[22,] 0.2 0.2 0.2
[23,] 0.2 0.2 0.2
[24,] 0.4 0.4 0.4
[25,] 0.4 0.4 0.4
[26,] 0.4 0.4 0.4
[27,] 0.2 0.2 0.2
[28,] 0.4 0.4 0.4
> listResults
$A
 [1] 0.4 0.6 0.6 0.6 0.6 0.8 0.8 0.8 0.6 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 0.8 0.6
[20] 0.4 0.2 0.2 0.2 0.4 0.4 0.4 0.2 0.4

$B
 [1] 0.4 0.6 0.6 0.6 0.6 0.8 0.8 0.8 0.6 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 0.8 0.6
[20] 0.4 0.2 0.2 0.2 0.4 0.4 0.4 0.2 0.4

$C
 [1] 0.4 0.6 0.6 0.6 0.6 0.8 0.8 0.8 0.6 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 0.8 0.6
[20] 0.4 0.2 0.2 0.2 0.4 0.4 0.4 0.2 0.4

> listResults2
$A
 [1] 0.4 0.6 0.6 0.6 0.6 0.8 0.8 0.8 0.6 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 0.8 0.6
[20] 0.4 0.2 0.2 0.2 0.4 0.4 0.4 0.2 0.4

$B
 [1] 0.4 0.6 0.6 0.6 0.6 0.8 0.8 0.8 0.6 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 0.8 0.6
[20] 0.4 0.2 0.2 0.2 0.4 0.4 0.4 0.2 0.4

$C
 [1] 0.4 0.6 0.6 0.6 0.6 0.8 0.8 0.8 0.6 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 0.8 0.6
[20] 0.4 0.2 0.2 0.2 0.4 0.4 0.4 0.2 0.4
Look Left
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