several questions regarding zoo::rollapply

Question

My understanding of rollapply's width option is that it specifies the window size on which the function will operate and by options specifies the shift size for this window. Here is my dataset:

> dataset <- as.vector(t(cbind(5:1, 1:5)))
> dataset
 [1] 5 1 4 2 3 3 2 4 1 5

And here are examples that confirms that I've written above:

> w3b3 <- rollapply(dataset, width = 3, by=3, FUN = print, align="left")
[1] 5 1 4
[1] 2 3 3
[1] 2 4 1

> w3b2 <- rollapply(dataset, width = 3, by=2, FUN = print, align="left")
[1] 5 1 4
[1] 4 2 3
[1] 3 3 2
[1] 2 4 1

> w2b3 <- rollapply(dataset, width = 2, by=3, FUN = print, align="left")
[1] 5 1
[1] 2 3
[1] 2 4

> w3b1 <- rollapply(dataset, width = 3, by=1, FUN = print, align="left")
[1] 5 1 4
[1] 1 4 2
[1] 4 2 3
[1] 2 3 3
[1] 3 3 2
[1] 3 2 4
[1] 2 4 1
[1] 4 1 5

# ACCORDING OT MAN WHEN NO VALUE IS USED THEN by=1 (SAME AS ABOVE)
> w3b1 <- rollapply(dataset, width = 3, FUN = print, align="left")
[1] 5 1 4
[1] 1 4 2
[1] 4 2 3
[1] 2 3 3
[1] 3 3 2
[1] 3 2 4
[1] 2 4 1
[1] 4 1 5

> w1b1 <- rollapply(dataset, width = 1, by=1, FUN = print, align="left")
[1] 5
[1] 1
[1] 4
[1] 2
[1] 3
[1] 3
[1] 2
[1] 4
[1] 1
[1] 5

Despite that I have several questions:

1) Why this one returns error while max(20) is working? Everything is the same as in last example except print is replaced by max:

> w1b1 <- rollapply(dataset, width = 1, by=1, FUN = max, align="left")
Error in if (is.na(a) || is.na(rval[i = 1]) || a == rval[i - 1]) max(xc[(i -  : 
  missing value where TRUE/FALSE needed

How can I debug those types *apply family function errors?

2) What is the purpose of using vector larger than 1 in with option and why following code prints one number to output in odd positions but assigns two numbers in odd positions to w12 variable?

> w12 <- rollapply(dataset, width = c(1,2), FUN = print, align="left")
[1] 5
[1] 1 4
[1] 4
[1] 2 3
[1] 3
[1] 3 2
[1] 2
[1] 4 1
[1] 1
> w12
      [,1] [,2]
 [1,]    5    5
 [2,]    1    4
 [3,]    4    4
 [4,]    2    3
 [5,]    3    3
 [6,]    3    2
 [7,]    2    2
 [8,]    4    1
 [9,]    1    1

# SAME AS ABOVE (ACCORDING TO MAN by IS USED ONLY IF width IS OF LENGTH 1)
> w12 <- rollapply(dataset, width = c(1,2), by=10, FUN = print, align="left")
[1] 5
[1] 1 4
[1] 4
[1] 2 3
[1] 3
[1] 3 2
[1] 2
[1] 4 1
[1] 1
> w12
      [,1] [,2]
 [1,]    5    5
 [2,]    1    4
 [3,]    4    4
 [4,]    2    3
 [5,]    3    3
 [6,]    3    2
 [7,]    2    2
 [8,]    4    1
 [9,]    1    1

3) What is the difference between passing vector and list to width argument (compared to previous output this is totally different)?

> rollapply(dataset, width = list(1,2), FUN = print, align="left")
[1] 1
[1] 2
[1] 2
[1] 3
[1] 3
[1] 4
[1] 4
[1] 5
[1] 5
[1] 1 2 2 3 3 4 4 5 5

4) what does by.column do? I was expecting that it has something to do with matrices so I've tried following:

> mtrx <- matrix(c(1:30), nrow=10)
> mtrx
      [,1] [,2] [,3]
 [1,]    1   11   21
 [2,]    2   12   22
 [3,]    3   13   23
 [4,]    4   14   24
 [5,]    5   15   25
 [6,]    6   16   26
 [7,]    7   17   27
 [8,]    8   18   28
 [9,]    9   19   29
[10,]   10   20   30

# THIS IS OK
> rollapply(mtrx, width = 2, by = 2, FUN = max, align = "left", by.column=T)
     [,1] [,2] [,3]
[1,]    2   12   22
[2,]    4   14   24
[3,]    6   16   26
[4,]    8   18   28
[5,]   10   20   30

# BUT WHAT IS THIS?
> rollapply(mtrx, width = 2, by = 2, FUN = max, align = "left", by.column=F)
[1] 22 24 26 28 30

you should try to limit your posts to one question each, for example, since I can give you some hints for some but not all of your questions: 1) not sure, but funny that min works without error; 2) variable windows; 3) not sure; 4) ??; 5) the last example is giving you the row max with a window of 2 so try again with min or median and see what happens — rawr
@rawr thank you for reply. Can you please try to further explain 2,4 (used to be 5 before correction) as answer? Thank you. — Wakan Tanka

G. Grothendieck G. Grothendieck · Accepted Answer · 2016-03-21T21:53:29

(1) is likely a bug.

(2) and (3) are explained in the help file:

If width is a plain numeric vector its elements are regarded as widths to be interpreted in conjunction with align whereas if width is a list its components are regarded as offsets. In the above cases if the length of width is 1 then width is recycled for every by-th point. If width is a list its components represent integer offsets such that the i-th component of the list refers to time points at positions i + width[[i]]. If any of these points are below 1 or above the length of index(data) then FUN is not evaluated for that point unless partial = TRUE and in that case only the valid points are passed.

so for the first w12 the specified width of c(1, 2) is recycled to the length of dataset so that alternate applications will be of width 1 and 2.

Thus the first w12 is the same as the following which uses a width of 1 and then a width of 2 and then 1 again, etc.

rollapply(dataset, c(1, 2, 1, 2, 1, 2, 1, 2, 1, 2), print)

If a list is specified for width then its components are regarded as offsets rather than widths. 1 means next value, 2 means value after next value, -1 means prior value and so on.

For the list example, it is recycled to list(1, 2, 1, 2, 1, 2, 1, 2, 1, 2) so for the first application it uses the next value, for the second application it uses the value after the next value, for the third application it uses the next value again and so on.

(4) by.column is defined in the help file:

by.column - logical. If TRUE, FUN is applied to each column separately.

The default is TRUE so rollapply by default it is applied to each column and then the result is cbind'ed together.; otherwise all columns are passed to the function at once.

several questions regarding zoo::rollapply

2 Answers