Referring to previous row in calculation

Question

I'm new to R and can't seem to get to grips with how to call a previous value of "self", in this case previous "b" b[-1].

b <- ( ( 1 / 14 ) * MyData$High + (( 13 / 14 )*b[-1]))

Obviously I need a NA somewhere in there for the first calculation, but I just couldn't figure this out on my own.

Adding example of what the sought after result should be (A=MyData$High):

  A  b
1 5  NA
2 10 0.7142...
3 15 3.0393...
4 20 4.6079...

Answer 1

1) for loop Normally one would just use a simple loop for this:

MyData <- data.frame(A = c(5, 10, 15, 20))


MyData$b <- 0
n <- nrow(MyData)
if (n > 1) for(i in 2:n) MyData$b[i] <- ( MyData$A[i] + 13 * MyData$b[i-1] )/ 14
MyData$b[1] <- NA

giving:

> MyData
   A         b
1  5        NA
2 10 0.7142857
3 15 1.7346939
4 20 3.0393586

2) Reduce It would also be possible to use Reduce. One first defines a function f that carries out the body of the loop and then we have Reduce invoke it repeatedly like this:

f <- function(b, A) (A + 13 * b) / 14
MyData$b <- Reduce(f, MyData$A[-1], 0, acc = TRUE)
MyData$b[1] <- NA

giving the same result.

This gives the appearance of being vectorized but in fact if you look at the source of Reduce it does a for loop itself.

3) filter Noting that the form of the problem is a recursive filter with coefficient 13/14 operating on A/14 (but with A[1] replaced with 0) we can write the following. Since filter returns a time series we use c(...) to convert it back to an ordinary vector. This approach actually is vectorized as the filter operation is performed in C.

MyData$b <- c(filter(replace(MyData$A, 1, 0)/14, 13/14, method = "recursive"))
MyData$b[1] <- NA

again giving the same result.

Note: All solutions assume that MyData has at least 1 row.