通过在R中的数据帧的行上应用Reduce函数来创建新列

Question

I have a data frame that contains IDs, Dates, and observed returns. It can be likened to this:

df <- data.frame(
  ID = gl(3, 10, labels = c("A", "B", "C")), 
  Date = factor(rep(2006, 2015, 3)), 
  lr = runif(30, -0.01, 0.01))

Now I want to use the following function to find the vectors of exponentially moving averages for each of the IDs and add them as a new column to my original dataframe:

Emean<-function(x){
    ema <- function(a,b) {lambda*a+(1-lambda)*b}
    Reduce(ema, x, accumulate=T)
}

So I want the resulting data frame to have columns ID, Date, lr, and mlr. The last column (mlr) will be calculated using above function; and (sorry for loose notation!) but this is the formula:

mlr_t=lambda*mlr_t-1 + (1-lambda)*lr_t

'_t' denotes the time.

Now as I said I want to apply my function to the rows grouped by IDs and add the result as a column to this data frame. The output of 'Reduce' cannot be added directly to that data frame and I have to manipulate it in several steps which is extremely time consuming in R.

I need a computationally efficient solution for doing what I said. In the actual data set I have +100K IDs and +250 dates for each ID.

Answer 1

As

mlr_0 = 0
mlr_1 = 0 + (1-lambda)*lr_1
mlr_2 = lambda * mlr_1 + (1-lambda)*lr_2
      = lambda * (1-lambda) * lr_1 + (1-lambda)*lr_2
mlr_3 = lambda * mlr_2 + (1-lambda)*lr_3
      = lambda^2 * (1-lambda) * lr_1 + lambda * (1-lambda) * lr_2 + (1-lambda)*lr_3
...
mlr_t = lambda^(t-1) * (1-lambda) * lr_1 + lambda^(t-2) * (1-lambda) * lr_2 + ...
      = \Sum_{i=1}^{t} lambda^(t-i) * (1-lambda)*lr_i

you can do something like this (using data.table )

setDT(df)
lambda <- 0.5
# This calculates the lambda^(t-i)
l <- function(i, lambda){ lambda^(i-seq_len(i)) }

# This calculates multiplies element wise and sums up the mlr_3
my_fun <- function(x, lr, lambda){
  sum((1-lambda) * c(0,lr)[1:x] * l(x, lambda))}

# Apply both function to the vector
df[, vapply(seq_len(.N), my_fun, numeric(1), lr, lambda)  ,by = ID]

Results in (with set.seed(42) )

    ID        V1
 1:  A 0.0000000
 2:  A 0.4574030
 3:  A 0.6972392
 4:  A 0.4916894
 5:  A 0.6610685
 6:  A 0.6514070
 7:  A 0.5852515
 8:  A 0.6609199
 9:  A 0.3977932
10:  A 0.5273928
11:  B 0.0000000
12:  B 0.2288709
...