Discounted Cumulative Sum in R

Question

I'm trying to calculate a discounted cumulative sum in which the later values are worth more.

Let's say I have the following dataset:

 dt <- data.table( "year" = c(79,80,81,82,83), "value" = c(5,2,6,8,9))  

> dt
   year value
1:   79     5
2:   80     2
3:   81     6
4:   82     8
5:   83     9

And I want the following output:

> dt2
year value     DCS    
1:   79     5  5.0000  
2:   80     2  6.5000 
3:   81     6 11.8500
4:   82     8 18.6650 
5:   83     9 25.7985 

The discounted cumulative sum (DCS) is calucuted by discounting the previous values at a 10% annual discount rate. So, for the first second line, the DCS value is given by 2 + 5*(0.9)^1. For the third line, the DCS is 6 + (0.9)^1*2 + (0.9)^2*5, and so on.

Formally, the discounted sum formula is given by:

Finally, a data.table solution is preferable if possible.

Answer 1

Here are 2 other options.

1) using Rcpp and then update by reference in data.table:

library(Rcpp)
cppFunction("
NumericVector dcs(NumericVector x, double disc) {
    int n = x.size();
    NumericVector res(n);
    res[0] = x[0];
    for (int i=1; i<n; i++) {
        res[i] += x[i] + res[i-1]*disc;
    }
    return res;
}")
dt[, DCS := dcs(value, 0.9)]

2) Or recursively in data.table:

s <- 0
dt[, dcs2 := {
       s <- value + s*0.9
       s
    }, 
    1L:nrow(dt)]

#or simply: s <- 0; dt[, dcs2 := s <- value + s*0.9, 1L:nrow(dt)]

output:

   year value     DCS    dcs2
1:   79     5  5.0000  5.0000
2:   80     2  6.5000  6.5000
3:   81     6 11.8500 11.8500
4:   82     8 18.6650 18.6650
5:   83     9 25.7985 25.7985

---

edit: in response to comment about grouping:

dt <- data.table(ID=c(1,1,2,2), value=1:4)
dt[, {
    n <- .N
    s <- 0;
    .SD[, {
        s <- value + s*0.9;
        s
      }, 
      1L:n]
  },  
  ID]

output:

   ID n  V1
1:  1 1 1.0
2:  1 2 2.9
3:  2 1 3.0
4:  2 2 6.7

Answer 2

Not a proper answer, but only a timing of the other answers. Hopefully this will be helpful to determine which option to choose:

Load Libraries

library(data.table)
library(Rcpp)

Create Dataset

set.seed(0L)
dt <- data.table(value = rpois(1e4, 100))

Create necessary functions

app_3 <- function(dt) {
  m <- matrix(0, nrow = nrow(dt), ncol = nrow(dt))
  v <- 0.9**(seq(nrow(dt)) - 1)
  m[lower.tri(m, diag = TRUE)] <- unlist(sapply(rev(seq_along(v)), function(k) head(v, k)))

  dt[, DCS3 := m %*% value]
}

system.time(
cppFunction("
NumericVector dcs(NumericVector x, double disc) {
    int n = x.size();
    NumericVector res(n);
    res[0] = x[0];
    for (int i=1; i<n; i++) {
        res[i] += x[i] + res[i-1]*disc;
    }
    return res;
}"))
#   user  system elapsed 
#   0.03    0.16   20.03 

Benchmark

res <- bench::mark(time_unit="s",
  app_1 = dt[, DCS1 := sapply(1:.N, function(k) sum(0.9**(k - 1:k)*head(value, k)))],
  app_2 = dt[, DCS2 := dt[, Reduce(function(x, y) 0.9 * x + y, as.list(value), accumulate = TRUE)]],
  app_3 = app_3(dt),

  dt_rcpp = dt[, DCS4 := dcs(value, 0.9)],
  dt_recursive = {s <- 0
  dt[, DCS5 := {
    s <- value + s*0.9
    s
  }, 1L:nrow(dt)]
  },

  min_time = 1
)

res

timings:

# A tibble: 5 x 13
  expression                   min   median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc total_time result       memory      time   gc        
  <bch:expr>                 <dbl>    <dbl>     <dbl> <bch:byt>    <dbl> <int> <dbl>      <dbl> <list>       <list>      <list> <list>    
1 app_1                   6.34     6.34         0.158    1.12GB    0.315     1     2      6.34  <df[,7] [10~ <df[,3] [5~ <bch:~ <tibble [~
2 app_2                   0.0109   0.0123      71.3    612.34KB   21.8      72    22      1.01  <df[,7] [10~ <df[,3] [2~ <bch:~ <tibble [~
3 app_3                   3.93     3.93         0.255     4.1GB    0.764     1     3      3.93  <df[,7] [10~ <df[,3] [2~ <bch:~ <tibble [~
4 dt_rcpp                 0.000308 0.000337  2681.     195.46KB    6.01   2679     6      0.999 <df[,7] [10~ <df[,3] [2~ <bch:~ <tibble [~
5 dt_recursive            0.00939  0.00972     99.2    294.52KB    6.94    100     7      1.01  <df[,7] [10~ <df[,3] [3~ <bch:~ <tibble [~

Another timings with 1e6 rows:

# A tibble: 3 x 13
  expression                  min  median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc total_time result        memory       time   gc        
  <bch:expr>                <dbl>   <dbl>     <dbl> <bch:byt>    <dbl> <int> <dbl>      <dbl> <list>        <list>       <list> <list>    
1 app_2                   1.52    1.52        0.659    53.5MB    6.59      1    10       1.52 <df[,5] [1,0~ <df[,3] [27~ <bch:~ <tibble [~
2 dt_rcpp                 0.00731 0.00942    89.9      15.3MB    0.899   100     1       1.11 <df[,5] [1,0~ <df[,3] [20~ <bch:~ <tibble [~
3 dt_recursive            0.902   0.905       1.10     22.9MB    1.66      2     3       1.81 <df[,5] [1,0~ <df[,3] [4,~ <bch:~ <tibble [~

^{Created on 2020-05-15 by the reprex package (v0.3.0)}

Answer 3

Maybe you can try the code below.

---

Approach 1

Follow the formula straightforwardly via using sum

dt[,DCS:=sapply(1:.N,function(k) sum(0.9**(k-1:k)*head(value,k)))]

---

Approach 2

Use Reduce from base R

dt[,Reduce(function(x,y) 0.9*x+y,as.list(value),accumulate = TRUE)]

---

Approach 3

• First, you can construct a matrix m that gives convolution-like coefficients

m <- matrix(0,nrow = nrow(dt),ncol = nrow(dt))
v <- 0.9**(seq(nrow(dt))-1)
m[lower.tri(m,diag = TRUE)] <- unlist(sapply(rev(seq_along(v)),function(k) head(v,k)))

or use shift to obtain matrix m (Thanks to @chinsoon12)

x <- 0L:(nrow(dt)-1L); 
m <- t(do.call(cbind, shift(0.9^x, x, fill=0)))

• Then you can run

dt[,DCS:=m%*%value]

---

Result

> dt
   year value     DCS
1:   79     5  5.0000
2:   80     2  6.5000
3:   81     6 11.8500
4:   82     8 18.6650
5:   83     9 25.7985