Discounted Cumulative Sum in R
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
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
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)