Reputation: 533
I need to optimize an R script. In particular, I need to speed up or delete some of the script's inundation cycles. I have defined many cycles of the type:
DT <- data.frame("x"=c(1:20),
"y"=c(20:1))
DT$vect[1] <- DT$y[1]
for (i in 2:20) {
DT$vect[i] <- DT$vect[i-1] * DT$x[i] - DT$x[i-1] * (1 + DT$y[i])
}
Since to calculate the value at position i
one needs to know that at position i-1
. I can not think of a better solution.
Does anyone know a smarter one?
Upvotes: 4
Views: 1050
Reputation: 206187
It might not be that much prettier, but you can use dplyr
and purrr
to do a reduce type function.
DT %>%
select(x,y) %>%
mutate(prevx=lag(x, default=-1)) %>%
transpose() %>%
accumulate(function(prev, xx) {
prev * xx$x - xx$prevx*(1+xx$y)
}, .init=-1/DT$x[1]) %>%
tail(-1)
# [1] 2.000000e+01 2.000000e+01 2.200000e+01 3.400000e+01 1.020000e+02
# [6] 5.320000e+02 3.634000e+03 2.897400e+04 2.606620e+05 2.606512e+06
# [11] 2.867152e+07 3.440582e+08 4.472756e+09 6.261858e+10 9.392787e+11
# [16] 1.502846e+13 2.554838e+14 4.598709e+15 8.737547e+16 1.747509e+18
We use the lag()
function to get both x[i]
and x[i-1]
on the same row. We use transpose
to get a list of named values that we can iterate over. Then accumulate()
allows use to keep plugging the output of a function back into itself as input and keeps track of the values along the way. Here we plug in the formula provided and use a special initial value that satisfied the initial conditions you gave of having the first value be equal to the first y value. Finally we trim off the dummy first value.
Upvotes: 3
Reputation: 26823
@MrFlick's solution is very nice, but if you are more comfortable with the for
loop and don't mind mixing in another language, you can give Rcpp a try. This type of loops are a prime example where C++ is more efficient:
#include <Rcpp.h>
using namespace Rcpp;
// [[Rcpp::export]]
NumericVector forLoop(DataFrame dt) {
int N = dt.nrow();
NumericVector x = dt["x"];
NumericVector y = dt["y"];
NumericVector vec(N, y(0));
for (int i = 1; i < N; ++i) {
vec(i) = vec(i-1) * x(i) - x(i-1) * (1 + y(i));
}
return vec;
}
/*** R
N <- 20000
DT <- data.frame("x"=c(1:N),
"y"=c(N:1))
DT$vect[1] <- DT$y[1]
system.time({
for (i in 2:N) {
DT$vect[i] <- DT$vect[i-1] * DT$x[i] - DT$x[i-1] * (1 + DT$y[i])
}
})
DT2 <- data.frame("x"=c(1:N),
"y"=c(N:1))
vect <- vector("numeric", length = N)
vect[1] <- DT2$y[1]
system.time({
for (i in 2:N) {
vect[i] <- vect[i-1] * DT2$x[i] - DT2$x[i-1] * (1 + DT2$y[i])
}
DT2$vect <- vect
})
all.equal(DT, DT2)
DT3 <- data.frame("x"=c(1:N),
"y"=c(N:1))
system.time({
vect <- forLoop(DT3)
DT3$vect <- vect
})
all.equal(DT, DT3)
*/
The original loop takes 1.5 s on my machine, while the C++ solution DT3
is "instantaneous". Between the two there is a minor optimization you can do in R: Do not write to a data.frame inside a loop. You are better of writing to a vector and adding that in the end. Here the output of profvis
for DT
and DT2
:
Still much slower than C++, though.
Upvotes: 2