Reputation: 77
I am currently trying to simplify this summation. I am new to R.
Lx = c(5050.0, 65.0, 25.0, 19.0, 17.5, 16.5, 15.5, 14.5, 13.5, 12.5, 6.0, 0.0)
Tx = c(sum(Lx[1:12]),sum(Lx[2:12]),sum(Lx[3:12]),sum(Lx[4:12]),
sum(Lx[5:12]),sum(Lx[6:12]),sum(Lx[7:12]),sum(Lx[8:12]),
sum(Lx[9:12]),sum(Lx[10:12]),sum(Lx[11:12]),sum(Lx[12:12]))
Upvotes: 6
Views: 1628
Reputation: 51974
Package spatstat.utils
provides a fast version ("under certain conditions") of the reverse cumulative sum, revcumsum
, which is based on computing sum(x[i:n])
with n = length(x)
(basically @Jan Brederecke's answer):
Lx = c(5050.0, 65.0, 25.0, 19.0, 17.5, 16.5, 15.5, 14.5, 13.5, 12.5, 6.0, 0.0)
# install.packages("spatstat.utils")
spatstat.utils::revcumsum(Lx)
# [1] 5255.0 205.0 140.0 115.0 96.0 78.5 62.0 46.5 32.0 18.5 6.0 0.0
Benchmark
x = c(5050.0, 65.0, 25.0, 19.0, 17.5, 16.5, 15.5, 14.5, 13.5, 12.5, 6.0, 0.0)
bm <- microbenchmark(
fRev(x),
fReduce(x),
fJan(x),
fEshita(x),
fsapply(x),
fRevcumsum(x),
times = 100L
)
autoplot(bm)
rev(cumsum(rev(Lx)))
and spatstat.utils::revcumsum(Lx)
seem like the fastest solutions.
Upvotes: 2
Reputation: 11
Please try the following code:
Lx = c(5050.0, 65.0, 25.0, 19.0, 17.5, 16.5, 15.5, 14.5, 13.5, 12.5, 6.0, 0.0)
l=length(Lx)
aa=list()
for(i in 1:l)
{
x=sum((Lx[i:l]))
aa=append(aa,x)
}
all the values after summation will be in the list "aa".
Upvotes: 1
Reputation: 34416
You can do:
rev(cumsum(rev(Lx)))
[1] 5255.0 205.0 140.0 115.0 96.0 78.5 62.0 46.5 32.0 18.5 6.0 0.0
Or alternatively, using Reduce()
:
Reduce(`+`, Lx, right = TRUE, accumulate = TRUE)
[1] 5255.0 205.0 140.0 115.0 96.0 78.5 62.0 46.5 32.0 18.5 6.0 0.0
Upvotes: 14
Reputation: 25323
A possible solution, using sapply
:
sapply(1:12, function(x) sum(Lx[x:12]))
#> [1] 5255.0 205.0 140.0 115.0 96.0 78.5 62.0 46.5 32.0 18.5
#> [11] 6.0 0.0
Upvotes: 2
Reputation:
Using a for loop:
Tx_new <- vector(length = length(Lx))
for (i in 1:length(Lx)) {
Tx_new[i] <- sum(Lx[i:length(Lx)])
}
Upvotes: 1