Combinatorical partial sum - r

Question

I am looking in R for a function partial.sum() which takes a vector of numbers and returns an ascending sorted vector of all partial sums:
test=c(2,5,10)
partial.sum(test)
# 2 5 7 10 12 15 17
## 2 is the sum of element 2
## 5 is the sum of element 5
## 7 is the sum of elements 2 & 5
## 10 is the sum of element 10
## 12 is the sum of elements 2 & 10
## 15 is the sum of elements 5 & 10
## 17 is the sum of elements 2 & 5 & 10

Answer 1

Here is one using recursion. (Not making claims about it being efficient either)
partial.sum <- function(x) {
slave <- function(x) {
if (length(x)) {
y <- Recall(x[-1])
c(y + 0, y + x[1])
} else 0
}
sort(unique(slave(x)[-1]))
}
partial.sum(c(2,5,10))
# [1] 2 5 7 10 12 15 17
Edit: well, turns out it is a little faster than I thought:
x <- 1:20
microbenchmark(flodel(x), dason(x), matthew(x), times = 10)
# Unit: milliseconds
# expr min lq median uq max neval
# flodel(x) 86.31128 86.9966 94.12023 125.1013 163.5824 10
# dason(x) 2407.27062 2461.2022 2633.73003 2846.2639 3031.7250 10
# matthew(x) 3084.59227 3191.7810 3304.36064 3693.8595 3883.2767 10
(I added sort and/or unique to dason and matthew's functions where appropriate for fair comparison.)

Answer 2

This probably doesn't scale too well and doesn't account for possible duplicates in the input vector or duplicates in the answer. You can use unique later if that is a concern for you.
partial.sum <- function(x){
n <- length(x)
# Something that will help get us every possible subset
# of the original vector
out <- do.call(expand.grid, replicate(n, c(T,F), simplify = F))
# Don't include the case where we don't grab any elements
out <- head(out, -1)
# ans <- apply(out, 1, function(row){sum(x[row])})
# As flodel points out the following will be faster than
# the previous line
ans <- data.matrix(out) %*% x
# If you want only unique value then add a call to unique here
ans <- sort(unname(ans))
ans
}

Answer 3

Here's an iterative approach using combn to produce combinations to sum. It works for vectors of of length greater than 1.
partial.sum <- function(x) {
sort(unique(unlist(sapply(seq_along(x), function(i) colSums(combn(x,i))))))
}
## [1] 2 5 7 10 12 15 17
To handle lengths less than 2, test for the length:
partial.sum <- function(x) {
if (length(x) > 1) {
sort(unique(unlist(sapply(seq_along(x), function(i) colSums(combn(x,i))))))
} else {
x
}
}
Some timings, out of rbenchmark, which don't entirely agree with flodel's results. I modified Dason's code, removing the comments and adding a call to unique. The version of my code is the first, without the if. flodel's code is a clear winner here.
> test <- 1:10
> benchmark(matthew(test), flodel(test), dason(test), replications=100)
test replications elapsed relative user.self sys.self user.child sys.child
3 dason(test) 100 0.180 12.857 0.175 0.004 0 0
2 flodel(test) 100 0.014 1.000 0.015 0.000 0 0
1 matthew(test) 100 0.244 17.429 0.242 0.001 0 0
> test <- 1:20
> benchmark(matthew(test), flodel(test), dason(test), replications=1)
test replications elapsed relative user.self sys.self user.child sys.child
3 dason(test) 1 5.231 98.698 5.158 0.058 0 0
2 flodel(test) 1 0.053 1.000 0.053 0.000 0 0
1 matthew(test) 1 2.184 41.208 2.180 0.000 0 0
> test <- 1:25
> benchmark(matthew(test), flodel(test), dason(test), replications=1)
test replications elapsed relative user.self sys.self user.child sys.child
3 dason(test) 1 288.957 163.345 264.068 23.859 0 0
2 flodel(test) 1 1.769 1.000 1.727 0.038 0 0
1 matthew(test) 1 75.712 42.799 74.745 0.847 0 0