Sum every nth points

Question

I have a vector and I need to sum every n numbers and return the results. This is the way I plan on doing it currently. Any better way to do this?

Answer 1

One way is to convert your vector to a matric then take the column sums:

colSums(matrix(v, nrow=n))
[1]  55 155 255 355 455 555 655 755 855 955

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Just be careful: this implicitly assumes that your input vector can in fact be reshaped to a matrix. If it can't, R will recycle elements of your vector to complete the matrix.

Answer 2

I will add one more way of doing it without any function from apply family

v <- 1:100
n <- 10

diff(c(0, cumsum(v)[slice.index(v, 1)%%n == 0]))
##  [1]  55 155 255 355 455 555 655 755 855 955

Answer 3

UPDATE:

If you want to sum every n consecutive numbers use colSums
If you want to sum every nth number use rowSums

as per Josh's comment, this will only work if n divides length(v) nicely.

rowSums(matrix(v, nrow=n))
 [1] 460 470 480 490 500 510 520 530 540 550

colSums(matrix(v, nrow=n))
 [1]  55 155 255 355 455 555 655 755 855 955

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Answer 4

Update

The olde version don't work. Here a ne awnser that use rep to create the grouping factor. No need to use cut:

n <- 5 
vv <- sample(1:1000,100)
seqs <- seq_along(vv)
tapply(vv,rep(seqs,each=n)[seqs],FUN=sum)

---

You can use tapply

tapply(1:100,cut(1:100,10),FUN=sum)

or to get a list

by(1:100,cut(1:100,10),FUN=sum)

EDIT

In case you have 1:92, you can replace your cut by this :

cut(1:92,seq(1,92,10),include.lowest=T)

Answer 5

Here are some of the main variants offered so far

f0 <- function(v, n) {
    sidx = seq.int(from=1, to=length(v), by=n)
    eidx = c((sidx-1)[2:length(sidx)], length(v))
    sapply(1:length(sidx), function(i) sum(v[sidx[i]:eidx[i]]))
}

f1 <- function(v, n, na.rm=TRUE) {    # 'tapply'
    unname(tapply(v, (seq_along(v)-1) %/% n, sum, na.rm=na.rm))
}

f2 <- function(v, n, na.rm=TRUE) {    # 'matrix'
    nv <- length(v)
    if (nv %% n)
        v[ceiling(nv / n) * n] <- NA
    colSums(matrix(v, n), na.rm=na.rm)
}

f3 <- function(v, n) {                # 'cumsum'
    nv = length(v)
    i <- c(seq_len(nv %/% n) * n, if (nv %% n) nv else NULL)
    diff(c(0L, cumsum(v)[i]))
}

Basic test cases might be

v = list(1:4, 1:5, c(NA, 2:4), integer())
n = 2

f0 fails with the final test, but this could probably be fixed

> f0(integer(), n)
Error in sidx[i]:eidx[i] : NA/NaN argument

The cumsum approach f3 is subject to rounding error, and the presence of an NA early in v 'poisons' later results

> f3(c(NA, 2:4), n)
[1] NA NA

In terms of performance, the original solution is not bad

> library(rbenchmark)
> cols <- c("test", "elapsed", "relative")
> v <- 1:100; n <- 10
> benchmark(f0(v, n), f1(v, n), f2(v, n), f3(v, n),
+           columns=cols)
      test elapsed relative
1 f0(v, n)   0.012     3.00
2 f1(v, n)   0.065    16.25
3 f2(v, n)   0.004     1.00
4 f3(v, n)   0.004     1.00

but the matrix solution f2 seems to be both fast and flexible (e.g., adjusting the handling of that trailing chunk of fewer than n elements)

> v <- runif(1e6); n <- 10
> benchmark(f0(v, n), f2(v, n), f3(v, n), columns=cols, replications=10)
      test elapsed relative
1 f0(v, n)   5.804   34.141
2 f2(v, n)   0.170    1.000
3 f3(v, n)   0.251    1.476

Answer 6

unname(tapply(v, (seq_along(v)-1) %/% n, sum))
# [1] 55 155 255 355 455 555 655 755 855 955