aggregate data frame by equal buckets - r

Question

I would like to aggregate an R data.frame by equal amounts of the cumulative sum of one of the variables in the data.frame. I googled quite a lot, but probably I don't know the correct terminology to find anything useful.
Suppose I have this data.frame:
> x <- data.frame(cbind(p=rnorm(100, 10, 0.1), v=round(runif(100, 1, 10))))
> head(x)
p v
1 10.002904 4
2 10.132200 2
3 10.026105 6
4 10.001146 2
5 9.990267 2
6 10.115907 6
7 10.199895 9
8 9.949996 8
9 10.165848 8
10 9.953283 6
11 10.072947 10
12 10.020379 2
13 10.084002 3
14 9.949108 8
15 10.065247 6
16 9.801699 3
17 10.014612 8
18 9.954638 5
19 9.958256 9
20 10.031041 7
I would like to reduce the x to a smaller data.frame where each line contains the weighted average of p, weighted by v, corresponding to an amount of n units of v. Something of this sort:
> n <- 100
> cum.v <- cumsum(x$v)
> f <- cum.v %/% n
> x.agg <- aggregate(cbind(v*p, v) ~ f, data=x, FUN=sum)
> x.agg$'v * p' <- x.agg$'v * p' / x.agg$v
> x.agg
f v * p v
1 0 10.039369 98
2 1 9.952049 94
3 2 10.015058 104
4 3 9.938271 103
5 4 9.967244 100
6 5 9.995071 69
First question, I was wondering if there is a better (more efficient approach) to the code above. The second, more important, question is how to correct the code above in order to obtain more precise bucketing. Namely, each row in x.agg should contain exacly 100 units of v, not just approximately as it is the case above. For example, the first row contains the aggregate of the first 17 rows of x which correspond to 98 units of v. The next row (18th) contains 5 units of v and is fully included in the next bucket. What I would like to achieve instead would be attribute 2 units of row 18th to the first bucket and the remaining 3 units to the following one.
Thanks in advance for any help provided.

Answer 1

Here's another method that does this with out repeating each p v times. And the way I understand it is, the place where it crosses 100 (see below)
18 9.954638 5 98
19 9.958256 9 107
should be changed to:
18 9.954638 5 98
19.1 9.958256 2 100 # ---> 2 units will be considered with previous group
19.2 9.958256 7 107 # ----> remaining 7 units will be split for next group
The code:
n <- 100
# get cumulative sum, an id column (for retrace) and current group id
x <- transform(x, cv = cumsum(x$v), id = seq_len(nrow(x)), grp = cumsum(x$v) %/% n)
# Paste these two lines in R to install IRanges
source("http://bioconductor.org/biocLite.R")
biocLite("IRanges")
require(IRanges)
ir1 <- successiveIRanges(x$v)
ir2 <- IRanges(seq(n, max(x$cv), by=n), width=1)
o <- findOverlaps(ir1, ir2)
# gets position where multiple of n(=100) occurs
# (where we'll have to do something about it)
pos <- queryHits(o)
# how much do the values differ from multiple of 100?
val <- start(ir2)[subjectHits(o)] - start(ir1)[queryHits(o)] + 1
# we need "pos" new rows of "pos" indices
x1 <- x[pos, ]
x1$v <- val # corresponding values
# reduce the group by 1, so that multiples of 100 will
# belong to the previous row
x1$grp <- x1$grp - 1
# subtract val in the original data x
x$v[pos] <- x$v[pos] - val
# bind and order them
x <- rbind(x1,x)
x <- x[with(x, order(id)), ]
# remove unnecessary entries
x <- x[!(duplicated(x$id) & x$v == 0), ]
x$cv <- cumsum(x$v) # updated cumsum
x$id <- NULL
require(data.table)
x.dt <- data.table(x, key="grp")
x.dt[, list(res = sum(p*v)/sum(v), cv = tail(cv, 1)), by=grp]
Running on your data:
# grp res cv
# 1: 0 10.037747 100
# 2: 1 9.994648 114
Running on #geektrader's data:
# grp res cv
# 1: 0 9.999680 100
# 2: 1 10.040139 200
# 3: 2 9.976425 300
# 4: 3 10.026622 400
# 5: 4 10.068623 500
# 6: 5 9.982733 562
Here's a benchmark on a relatively big data:
set.seed(12345)
x <- data.frame(cbind(p=rnorm(1e5, 10, 0.1), v=round(runif(1e5, 1, 10))))
require(rbenchmark)
benchmark(out <- FN1(x), replications=10)
# test replications elapsed relative user.self
# 1 out <- FN1(x) 10 13.817 1 12.586
It takes about 1.4 seconds on 1e5 rows.

Answer 2

If you are looking for precise bucketing, I am assuming value of p is same for 2 "split" v
i.e. in your example, value of p for 2 units of row 18th that go in first bucket is 9.954638
With above assumption, you can do following for not super large datasets..
> set.seed(12345)
> x <- data.frame(cbind(p=rnorm(100, 10, 0.1), v=round(runif(100, 1, 10))))
> z <- unlist(mapply(function(x,y) rep(x,y), x$p, x$v, SIMPLIFY=T))
this creates a vector with each value of p repeated v times for each row and result is combined into single vector using unlist.
After this aggregation is trivial using aggregate function
> aggregate(z, by=list((1:length(z)-0.5)%/%100), FUN=mean)
Group.1 x
1 0 9.999680
2 1 10.040139
3 2 9.976425
4 3 10.026622
5 4 10.068623
6 5 9.982733