Sum of most recent values across groups

Question

For each row of my data I\'d like to compute the sum of most recent value for each group:

dt = data.table(group = c(\'a\',\'b\',\'a\',\         


        

Answer 1

I would create a column for each group showing the latest value for that group. Then just sum those columns:

library(zoo)
result <- rep(0, nrow(dt))
for(g in dt[, unique(group)]) {
  result <- result + dt[, na.fill(na.locf(ifelse(group==g, 1, NA)*value, na.rm=F), 0)]
}

all(dt[, desired] == result)

Answer 2

using dplyr, works for many groups, but data must not be data table.

library(dplyr)
library(tidyr)
library(zoo)
dt %>% 
  mutate(row_number = row_number()) %>%
  spread(group, value) %>%
  arrange(row_number) %>%
  mutate_each(funs(na.locf(., na.rm = FALSE))) %>%
  mutate(answer = rowSums(.[,-1:-2], na.rm = T))

Using the above function on example data (notice data.frame() not data.table():

dt = data.frame(group = c('a','b','a','a','b','a'),
                value = c(10, 5, 20, 15, 15, 10),
                desired = c(10, 15, 25, 20, 30, 25))
  desired row_number  a  b answer
1      10          1 10 NA     10
2      15          2 10  5     15
3      25          3 20  5     25
4      20          4 15  5     20
5      30          5 15 15     30
6      25          6 10 15     25

dt = data.frame(group = c('a','b','c','a','a','b','c','a'),
                value = c(10, 5, 20, 25, 15, 15, 30, 10),
                desired = c(10, 15, 35, 50, 40, 50, 60, 55))

  desired row_number  a  b  c answer
1      10          1 10 NA NA     10
2      15          2 10  5 NA     15
3      35          3 10  5 20     35
4      50          4 25  5 20     50
5      40          5 15  5 20     40
6      50          6 15 15 20     50
7      60          7 15 15 30     60
8      55          8 10 15 30     55

Answer 3

Even simpler logic from @eddi (under comments) reducing the roundabout one shown below:

dt[, incr := diff(c(0, value)), by = group][, ans := cumsum(incr)]

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Not sure how it extends to more groups, but here's on an example data with 3 groups:

# I hope I got the desired output correctly
require(data.table)
dt = data.table(group = c('a','b','c','a','a','b','c','a'),
                value = c(10, 5, 20, 25, 15, 15, 30, 10),
                desired = c(10, 15, 35, 50, 40, 50, 60, 55))

Add an rleid:

dt[, id := rleid(group)]

Extract the last row for each group, id:

last = dt[, .(value=value[.N]), by=.(group, id)]

last will have unique id. Now the idea is to get the increment for each id, and then join+update back.

last = last[, incr := value - shift(value, type="lag", fill=0L), by=group
          ][, incr := cumsum(incr)-value][]

Join + update now:

dt[last, ans := value + i.incr, on="id"][, id := NULL][]
#    group value desired ans
# 1:     a    10      10  10
# 2:     b     5      15  15
# 3:     c    20      35  35
# 4:     a    25      50  50
# 5:     a    15      40  40
# 6:     b    15      50  50
# 7:     c    30      60  60
# 8:     a    10      55  55

I'm not yet sure where/if this breaks.. will look at it carefully now. I wrote it immediately so that there are more eyes on it.

---

Comparing on 500 groups with 10,000 rows with David's solution:

require(data.table)
set.seed(45L)
groups = apply(matrix(sample(letters, 500L*10L, TRUE), ncol=10L), 1L, paste, collapse="")
uniqueN(groups) # 500L
N = 1e4L
dt = data.table(group=sample(groups, N, TRUE), value = sample(100L, N, TRUE))

arun <- function(dt) {

    dt[, id := rleid(group)]
    last = dt[, .(value=value[.N]), by=.(group, id)]
    last = last[, incr := value - shift(value, type="lag", fill=0L), by=group
              ][, incr := cumsum(incr)-value][]
    dt[last, ans := value + i.incr, on="id"][, id := NULL][]
    dt$ans
}

david <- function(dt) {
    dt[, indx := .I]
    res <- dcast(dt, indx ~ group)
    for (j in names(res)[-1L]) 
        set(res, j = j, value = res[!is.na(res[[j]])][res, on = "indx", roll = TRUE][[j]])
    rowSums(as.matrix(res)[, -1], na.rm = TRUE)

}

system.time(ans1 <- arun(dt))  ## 0.024s
system.time(ans2 <- david(dt)) ## 38.97s 
identical(ans1, as.integer(ans2))
# [1] TRUE