Aggregate column - how to handle uneven dataframe - r

Question

I want to aggregate per Chr every third row (sum of three rows). However, since my df is not divisible by 3, I am not sure how to handle the last remaining rows, which could be just 1 or 2 rows. If two rows remain, I would like to just sum those 2 remaining rows.
Input
data.frame(Chr = c("chr1","chr1","chr1","chr1","chr1","chr2","chr2","chr2","chr2","chr2","chr3"),
value = c(1,3,1,3,5,6,3,1,3,5,0),
seq = c(1,2,3,4,5,1,2,3,4,5,6))
Output (using dplyr mutate, keeping all columns)
data.frame(Chr = c("chr1","chr1","chr1","chr1","chr1","chr2","chr2","chr2","chr2","chr2","chr3"),
value = c(1,3,1,3,5,6,3,1,3,5,0),
seq = c(1,2,3,4,5,1,2,3,4,5,6),
agg = c(5,5,5,8,8,10,10,10,8,8,8))

Answer 1

Here is a similar yet other approach: Trying out the new .by argument.
I tried to use it also for the last mutate but it was not possible to combine cumsum(.. with .by:
library(dplyr)
df %>%
mutate(group = as.integer(gl(n(),3,n())), .by=Chr) %>%
mutate(id = row_number(), .by = c(Chr, group)) %>%
group_by(Chr, sumgroup = cumsum(id == 1)) %>%
mutate(agg = sum(value)) %>%
ungroup() %>%
select(Chr, value, seq, agg)
Chr value seq agg
<chr> <dbl> <dbl> <dbl>
1 chr1 1 1 5
2 chr1 3 2 5
3 chr1 1 3 5
4 chr1 3 4 8
5 chr1 5 5 8
6 chr2 6 1 10
7 chr2 3 2 10
8 chr2 1 3 10
9 chr2 3 4 8
10 chr2 5 5 8
11 chr3 0 6 0

Answer 2

An option with dplyr & ave
library(dplyr) # >= 1.1.0
df1 %>%
mutate(agg = ave(value, as.integer(gl(n(), 3, n())), FUN = sum), .by = Chr)
-output
Chr value seq agg
1 chr1 1 1 5
2 chr1 3 2 5
3 chr1 1 3 5
4 chr1 3 4 8
5 chr1 5 5 8
6 chr2 6 1 10
7 chr2 3 2 10
8 chr2 1 3 10
9 chr2 3 4 8
10 chr2 5 5 8
11 chr3 0 6 0
Or with data.table
library(data.table)
setDT(df1)[, agg := .SD[, rep(sum(value), .N),
as.integer(gl(.N, 3, .N))]$V1, Chr]
-output
> df1
Chr value seq agg
1: chr1 1 1 5
2: chr1 3 2 5
3: chr1 1 3 5
4: chr1 3 4 8
5: chr1 5 5 8
6: chr2 6 1 10
7: chr2 3 2 10
8: chr2 1 3 10
9: chr2 3 4 8
10: chr2 5 5 8
11: chr3 0 6 0

Answer 3

You can use (row_number()-1) %/% 3 to group per 3 observations.
library(dplyr)
df %>%
mutate(Grp = (row_number()-1) %/% 3 + 1, .by = Chr) %>%
mutate(agg = sum(value), .by = c(Chr, Grp))
# Chr value seq Grp agg
# 1 chr1 1 1 1 5
# 2 chr1 3 2 1 5
# 3 chr1 1 3 1 5
# 4 chr1 3 4 2 8
# 5 chr1 5 5 2 8
# 6 chr2 6 1 1 10
# 7 chr2 3 2 1 10
# 8 chr2 1 3 1 10
# 9 chr2 3 4 2 8
# 10 chr2 5 5 2 8
# 11 chr3 0 6 1 0
If the Grp column is not needed, you can drop it with select(-Grp).

Answer 4

Diving the row number (within Chr groups) by 3 and rounding up this ratio gives 1 for the first 3 values, 2 for the next 3 values, etc. You can then group by Chr and this variable to compute your sums:
d %>%
group_by(Chr) %>%
mutate(Chr_group = ceiling(row_number()/3)) %>%
group_by(Chr, Chr_group) %>%
mutate(agg = sum(value)) %>%
ungroup()