Reputation: 2675
I have a dataset like the following:
library(data.table)
dt1 <- data.table(urn = c(rep("a", 5), rep("b", 4)),
amount = c(10, 12, 23, 15, 19, 42, 11, 5, 10),
date = as.Date(c("2016-01-01", "2017-01-02", "2017-02-04",
"2017-04-19", "2018-02-11", "2016-02-14",
"2017-05-06", "2017-05-12", "2017-12-12")))
dt1
# urn amount date
# 1: a 10 2016-01-01
# 2: a 12 2017-01-02
# 3: a 23 2017-02-04
# 4: a 15 2017-04-19
# 5: a 19 2018-02-11
# 6: b 42 2016-02-14
# 7: b 11 2017-05-06
# 8: b 5 2017-05-12
# 9: b 10 2017-12-12
I am trying to determine the cumulative value for a group over the preceding 12 months. I know I can use shift
with data.table
to scan backwards or forwards, the biggest challenge I can't get my head around is how to know how many records to sum when the number can change based on how many records each urn
has.
The type of results I am looking for are:
dt1
# urn amount date summed12m
# 1: a 10 2016-01-01 10
# 2: a 12 2017-01-02 12
# 3: a 23 2017-02-04 35
# 4: a 15 2017-04-19 50
# 5: a 19 2018-02-11 34
# 6: b 42 2016-02-14 42
# 7: b 11 2017-05-06 11
# 8: b 5 2017-05-12 16
# 9: b 10 2017-12-12 26
I'm preferably looking for a data.table
solution due to the volume of my data, but am open to other options too if it is likely to be efficient over a table with about 12M records.
Upvotes: 9
Views: 1622
Reputation: 11955
dt1[, summed12m := {
date_diff <- date - dt1$date
sum(dt1$amount[date_diff >= 0 & date_diff <= 365 & urn==dt1$urn])
},
by = list(date, urn)]
Output is:
urn amount date summed12m
1: a 10 2016-01-01 10
2: a 12 2017-01-02 12
3: a 23 2017-02-04 35
4: a 15 2017-04-19 50
5: a 19 2018-02-11 34
6: b 42 2016-02-14 42
7: b 11 2017-05-06 11
8: b 5 2017-05-12 16
9: b 10 2017-12-12 26
Sample data:
dt1 <- structure(list(urn = c("a", "a", "a", "a", "a", "b", "b", "b",
"b"), amount = c(10, 12, 23, 15, 19, 42, 11, 5, 10), date = structure(c(16801,
17168, 17201, 17275, 17573, 16845, 17292, 17298, 17512), class = "Date")), .Names = c("urn",
"amount", "date"), row.names = c(NA, -9L), class = c("data.table",
"data.frame"), .internal.selfref = <pointer: 0x00000000027b0788>)
Upvotes: 2
Reputation: 42544
As alternative to foverlaps()
, this also can be solved by aggregating in a non-equi join:
library(lubridate)
dt1[, summed12m := dt1[.(urn, date, date %m-% months(12)),
on = .(urn = V1, date <= V2, date >= V3),
sum(amount), by = .EACHI]$V1][]
urn amount date summed12m 1: a 10 2016-01-01 10 2: a 12 2017-01-02 12 3: a 23 2017-02-04 35 4: a 15 2017-04-19 50 5: a 19 2018-02-11 34 6: b 42 2016-02-14 42 7: b 11 2017-05-06 11 8: b 5 2017-05-12 16 9: b 10 2017-12-12 26
lubridate
is used for date arithmetic to avoid mishaps in case one of the dates is February, 29.
The essential part is the non-equi join
dt1[.(urn, date, date %m-% months(12)),
on = .(urn = V1, date <= V2, date >= V3),
sum(amount), by = .EACHI]
urn date date V1 1: a 2016-01-01 2015-01-01 10 2: a 2017-01-02 2016-01-02 12 3: a 2017-02-04 2016-02-04 35 4: a 2017-04-19 2016-04-19 50 5: a 2018-02-11 2017-02-11 34 6: b 2016-02-14 2015-02-14 42 7: b 2017-05-06 2016-05-06 11 8: b 2017-05-12 2016-05-12 16 9: b 2017-12-12 2016-12-12 26
of which the last column is picked to create the new summed12m
column in dt1
.
The OP has asked where V1
, V2
, and V3
come from.
The expression .(urn, date, date %m-% months(12))
creates a new data.table on the fly. (.()
is an data.table
abbreviation for list()
). As no column names have been specified, data.table
creates default column names V1
, V2
, etc.
Less sloppily, the expression can be re-written with explicitely named columns
dt1[.(urn = urn, end = date, start = date %m-% months(12)),
on = .(urn, date <= end, date >= start),
sum(amount), by = .EACHI]
Upvotes: 12
Reputation: 25225
This is shouting for foverlaps
. My first time using foverlaps
, hence I am quite sure a few experts here can better wield the function. Here it goes:
dt1[, date2 := date]
rng <- dt1[, .(urn, enddate=date,
startdate=as.Date(paste(year(date)-1, month(date), mday(date), sep="-")))]
setkey(rng, urn, startdate, enddate)
foverlaps(dt1, rng, by.x=c("urn","date","date2"), type="within")[,
sum(amount), by=.(urn, enddate)]
# urn enddate V1
# 1: a 2016-01-01 10
# 2: a 2017-01-02 12
# 3: a 2017-02-04 35
# 4: a 2017-04-19 50
# 5: a 2018-02-11 34
# 6: b 2016-02-14 42
# 7: b 2017-05-06 11
# 8: b 2017-05-12 16
# 9: b 2017-12-12 26
Further reading:
Upvotes: 5