Rolling Sum by Another Variable in R

Question

I want to get the rolling 7-day sum by ID. Suppose my data looks like this:

data<-as.data.frame(matrix(NA,42,3))
data$V1<-seq(as.Date("2014-05-01"),as.Date("2014-09-01"),by=3)
data$V2<-rep(1:6,7)
data$V3<-rep(c(1,2),21)
colnames(data)<-c("Date","USD","ID")

         Date USD ID
1  2014-05-01   1  1
2  2014-05-04   2  2
3  2014-05-07   3  1
4  2014-05-10   4  2
5  2014-05-13   5  1
6  2014-05-16   6  2
7  2014-05-19   1  1
8  2014-05-22   2  2
9  2014-05-25   3  1
10 2014-05-28   4  2

How can I add a new column that will contain the rolling 7-day sum by ID?

Answer 1

If your data is big, you might want to check out this solution which uses data.table. It is pretty fast. If you need more speed, you can always change mapply to mcmapply and use multiple cores.

#Load data.table and convert to data.table object
require(data.table)
setDT(data)[,ID2:=.GRP,by=c("ID")]

#Build reference table
Ref <- data[,list(Compare_Value=list(I(USD)),Compare_Date=list(I(Date))), by=c("ID2")]

#Use mapply to get last seven days of value by id
data[,Roll.Val := mapply(RD = Date,NUM=ID2, function(RD, NUM) {
                  d <- as.numeric(Ref$Compare_Date[[NUM]] - RD)
                  sum((d <= 0 & d >= -7)*Ref$Compare_Value[[NUM]])})]

Answer 2

1) Assuming you mean every successive overlapping 7 rows for that ID:

library(zoo)

transform(data, roll = ave(USD, ID, FUN = function(x) rollsumr(x, 7, fill = NA)))

2) If you really did mean 7 days and not 7 rows then try this:

library(zoo)

z <- read.zoo(data)
z0 <- merge(z, zoo(, seq(start(z), end(z), "day")), fill = 0) # expand to daily
roll <- function(x) rollsumr(x, 7, fill = NA)
transform(data, roll = ave(z0$USD, z0$ID, FUN = roll)[time(z)])

Updated Added (2) and made some improvements.

Answer 3

Dataset provided by OP does not expose the complexity of the task. In terms of addressing OP question so far only Mike's answer was the correct one.
In fact for a 8 rolling days, instead of 7 rolling days, due to d <= 0 & d >= -7.
zoo solution by @G. Grothendieck is almost valid, only if merge would be made to each group of ID.
Below second data.table solution, this time valid results, using dev RcppRoll which allows na.rm=TRUE.
And slightly formatted Mike's solution output.

data<-as.data.frame(matrix(NA,42,3))
data$V1<-seq(as.Date("2014-05-01"),as.Date("2014-09-01"),by=3)
data$V2<-rep(1:6,7)
data$V3<-rep(c(1,2),21)
colnames(data)<-c("Date","USD","ID")

library(microbenchmark)
library(RcppRoll) # install_github("kevinushey/RcppRoll")
library(data.table) # install_github("Rdatatable/data.table")
correct_jan_dt = function(n, partial=TRUE){
  DT = as.data.table(data) # this can be speedup by setDT()
  date.range = DT[,range(Date)]
  all.dates = seq.Date(date.range[1],date.range[2],by=1)
  setkey(DT,ID,Date)
  r = DT[CJ(unique(ID),all.dates)][, c("roll") := as.integer(roll_sumr(USD, n, normalize = FALSE, na.rm = TRUE)), by="ID"][!is.na(USD)]
  # This could be simplified when `partial` arg will be implemented in [kevinushey/RcppRoll](https://github.com/kevinushey/RcppRoll)
  if(isTRUE(partial)){
    r[is.na(roll), roll := cumsum(USD), by="ID"][]
  }
  return(r[order(Date,ID)])
}
correct_mike_dt = function(){
  data = as.data.table(data)[,ID2:=.GRP,by=c("ID")]
  #Build reference table
  Ref <- data[,list(Compare_Value=list(I(USD)),Compare_Date=list(I(Date))), by=c("ID2")]
  #Use mapply to get last seven days of value by id
  data[, c("roll") := mapply(RD = Date,NUM=ID2, function(RD, NUM){
    d <- as.numeric(Ref$Compare_Date[[NUM]] - RD)
    sum((d <= 0 & d >= -7)*Ref$Compare_Value[[NUM]])})][,ID2:=NULL][]
}
identical(correct_mike_dt(), correct_jan_dt(n=8,partial=TRUE))
# [1] TRUE
microbenchmark(unit="relative", times=5L, correct_mike_dt(), correct_jan_dt(8))
# Unit: relative
#               expr      min       lq     mean   median       uq      max neval
#  correct_mike_dt() 274.0699 273.9892 267.2886 266.6009 266.2254 256.7296     5
#  correct_jan_dt(8)   1.0000   1.0000   1.0000   1.0000   1.0000   1.0000     5

Looking forward for update from @Khashaa.

Edit (20150122.2): Below benchmarks do not answer OP question.

Timing on a bigger (still very tiny) dataset, 5439 rows:

library(zoo)
library(data.table)
library(dplyr)
library(RcppRoll)
library(microbenchmark)
data<-as.data.frame(matrix(NA,5439,3))
data$V1<-seq(as.Date("1970-01-01"),as.Date("2014-09-01"),by=3)
data$V2<-sample(1:6,5439,TRUE)
data$V3<-sample(c(1,2),5439,TRUE)
colnames(data)<-c("Date","USD","ID")
zoo_f = function(){
    z <- read.zoo(data)
    z0 <- merge(z, zoo(, seq(start(z), end(z), "day")), fill = 0) # expand to daily
    roll <- function(x) rollsumr(x, 7, fill = NA)
    transform(data, roll = ave(z0$USD, z0$ID, FUN = roll)[time(z)])
}
dt_f = function(){
    DT = as.data.table(data) # this can be speedup by setDT()
    date.range = DT[,range(Date)]
    all.dates = seq.Date(date.range[1],date.range[2],by=1)
    setkey(DT,Date)
    DT[.(all.dates)
       ][order(Date), c("roll") := rowSums(setDT(shift(USD, 0:6, NA, "lag")),na.rm=FALSE), by="ID"
         ][!is.na(ID)]
}
dp_f = function(){
  data %>% group_by(ID) %>% 
    mutate(roll=roll_sum(c(rep(NA,6), USD), 7))
} 
dt2_f = function(){
  # this can be speedup by setDT()
  as.data.table(data)[, c("roll") := roll_sum(c(rep(NA,6), USD), 7), by="ID"][]
}
identical(as.data.table(zoo_f()),dt_f())
# [1] TRUE
identical(setDT(as.data.frame(dp_f())),dt_f())
# [1] TRUE
identical(dt2_f(),dt_f())
# [1] TRUE
microbenchmark(unit="relative", times=20L, zoo_f(), dt_f(), dp_f(), dt2_f())
# Unit: relative
#     expr        min         lq       mean     median         uq        max neval
#  zoo_f() 140.331889 141.891917 138.064126 139.381336 136.029019 137.730171    20
#   dt_f()  14.917166  14.464199  15.210757  16.898931  16.543811  14.221987    20
#   dp_f()   1.000000   1.000000   1.000000   1.000000   1.000000   1.000000    20
#  dt2_f()   1.536896   1.521983   1.500392   1.518641   1.629916   1.337903    20

Yet I'm not sure if my data.table code is already optimal.

Above functions did not answer OP question. Read the top of post for update. Mike's solution was the correct one.