Rolling / moving avg by group
grouping, moving-average, r
Solution
The only thing missing is a `do.call(rbind,d2_roll_mean)`. Add original data:
cbind(d,do.call(rbind,d2_roll_mean))
EDIT: I ran this through `system.time()` for a bigger example, and it does take its sweet time:
set.seed(31)
dd <- matrix(sample(seq(1:20),20000*500,replace=TRUE),ncol=500)
du <- sample(seq(1:350),20000,replace=TRUE)
d <- cbind(du,dd)
d <- d[order(d[,1]),]
system.time(d2_roll_mean <- by(d[,-1], d[,1], doApply))
User System elapsed
399.60 0.57 409.91
`by()` and `apply()` are not the fastest functions. It may actually be faster to walk through the columns using a `for` loop and doing this by brute force, relying on the fact that `d` is sorted by ID.
Problem
How to generate rolling mean with grouped data. Here's the data ``` set.seed(31) dd<-matrix(sample(seq(1:20),30,replace=TRUE),ncol=3) ``` Add a group identifier, and sort by group identifier ``` du<-sample(seq(1:4),10,replace=TRUE) d<-cbind(du,dd) d<-d[order(d[,1]),] ``` This gives the rolling mean but ignores group bounderis ``` d_roll_mean <- apply(d[,2:4], 2, function(x) { rollapply(zoo(x), 3, mean, partial=TRUE, align='right') } ) ``` This gives the results below ``` # cbind(d,d_roll_mean) # [1,] 1 3 3 12 3.000000 3.000000 12.000000 # [2,] 2 10 13 8 6.500000 8.000000 10.000000 # [3,] 2 17 2 17 10.000000 6.000000 12.333333 # [4,] 3 14 6 3 13.666667 7.000000 9.333333 # [5,] 3 6 20 1 12.333333 9.333333 7.000000 # [6,] 3 1 16 19 7.000000 14.000000 7.666667 # [7,] 3 19 2 11 8.666667 12.666667 10.333333 # [8,] 4 12 1 9 10.666667 6.333333 13.000000 # [9,] 4 10 13 12 13.666667 5.333333 10.666667 # [10,] 4 8 20 7 10.000000 11.333333 9.333333 ``` Here's the goal, rolling mean by group boundary ``` # Desired # [1,] 1 3 3 12 3.000000 3.000000 12.000000 # [2,] 2 10 13 8 10.000000 13.000000 8.000000 # [3,] 2 17 2 17 13.500000 7.500000 12.500000 # [4,] 3 14 6 3 14.000000 6.000000 3.000000 # [5,] 3 6 20 1 10.000000 13.000000 2.000000 # [6,] 3 1 16 19 7.000000 14.000000 7.666667 # [7,] 3 19 2 11 8.666667 12.666667 10.333333 # [8,] 4 12 1 9 12.000000 1.000000 9.000000 # [9,] 4 10 13 12 11.000000 7.000000 10.500000 # [10,] 4 8 20 7 10.000000 8.000000 9.333333 ``` This is close, but generates a list by factor, instead of a matrix ``` doApply <- function(x) { apply(x, 2, function(y) { rollapply(zoo(y), 3, mean, partial=TRUE, align='right') }) } d2_roll_mean <- by(d[,2:4], d[,1], doApply) ``` So there are some answers to the question, here's how they compare in execution time ``` set.seed(31) nrow=20000 ncol=600 nun=350 nValues = 20 dd<-matrix(sample(seq(1:nValues),nrow*ncol,replace=TRUE),ncol=ncol) du<-sample(seq(1:nun),nrow,replace=TRUE) d<-cbind(du,dd) d<-d[order(d[,1]),] library(zoo) doApply <- function(x) { apply(x, 2, function(y) { rollapply(zoo(y), 3, mean, partial=TRUE, align='right') }) } library(data.table) library(caTools) fun1<-function(d) {by(d[,-1], d[,1], doApply)} fun2<- function(d){ DT <- data.table(d, key='du') DT[, lapply(.SD, function(y) runmean(y, 3, alg='fast',align='right')), by=du] } system.time(d2_roll_mean <- fun1(d)) system.time(d2_roll_mean2 <- fun2(d)) ``` The timing indicates using data tables is about 10 times faster than rollapply. ``` user system elapsed fun1 1048.910 0.378 1049.158 fun2 107.296 0.097 107.392 ``` I don't get equality, but by inspection they seem the same... ``` d2a<-do.call(rbind,d2_roll_mean) d2b<-cbind(1,d2a) d2c<-data.table(d2b) setnames(d2c,names(d2c),names(d2_roll_mean2)) all.equal(d2c,d2_roll_mean2) ``` The output of all equal is ``` [1] "Attributes: < Length mismatch: comparison on first 1 components >" [2] "Component “du”: Mean relative difference: 175.6631" ``` When the above approach was applied to data, the following error was generated ``` Error in `[<-`(`*tmp*`, (k2 + 1):n, , value = 2) : subscript out of bounds ``` This error was the result of some factors have too few rows. Those rows were removed, and the process worked. Ref: How to drop factors that have fewer than n members