Calculating moving average

Question

I'm trying to use R to calculate the moving average over a series of values in a matrix. There doesn't seem to be a built-in function in R that will allow me to calculate moving averages. Do any packages provide one? Or do I need to write my own?

Answer 1

Another useful function if you want the two ends of series not to be NA but to be recursively calculated moving averages:

smoothing = function(x, k=1) {
  sapply(seq_along(x), function(i) {
    i.min = max(i-k, 1)
    i.max = min(i+k, length(x))
    mean(x[i.min:i.max], na.rm=TRUE)
  })
}

Example:

x = 1:10/2

[1] 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

smoothing(x, 2)

[1] 1.00 1.25 1.50 2.00 2.50 3.00 3.50 4.00 4.25 4.50

Answer 2

In data.table 1.12.0 new frollmean function has been added to compute fast and exact rolling mean carefully handling NA, NaN and +Inf, -Inf values.

As there is no reproducible example in the question there is not much more to address here.

You can find more info about ?frollmean in manual, also available online at ?frollmean.

Examples from manual below:

library(data.table)
d = as.data.table(list(1:6/2, 3:8/4))

# rollmean of single vector and single window
frollmean(d[, V1], 3)

# multiple columns at once
frollmean(d, 3)

# multiple windows at once
frollmean(d[, .(V1)], c(3, 4))

# multiple columns and multiple windows at once
frollmean(d, c(3, 4))

## three above are embarrassingly parallel using openmp

Answer 3

I use aggregate along with a vector created by rep(). This has the advantage of using cbind() to aggregate more than 1 column in your dataframe at time. Below is an example of a moving average of 60 for a vector (v) of length 1000:

v=1:1000*0.002+rnorm(1000)
mrng=rep(1:round(length(v)/60+0.5), length.out=length(v), each=60)
aggregate(v~mrng, FUN=mean, na.rm=T)

Note the first argument in rep is to simply get enough unique values for the moving range, based on the length of the vector and the amount to be averaged; the second argument keeps the length equal to the vector length, and the last repeats the values of the first argument the same number of times as the averaging period.

In aggregate you could use several functions (median, max, min) - mean shown for example. Again, could could use a formula with cbind to do this on more than one (or all) columns in a dataframe.

Answer 4

EDIT: took great joy in adding the side parameter, for a moving average (or sum, or ...) of e.g. the past 7 days of a Date vector.

---

For people just wanting to calculate this themselves, it's nothing more than:

# x = vector with numeric data
# w = window length
y <- numeric(length = length(x))

for (i in seq_len(length(x))) {
  ind <- c((i - floor(w / 2)):(i + floor(w / 2)))
  ind <- ind[ind %in% seq_len(length(x))]
  y[i] <- mean(x[ind])
}

y

But it gets fun to make it independent of mean(), so you can calculate any 'moving' function!

# our working horse:
moving_fn <- function(x, w, fun, ...) {
  # x = vector with numeric data
  # w = window length
  # fun = function to apply
  # side = side to take, (c)entre, (l)eft or (r)ight
  # ... = parameters passed on to 'fun'
  y <- numeric(length(x))
  for (i in seq_len(length(x))) {
    if (side %in% c("c", "centre", "center")) {
      ind <- c((i - floor(w / 2)):(i + floor(w / 2)))
    } else if (side %in% c("l", "left")) {
      ind <- c((i - floor(w) + 1):i)
    } else if (side %in% c("r", "right")) {
      ind <- c(i:(i + floor(w) - 1))
    } else {
      stop("'side' must be one of 'centre', 'left', 'right'", call. = FALSE)
    }
    ind <- ind[ind %in% seq_len(length(x))]
    y[i] <- fun(x[ind], ...)
  }
  y
}

# and now any variation you can think of!
moving_average <- function(x, w = 5, side = "centre", na.rm = FALSE) {
  moving_fn(x = x, w = w, fun = mean, side = side, na.rm = na.rm)
}

moving_sum <- function(x, w = 5, side = "centre", na.rm = FALSE) {
  moving_fn(x = x, w = w, fun = sum, side = side, na.rm = na.rm)
}

moving_maximum <- function(x, w = 5, side = "centre", na.rm = FALSE) {
  moving_fn(x = x, w = w, fun = max, side = side, na.rm = na.rm)
}

moving_median <- function(x, w = 5, side = "centre", na.rm = FALSE) {
  moving_fn(x = x, w = w, fun = median, side = side, na.rm = na.rm)
}

moving_Q1 <- function(x, w = 5, side = "centre", na.rm = FALSE) {
  moving_fn(x = x, w = w, fun = quantile, side = side, na.rm = na.rm, 0.25)
}

moving_Q3 <- function(x, w = 5, side = "centre", na.rm = FALSE) {
  moving_fn(x = x, w = w, fun = quantile, side = side, na.rm = na.rm, 0.75)
}

Answer 5

You may calculate the moving average of a vector x with a window width of k by:

apply(embed(x, k), 1, mean)

Answer 6

Here is a simple function with filter demonstrating one way to take care of beginning and ending NAs with padding, and computing a weighted average (supported by filter) using custom weights:

wma <- function(x) { 
  wts <- c(seq(0.5, 4, 0.5), seq(3.5, 0.5, -0.5))
  nside <- (length(wts)-1)/2
  # pad x with begin and end values for filter to avoid NAs
  xp <- c(rep(first(x), nside), x, rep(last(x), nside)) 
  z <- stats::filter(xp, wts/sum(wts), sides = 2) %>% as.vector 
  z[(nside+1):(nside+length(x))]
}

Answer 7

vector_avg <- function(x){
  sum_x = 0
  for(i in 1:length(x)){
    if(!is.na(x[i]))
      sum_x = sum_x + x[i]
  }
  return(sum_x/length(x))
}

Answer 8

Here is example code showing how to compute a centered moving average and a trailing moving average using the rollmean function from the zoo package.

library(tidyverse)
library(zoo)

some_data = tibble(day = 1:10)
# cma = centered moving average
# tma = trailing moving average
some_data = some_data %>%
    mutate(cma = rollmean(day, k = 3, fill = NA)) %>%
    mutate(tma = rollmean(day, k = 3, fill = NA, align = "right"))
some_data
#> # A tibble: 10 x 3
#>      day   cma   tma
#>    <int> <dbl> <dbl>
#>  1     1    NA    NA
#>  2     2     2    NA
#>  3     3     3     2
#>  4     4     4     3
#>  5     5     5     4
#>  6     6     6     5
#>  7     7     7     6
#>  8     8     8     7
#>  9     9     9     8
#> 10    10    NA     9

Answer 9

The slider package can be used for this. It has an interface that has been specifically designed to feel similar to purrr. It accepts any arbitrary function, and can return any type of output. Data frames are even iterated over row wise. The pkgdown site is here.

library(slider)

x <- 1:3

# Mean of the current value + 1 value before it
# returned as a double vector
slide_dbl(x, ~mean(.x, na.rm = TRUE), .before = 1)
#> [1] 1.0 1.5 2.5


df <- data.frame(x = x, y = x)

# Slide row wise over data frames
slide(df, ~.x, .before = 1)
#> [[1]]
#>   x y
#> 1 1 1
#> 
#> [[2]]
#>   x y
#> 1 1 1
#> 2 2 2
#> 
#> [[3]]
#>   x y
#> 1 2 2
#> 2 3 3

The overhead of both slider and data.table's frollapply() should be pretty low (much faster than zoo). frollapply() looks to be a little faster for this simple example here, but note that it only takes numeric input, and the output must be a scalar numeric value. slider functions are completely generic, and you can return any data type.

library(slider)
library(zoo)
library(data.table)

x <- 1:50000 + 0L

bench::mark(
  slider = slide_int(x, function(x) 1L, .before = 5, .complete = TRUE),
  zoo = rollapplyr(x, FUN = function(x) 1L, width = 6, fill = NA),
  datatable = frollapply(x, n = 6, FUN = function(x) 1L),
  iterations = 200
)
#> # A tibble: 3 x 6
#>   expression      min   median `itr/sec` mem_alloc `gc/sec`
#>   <bch:expr> <bch:tm> <bch:tm>     <dbl> <bch:byt>    <dbl>
#> 1 slider      19.82ms   26.4ms     38.4    829.8KB     19.0
#> 2 zoo        177.92ms  211.1ms      4.71    17.9MB     24.8
#> 3 datatable    7.78ms   10.9ms     87.9    807.1KB     38.7

Answer 10

One can use runner package for moving functions. In this case mean_run function. Problem with cummean is that it doesn't handle NA values, but mean_run does. runner package also supports irregular time series and windows can depend on date:

library(runner)
set.seed(11)
x1 <- rnorm(15)
x2 <- sample(c(rep(NA,5), rnorm(15)), 15, replace = TRUE)
date <- Sys.Date() + cumsum(sample(1:3, 15, replace = TRUE))

mean_run(x1)
#>  [1] -0.5910311 -0.2822184 -0.6936633 -0.8609108 -0.4530308 -0.5332176
#>  [7] -0.2679571 -0.1563477 -0.1440561 -0.2300625 -0.2844599 -0.2897842
#> [13] -0.3858234 -0.3765192 -0.4280809

mean_run(x2, na_rm = TRUE)
#>  [1] -0.18760011 -0.09022066 -0.06543317  0.03906450 -0.12188853 -0.13873536
#>  [7] -0.13873536 -0.14571604 -0.12596067 -0.11116961 -0.09881996 -0.08871569
#> [13] -0.05194292 -0.04699909 -0.05704202

mean_run(x2, na_rm = FALSE )
#>  [1] -0.18760011 -0.09022066 -0.06543317  0.03906450 -0.12188853 -0.13873536
#>  [7]          NA          NA          NA          NA          NA          NA
#> [13]          NA          NA          NA

mean_run(x2, na_rm = TRUE, k = 4)
#>  [1] -0.18760011 -0.09022066 -0.06543317  0.03906450 -0.10546063 -0.16299272
#>  [7] -0.21203756 -0.39209010 -0.13274756 -0.05603811 -0.03894684  0.01103493
#> [13]  0.09609256  0.09738460  0.04740283

mean_run(x2, na_rm = TRUE, k = 4, idx = date)
#> [1] -0.187600111 -0.090220655 -0.004349696  0.168349653 -0.206571573 -0.494335093
#> [7] -0.222969541 -0.187600111 -0.087636571  0.009742884  0.009742884  0.012326968
#> [13]  0.182442234  0.125737145  0.059094786

One can also specify other options like lag, and roll only at specific indexes. More in package and function documentation.

Answer 11

Or you can simply calculate it using filter, here's the function I use:

ma <- function(x, n = 5){filter(x, rep(1 / n, n), sides = 2)}

If you use dplyr, be careful to specify stats::filter in the function above.

Answer 12

Using cumsum should be sufficient and efficient. Assuming you have a vector x and you want a running sum of n numbers

cx <- c(0,cumsum(x))
rsum <- (cx[(n+1):length(cx)] - cx[1:(length(cx) - n)]) / n

As pointed out in the comments by @mzuther, this assumes that there are no NAs in the data. to deal with those would require dividing each window by the number of non-NA values. Here's one way of doing that, incorporating the comment from @Ricardo Cruz:

cx <- c(0, cumsum(ifelse(is.na(x), 0, x)))
cn <- c(0, cumsum(ifelse(is.na(x), 0, 1)))
rx <- cx[(n+1):length(cx)] - cx[1:(length(cx) - n)]
rn <- cn[(n+1):length(cx)] - cn[1:(length(cx) - n)]
rsum <- rx / rn

This still has the issue that if all the values in the window are NAs then there will be a division by zero error.

Answer 13

Though a bit slow but you can also use zoo::rollapply to perform calculations on matrices.

reqd_ma <- rollapply(x, FUN = mean, width = n)

where x is the data set, FUN = mean is the function; you can also change it to min, max, sd etc and width is the rolling window.

Answer 14

You could use RcppRoll for very quick moving averages written in C++. Just call the roll_mean function. Docs can be found here.

Otherwise, this (slower) for loop should do the trick:

ma <- function(arr, n=15){
  res = arr
  for(i in n:length(arr)){
    res[i] = mean(arr[(i-n):i])
  }
  res
}

Answer 15

• Rolling Means/Maximums/Medians in the zoo package (rollmean)
• MovingAverages in TTR
• ma in forecast

Answer 16

In order to complement the answer of cantdutchthis and Rodrigo Remedio;

moving_fun <- function(x, w, FUN, ...) {
  # x: a double vector
  # w: the length of the window, i.e., the section of the vector selected to apply FUN
  # FUN: a function that takes a vector and return a summarize value, e.g., mean, sum, etc.
  # Given a double type vector apply a FUN over a moving window from left to the right, 
  #    when a window boundary is not a legal section, i.e. lower_bound and i (upper bound) 
  #    are not contained in the length of the vector, return a NA_real_
  if (w < 1) {
    stop("The length of the window 'w' must be greater than 0")
  }
  output <- x
  for (i in 1:length(x)) {
     # plus 1 because the index is inclusive with the upper_bound 'i'
    lower_bound <- i - w + 1
    if (lower_bound < 1) {
      output[i] <- NA_real_
    } else {
      output[i] <- FUN(x[lower_bound:i, ...])
    }
  }
  output
}

# example
v <- seq(1:10)

# compute a MA(2)
moving_fun(v, 2, mean)

# compute moving sum of two periods
moving_fun(v, 2, sum)

Answer 17

In fact RcppRoll is very good.

The code posted by cantdutchthis must be corrected in the fourth line to the window be fixed:

ma <- function(arr, n=15){
  res = arr
  for(i in n:length(arr)){
    res[i] = mean(arr[(i-n+1):i])
  }
  res
}

Another way, which handles missings, is given here.

A third way, improving cantdutchthis code to calculate partial averages or not, follows:

  ma <- function(x, n=2,parcial=TRUE){
  res = x #set the first values

  if (parcial==TRUE){
    for(i in 1:length(x)){
      t<-max(i-n+1,1)
      res[i] = mean(x[t:i])
    }
    res

  }else{
    for(i in 1:length(x)){
      t<-max(i-n+1,1)
      res[i] = mean(x[t:i])
    }
    res[-c(seq(1,n-1,1))] #remove the n-1 first,i.e., res[c(-3,-4,...)]
  }
}

Answer 18

The caTools package has very fast rolling mean/min/max/sd and few other functions. I've only worked with runmean and runsd and they are the fastest of any of the other packages mentioned to date.