为了配合坎迪奇西斯和罗德里戈·雷麦黛奥的回答;

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)

可以使用runner包来移动函数。在本例中是mean_run函数。cummean的问题是它不处理NA值，但mean_run可以。Runner包还支持不规则时间序列，Windows可以依赖于日期:

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

还可以指定其他选项，如延迟和仅在特定索引处滚动。更多内容在包和函数文档中。

使用费用应充分、有效。假设你有一个向量x，你想要n个数的和

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

正如@mzuther在评论中指出的那样，这假设数据中没有NAs。要处理这些问题，需要将每个窗口除以非na值的数量。这里有一种方法，结合@里卡多·克鲁兹的评论:

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

这仍然有一个问题，如果窗口中的所有值都是NAs，那么将会有一个零误差的除法。

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))
}

下面是一个简单的带有过滤器的函数，演示了一种方法来处理带有填充的开始和结束NAs，并使用自定义权重计算加权平均值(由过滤器支持):

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))]
}

caTools包具有非常快速的滚动mean/min/max/sd和其他一些功能。我只使用过runmean和runsd，它们是迄今为止提到的其他包中最快的。