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)

事实上，RcppRoll非常好。

cantdutchthis发布的代码必须在窗口的第四行进行修正:

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

这里给出了另一种处理缺失的方法。

第三种方法，改进cantdutch这段代码来计算部分平均与否，如下:

  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,...)]
  }
}

或者你可以简单地计算它使用过滤器，这是我使用的函数:

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

如果使用dplyr，请注意在上面的函数中指定stats::filter。

编辑:非常喜欢添加侧参数，例如，一个日期向量的过去7天的移动平均值(或总和，或…)。

对于那些只想自己计算的人来说，它无非是:

# 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

但是让它独立于mean()会很有趣，所以你可以计算任何“移动”函数!

# 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)
}

下面的示例代码展示了如何使用zoo包中的rollmean函数计算居中移动平均和尾随移动平均。

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

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。