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)
}

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,...)]
  }
}

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。

你可以通过以下方法计算窗口宽度为k的向量x的移动平均值:

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

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。

您可以使用RcppRoll来实现用c++编写的快速移动平均线。只需调用roll_mean函数。文档可以在这里找到。

否则，这个(较慢的)for循环应该可以做到:

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