我试着用R来计算矩阵中一系列值的移动平均值。R中似乎没有一个内置函数可以让我计算移动平均线。有任何软件包提供这样的服务吗?还是需要我自己写?


当前回答

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
}

其他回答

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)

在数据。表1.12.0增加了新的滚动平均值函数,以计算快速和准确的滚动平均值,仔细处理NA, NaN和+Inf, -Inf值。

由于在这个问题中没有可重复的例子,所以在这里没有更多的问题要解决。

你可以在手册中找到更多关于?frollmean的信息,也可以在?frollmean网站上找到。

下面是手册中的例子:

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

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))]
}

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。

滑块包可以用于此。它有一个专门设计的界面,感觉类似呜呜声。它接受任何任意函数,并可以返回任何类型的输出。数据帧甚至按行迭代。pkgdown网站在这里。

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

滑块和数据的开销。Table的frollapply()应该非常低(比zoo快得多)。对于这个简单的示例,Frollapply()看起来稍微快一些,但请注意,它只接受数字输入,并且输出必须是标量数值。滑块函数是完全通用的,你可以返回任何数据类型。

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