用锥形窗口计算运行平均值

Question

给出(虚拟)向量

index=log(seq(10,20,by=0.5))

我想计算具有居中窗口的运行平均值和每端的锥形窗口,即第一个条目保持不变,第二个是窗口大小的平均值3,依此类推,直到达到指定的窗口大小.

这里给出的答案:计算移动平均线,似乎都会产生一个较短的向量,切断窗口太大的起点和终点,例如:

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

ma(index)

Time Series:
Start = 1 
End = 21 
Frequency = 1 
[1]       NA       NA 2.395822 2.440451 2.483165 2.524124 2.563466 2.601315
[9] 2.637779 2.672957 2.706937 2.739798 2.771611 2.802441 2.832347 2.861383
[17] 2.889599 2.917039 2.943746       NA       NA

同样的

rollmean(index,5)

来自动物园的包裹

有没有一种快速的方法来实现锥形窗口而不需要编写循环编码？

Answer 1

由于rollapply可以说是相当缓慢的,它往往是值得写一个简单的定制功能......

tapermean <- function(x, width=5){
                 taper <- pmin(width,
                               2*(seq_along(x))-1,
                               2*rev(seq_along(x))-1) #set taper pattern
                 lower <- seq_along(x)-(taper-1)/2    #lower index for each mean
                 upper <- lower+taper                 #upper index for each mean
                 x <- c(0, cumsum(x))                 #sum x once
                 return((x[upper]-x[lower])/taper)}   #return means

这比rollapply解决方案快200倍以上......

library(microbenchmark)
index <- log(seq(10,200,by=0.5)) #longer version for testing
w <- c(seq(1,5,2),rep(5,length(index)-5-1),seq(5,1,-2)) #as in Scarabees answer

microbenchmark(tapermean(index),
               rollapply(index,w,mean))

Unit: microseconds
                   expr       min         lq       mean     median        uq       max neval
       tapermean(index)   185.562   193.9405   246.4123   210.6965   284.548   590.197   100
rollapply(index,w,mean) 48213.027 49681.0715 52053.7352 50583.4320 51756.378 97187.538   100

我休息一下!

Answer 2

所述width的参数zoo::rollapply可以是一个数值向量.

因此,在您的示例中,您可以使用:

rollapply(index, c(1, 3, 5, rep(5, 15), 5, 3, 1), mean)
#  [1] 2.302585 2.350619 2.395822 2.440451 2.483165 2.524124 2.563466 2.601315 2.637779 2.672957 2.706937 2.739798 2.771611 2.802441 2.832347 2.861383
# [17] 2.889599 2.917039 2.943746 2.970195 2.995732

如果n是一个奇数,一般的解决方案是:

w <- c(seq(1, n, 2), rep(n, length(index) - n - 1), seq(n, 1, -2))
rollapply(index, w, mean)

编辑:如果您关心性能,可以使用自定义Rcpp功能:

library(Rcpp)

cppFunction("NumericVector fasttapermean(NumericVector x, const int window = 5) {
  const int n = x.size();
  NumericVector y(n);

  double s = x[0];
  int w = 1;

  for (int i = 0; i < n; i++) {
    y[i] = s/w;
    if (i < window/2) {
      s += x[i + (w+1)/2] + x[i + (w+3)/2];
      w += 2;
    } else if (i > n - window/2 - 2) {
      s -= x[i - (w-1)/2] + x[i - (w-3)/2];
      w -= 2;
    } else {
      s += x[i + (w+1)/2] - x[i - (w-1)/2];
    }
  }

  return y;
}")

新基准:

n <- 5
index <- log(seq(10, 200, by = .5))
w <- c(seq(1, n, 2), rep(n, length(index) - n - 1), seq(n, 1, -2))

bench::mark(
  fasttapermean(index),
  tapermean(index),
  zoo::rollapply(index, w, mean)
)
# # A tibble: 3 x 14
#   expression                          min     mean   median      max `itr/sec` mem_alloc  n_gc n_itr total_time result      memory              time     gc
#   <chr>                          <bch:tm> <bch:tm> <bch:tm> <bch:tm>     <dbl> <bch:byt> <dbl> <int>   <bch:tm> <list>      <list>              <list>   <list>
# 1 fasttapermean(index)              4.7us   5.94us   5.56us   67.6us  168264.     5.52KB     0 10000     59.4ms <dbl [381]> <Rprofmem [2 x 3]>  <bch:tm> <tibble [10,000 x 3]>
# 2 tapermean(index)                 53.9us  79.68us  91.08us  405.8us   12550.    37.99KB     3  5951    474.2ms <dbl [381]> <Rprofmem [16 x 3]> <bch:tm> <tibble [5,954 x 3]>
# 3 zoo::rollapply(index, w, mean)   12.8ms  15.42ms  14.31ms   29.2ms      64.9  100.58KB     8    23    354.7ms <dbl [381]> <Rprofmem [44 x 3]> <bch:tm> <tibble [31 x 3]>

但是,如果您关心(极端)精度,则应使用该rollapply方法,因为meanR 的内置算法比天真的求和除法更准确.

另请注意,该rollapply方法是唯一允许您na.rm = TRUE根据需要使用的方法.