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[英]Calculate rolling sum product

If I have following variables如果我有以下变量

x <- data.frame(ret = c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15) )
k <- 4

and I want to get y such that我想让你这样

y[i,1] = x[i,1]*(1/k) + x[i+1,1]*(2/k) + x[i+2,1]*(3/k) + x[i+3,1]*(k/k)
.
.
.

till i = nrow(x) - k + 1

how can I achieve this?我怎样才能做到这一点?

It is basically sum of last k values but it is multiplied by n/k where n is the index of last k elements.它基本上是最后 k 个值的总和,但它乘以 n/k,其中 n 是最后 k 个元素的索引。

for the given x as input the output will have following values对于给定的 x 作为输入,output 将具有以下值

y
7.5   <- y[1,1] = (x[1,1] * 0.25 + x[2,1] *0.5 + x[3,1]  *0.75 +  1 * x[4,1])
10
12.5
15
17.5
20
22.5
25
27.5
30
32.5
35

Use rollapply with the indicated function:将 rollapply 与指示的 function 一起使用:

library(zoo)

wsum <- function(x, k) sum(seq(k) * x) / k 
transform(x, ret = rollapply(ret, k, wsum, k = k, align = "left", fill = NA))

Update更新

An alternative that allows us to omit the k = k argument is:允许我们省略 k = k 参数的替代方法是:

wsum <- function(x, k = length(x)) sum(seq(k) * x) / k 
transform(x, ret = rollapply(ret, k, wsum, align = "left", fill = NA))

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