R:是否可以在同一列上使用mutate + lag? [英] R: Is it possible to use mutate+lag with the same column?

问题描述

我正在尝试在R中复制以下公式:

I'm trying to replicate the following formula in R:

Xt = Xt-1 * b + Zt *(1-b)

Xt = Xt-1 * b + Zt * (1-b)

我正在使用以下代码

t %>% 
  mutate(x= ifelse(week == 1, z, NaN)) %>% # Initial value for the first lag
  mutate(x= ifelse(week != 1, lag(x,1 ,default = 0) * b + z, z)

但是我从第二个元素中得到了所有的NaN.

But I get all NaN except from the second element.

     z      b           x
     (dbl)  (dbl)    (dbl)
1  168.895  0.9      168.8950
2   20.304  0.9      131.7472
3   14.943  0.9         NA
4   11.028  0.9         NA
5    8.295  0.9         NA
6    8.024  0.9         NA
7    6.872  0.9         NA
8    7.035  0.9         NA
9    4.399  0.9         NA
10   4.158  0.9         NA

这在excel中非常简单，但是我必须在R中完成，您有什么方法吗?

This is fairly simple in excel but I must do it in R, Do you have any approaches?

可复制的示例:

set.seed(2)
t = data.frame(week = seq(1:52),
               z = runif(52, 0, 100),
               b = 0.2)

推荐答案

由于@Frank和@docendo discimus，我发现解决方案正在运行以下循环

I found the solution running the following loop, thanks to @Frank and @docendo discimus

for (row in 2:dim(t)[1]) {
    t[row,] <- mutate(t[1:row,], x= lag(x,1) * b + z * (1 - b))[row,]
  }

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