2
votes

I have the following dataset:

date <- StockData[1:10,2]
x <- c(2,4,2,3,5,1,6,2,3,4)
df <- as.data.frame(cbind(x,date))
df

   x       date
1  2 2019-06-28
2  4 2019-07-01
3  2 2019-07-02
4  3 2019-07-03
5  5 2019-07-04
6  1 2019-07-05
7  6 2019-07-08
8  2 2019-07-09
9  3 2019-07-10
10 4 2019-07-11

Now I want to create a new column with the 5 (n) days ahead rolling average, so the dataset looks as following:

   x       date five.day.average
1  2 2019-06-28              3.2
2  4 2019-07-01              3
3  2 2019-07-02              3.4
4  3 2019-07-03              3.4
5  5 2019-07-04              3.4
6  1 2019-07-05              3.2
7  6 2019-07-08              NA
8  2 2019-07-09              NA
9  3 2019-07-10              NA
10 4 2019-07-11              NA

In other words, I want a formula/function that does the same thing as rollmean in package zoo, but instead of back in time, future in time ( since k in rollmean cannot be negative).

Thank you in advance!

2

2 Answers

3
votes

An option is frollmean from data.table

library(data.table)
setDT(df)[, five.day.average := frollmean(x, 5, align = 'left')]
df
#    x       date five.day.average
# 1: 2 2019-06-28              3.2
# 2: 4 2019-07-01              3.0
# 3: 2 2019-07-02              3.4
# 4: 3 2019-07-03              3.4
# 5: 5 2019-07-04              3.4
# 6: 1 2019-07-05              3.2
# 7: 6 2019-07-08               NA
# 8: 2 2019-07-09               NA
# 9: 3 2019-07-10               NA
#10: 4 2019-07-11               NA

Or with rollmean from zoo

rollmean(df$x, 5, align = 'left', fill = NA)
#[1] 3.2 3.0 3.4 3.4 3.4 3.2  NA  NA  NA  NA

data

df <- structure(list(x = c(2L, 4L, 2L, 3L, 5L, 1L, 6L, 2L, 3L, 4L), 
    date = c("2019-06-28", "2019-07-01", "2019-07-02", "2019-07-03", 
    "2019-07-04", "2019-07-05", "2019-07-08", "2019-07-09", "2019-07-10", 
    "2019-07-11")), class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6", "7", "8", "9", "10"))
3
votes

In base you can use filter to get a rolling average. To make it looking ahead rev can be used.

rev(filter(rev(df$x), rep(1/5, 5), sides=1))
# [1] 3.2 3.0 3.4 3.4 3.4 3.2  NA  NA  NA  NA