CODEWITHSUNDEEP
rdplyr
BillyJean
BillyJean

Reputation: 1587

Find average of 4 last elements

My dataset has the following form:

df<- data.frame(c("a", "a", "a", "a", "a", "a", "a", "a", "b", "b", "b", "b", "b", "b", "b", "b"),
                c(1,    1,   1,   1,   2,   2,   2,   2,   1,   1,    1,  1,   2,    2,   2,   2),
                c(1,    2,   3,   4,   1,   2,   3,   4,   1,   2,   3 , 4,  1,    2,   3,   4),
                c(25,   75,  20,  40,  60,  50,  20,  10,  20,  30,  40,  60, 25,   75,  20,  40))
colnames(df)<-c("car", "year", "mnth", "val")

For clarity I show it here as well: 

   car year mnth val
1    a    1    1  25
2    a    1    2  75
3    a    1    3  20
4    a    1    4  40
5    a    2    1  60
6    a    2    2  50
7    a    2    3  20
8    a    2    4  10
9    b    1    1  20
10   b    1    2  30
11   b    1    3  40
12   b    1    4  60
13   b    2    1  25
14   b    2    2  75
15   b    2    3  20
16   b    2    4  40

I would like to add a new column tmp to df where, for a particular row, the value of tmp should be the average of df$val and the 3 preceeding values. Some examples of tmp are shown here

#row 3: mean(25,75,20)=40
#row 4: mean(25,75,20,40)=40
#row 5: mean(75,20,40,60)=48.75
#row 16: mean(25,75,20,40)=40

Is there an efficient way to do this in R without using for-loops?

Upvotes: 1

Views: 88

Answers (4)

d.b
d.b

Reputation: 32548

For each value, calculate the mean of a rolling window which includes the value as well as preceding 3 values (from index i-3 up to index i in the solution below). For cases when i-3 is negative, you can just use 0 (max((i-3),0))

sapply(seq_along(df$val), function(i)
      mean(df$val[max((i-3),0):i], na.rm = TRUE))
#[1] 25.00 50.00 40.00 40.00 48.75 42.50 42.50 35.00 25.00
#[10] 20.00 25.00 37.50 38.75 50.00 45.00 40.00

Also consider rollmean of zoo

library(zoo)
c(rep(NA,3), rollmean(x = df$val, k = 4))
#[1]    NA    NA    NA 40.00 48.75 42.50 42.50 35.00 25.00 20.00 25.00
#[12] 37.50 38.75 50.00 45.00 40.00
#FURTHER TWEAKING MAY BE NECESSARY

Upvotes: 1

Umberto
Umberto

Reputation: 1421

Or simply so 

library(dplyr)
df$tmp <- (df$val+lag(df$val,1)+lag(df$val,2)+lag(df$val,3))/4

This does not use any loop. It simply shift the list and sum the shifted lists.

For example if you define

a <- c(1,2,3,4,5)

then

lag(a)

is

NA  1  2  3  4

I hope it can help you.

Upvotes: 2

David Arenburg
David Arenburg

Reputation: 92292

Here's (somewhat) vectorized solution using data.table::shift

library(data.table)
colMeans(do.call(rbind, shift(df$val, 0:3)), na.rm = TRUE)
## [1] 25.00 50.00 40.00 40.00 48.75 42.50 42.50 35.00 25.00 20.00 25.00 37.50 38.75 50.00 45.00 40.00

Or as @Frank suggested

rowMeans(setDF(shift(df$val, 0:3)), na.rm = TRUE)

Upvotes: 4

Jaime Caffarel
Jaime Caffarel

Reputation: 2469

You could also use data.table

library(data.table)

setDT(df)
df[, tmp := (val + shift(val,1,type="lag") + shift(val,2,type="lag") + shift(val,3,type="lag"))/4]

Upvotes: 1

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