CODEWITHSUNDEEP
rvectorization
Todd Shannon
Todd Shannon

Reputation: 567

Incrementally mean for list of vectors

I have a vector that I want to take the incremental mean of a vector. 

a <- 2 4 6 2 4 0 1 0 0 1

This works: 

for(i in seq_along(a)) {
    print(mean(a[1:seq_along(a)[[i]]]))
    }
[1] 2
[1] 3
[1] 4
[1] 3.5
[1] 3.6
[1] 3
[1] 2.714286
[1] 2.375
[1] 2.111111
[1] 2

But now I want to extend that to a list of vectors and I'm stuck. 

b <- list(x = rnorm(10, mean = 5), x2 = rnorm(10, mean = 20))
b
$x
 [1] 4.893252 5.129610 4.599701 5.409024 4.666844 5.787243 5.697621 2.968771 5.216302 6.268629

$x2
 [1] 19.50947 22.14797 20.80683 19.47857 21.24126 18.36233 20.57424 19.68233 20.67508 19.83930

When I try this, I get the following: 

for(i in seq_along(b)) {
     print(mean(b[[i]][1:seq_along(b[[i]])[[i]]]))
}
[1] 4.893252
[1] 20.82872

It should return two vectors showing the mean at each index like the previous example, not the total mean of the vector itself

I have no idea how to proceed. Thanks in advance for the help.

Upvotes: 0

Views: 110

Answers (2)

MrFlick
MrFlick

Reputation: 206242

Rather than a for loop, you'll be better off just using cumsum() with division to calculate the cumulative mean. For example

cummean <- function(x) cumsum(x)/seq_along(x)

Then you can just do

cummean(a)

or with a list

lapply(b, cummean)

Upvotes: 6

zack
zack

Reputation: 5405

This work for you?

set.seed(1)
b <- list(x = rnorm(10, mean = 5), x2 = rnorm(10, mean = 20))
lapply(b, function(x) cumsum(x) / seq_along(x))

Upvotes: 0

Related Questions