How to apply a function to every consecutive n elements in a vector

Question

I am trying to convert quarterly returns to yearly returns. Given a vector of quarterly returns, how can this be done?

I am fairly new to R-programming, so I haven\'t r

Answer 1

An idea via base R,

sapply(split(a, rep(c(FALSE, TRUE), each = length(a) / 2)), 
             function(i)((1 + i[1]) * (1 + i[2]) * (1 + i[3]) * (1 + i[4])) - 1)

#    FALSE      TRUE 
#0.2578742 0.1800339

Answer 2

An option:

    c(by(a, rep(1:(length(a)/4), each = 4), FUN = function(x) 

  (1 + x[1]) * (1 + x[2]) * (1 + x[3]) * (1 + x[4])-1

  )
))

        1         2 
0.2578742 0.1800339 

Answer 3

As they are quarterly returns divide the vector into groups of 4 and calculate for each group.

tapply(a, gl(length(a)/4, 4), function(x) prod(1 + x) - 1)
#    1      2 
#0.2579 0.1800 

The grouping part and calculation part can be done in variety of ways. For example, the above can also be done using split + sapply

sapply(split(a, gl(length(a)/4, 4)), function(x) prod(1 + x) - 1)

Or grouping can also be done using rep

tapply(a,rep(seq_along(a), each = 4, length.out = length(a)), 
         function(x) prod(1 + x) - 1)

Answer 4

1) ts These are really intended to represent time series so it makes sense to use a time series representation:

aggregate(ts(1+a, freq = 4), 1, prod) - 1
## Time Series:
## Start = 1 
## End = 2 
## Frequency = 1 
[1] 0.2578742 0.1800339

We could also use the start= argument if knew the year that the series starts, e.g.

aggregate(ts(1+a, freq = 4, start = 2000), 1, prod) - 1

2) rollapply This takes the product of every 4 values skipping ahead 4 each time (instead of a sliding window moving ahead by 1).

library(zoo)
rollapply(1 + a, 4, by = 4, prod) - 1
## [1] 0.2578742 0.1800339

3) aggregate.zoo Suppose we have a yeaqqtr vector yq that gives the year and quarter of each point. yearqtr renders as shown below and internally is represented as year + fraction where the fraction is 0, 1/4, 2/4 and 3/4 for Q1, Q2, Q3 and Q4. Given that we can aggregate over it:

library(zoo)

yq <- as.yearqtr(2000) + 0:7/4
yq
## [1] "2000 Q1" "2000 Q2" "2000 Q3" "2000 Q4" "2001 Q1" "2001 Q2" "2001 Q3"
## [8] "2001 Q4"

aggregate(zoo(1 + a), as.integer(yq), prod) - 1
##      2000      2001 
## 0.2578742 0.1800339

3a) Using the same yq we can tapply over 1+a :

library(zoo)

tapply(1 + a, as.integer(yq), prod) - 1
##      2000      2001 
## 0.2578742 0.1800339 

Answer 5

Using apply:

apply(matrix(a, nrow = 4), 2, function(x) prod(1 + x) - 1)
#[1] 0.2578742 0.180033