Question

I have a zoo object called pp with daily data and 77 columns that looks like this:

            X02R X03N X04K X04N X04R X06I X06N X08J X08P X09O X11O X12L X14N X15G X16K  (...)
1961-01-01  8.3  5.2  3.2  0.0  8.7  5.2 15.0  7.2 11.5 13.0  0.0  4.9  0.0  2.9  6.0   
1961-01-02  1.1  3.2 10.0  0.0  0.0  3.5  0.0  8.7  0.4  1.2  0.0  0.4  0.0  3.2  0.2    
1961-01-03 12.0  4.2 50.5  0.0  9.0 38.5 15.0 31.7  1.7  8.7  9.0 69.2  4.2 22.2  9.2 
(...)  

I want to use apply.monthly to each of the columns, so in the end I will still have 77 columns but with monthly data instead of daily data. I tried
apply.monthly(pp, FUN=sum) but the result is a zoo object with just one column (I think is adding all the columns).

I also tried a loop:

for (i in 1:77)
{
mensal<-apply.monthly(pp[,i], FUN=sum)
}
but it also results in just one column (the 77th column). I might be able to make the loop work with some trial and error but it takes ages to compute ( I have 17897 rows and 77 columns) and I guess there is a simpler way of doing this without using loops… So if you know how, please help. Thanks!

Answer 1

In order for apply.monthly to return an object with more than one column, you have to use a function that operates by column (or apply a function that doesn’t).

library(quantmod)
getSymbols("SPY")
zSPY <- as.zoo(SPY)
# sum doesn't operate by column; it sums everything to one value
sum(zSPY)
spy.sum <- apply.monthly(zSPY, sum)
# colSums operates by column
spy.colSums <- apply.monthly(zSPY, colSums)
# use apply to operate by column
spy.apply.sum <- apply.monthly(zSPY, apply, 2, sum)