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!
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)
Try this:
> library(zoo)
>
> # test data
> z <- zooreg(cbind(a = 1:365, b = 1:365), Sys.Date())
> head(z)
a b
2011-09-02 1 1
2011-09-03 2 2
2011-09-04 3 3
2011-09-05 4 4
2011-09-06 5 5
2011-09-07 6 6
>
> aggregate(z, as.yearmon)
a b
Sep 2011 435 435
Oct 2011 1395 1395
Nov 2011 2265 2265
Dec 2011 3286 3286
Jan 2012 4247 4247
Feb 2012 4843 4843
Mar 2012 6107 6107
Apr 2012 6825 6825
May 2012 7998 7998
Jun 2012 8655 8655
Jul 2012 9889 9889
Aug 2012 10850 10850
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