Hello,
Inline.
Às 17:33 de 12/09/19, Bert Gunter escreveu:
But she wants *monthly* averages, Rui.
Thanks, my mistake.
Ergo ave() or tidyData
equivalent, right?
Maybe. But ave() returns as many values as the input length, this seems
more suited for tapply or aggregate.
I will first create an example data set.
set.seed(1234)
start <- as.Date("03-01-1994", "%d-%m-%Y")
end <- as.Date("29-12-2000", "%d-%m-%Y")
date <- seq(start, end, by = "day")
date <- date[as.integer(format(date, "%u")) %in% 1:5]
df1 <- data.frame(date,
CountryA = rnorm(length(date)),
CountryB = rnorm(length(date)))
Now the averages by month
month <- zoo::as.yearmon(df1[[1]])
aggA <- aggregate(CountryA ~ month, df1, mean)
aggB <- aggregate(CountryB ~ month, df1, mean)
MonthReturns <- merge(aggA, aggB)
head(MonthReturns)
Final clean up.
rm(date, month, aggA, aggB)
Hope this helps,
Rui Barradas
-- Bert
On Thu, Sep 12, 2019 at 8:41 AM Rui Barradas <ruipbarra...@sapo.pt
<mailto:ruipbarra...@sapo.pt>> wrote:
Hello,
Please include data, say
dput(head(data, 20)) # post the output of this
But, is the problem as simple as
rowMeans(data[2:3], na.rm = TRUE)
?
Hope this helps,
Rui Barradas
Às 15:53 de 12/09/19, Subhamitra Patra escreveu:
> Dear R-users,
>
> I have daily data from 03-01-1994 to 29-12-2000. In my datafile,
he first
> column is date and the second and third columns are the returns
of the
> country A, and B. Here, the date column is same for both
countries. I want
> to calculate the monthly average of both country's returns by
using a loop,
> and then, I want to export the results into excel.
>
> Please help me in this regard.
>
> Please find the attached datasheet.
>
> Thank you.
>
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