See ?which.max :
library(zoo)
z - read.zoo(myfile.dat, sep = ,, format = %m/%d/%Y)
f - function(x) time(x)[which.max(x)]
ix - tapply(z, floor(as.yearmon(time(z))), f)
z[ix]
On Tue, Feb 17, 2009 at 3:58 AM, CJ Rubio cjru...@kongju.ac.kr wrote:
i have the following code - assimilating the maximum annual discharge each
year ffrom a daily discharge record from year 1989-2005.
m - read.table(D:/documents/5 stations/01014000.csv, sep =,)
z - zoo(m[,4],as.Date(as.character(m[,3]), %m/%d/%Y))
x - aggregate(z, floor(as.numeric(as.yearmon(time(z, max) #code
suggested by Gabor Grothendieck
which gives me the maximum discharge each year and the year itself.. what i
am trying now is to produce the date when the maximum discharge was observed
in the pattern -mm-dd, like:
1988 11/9/1988 18600
1989 5/8/1989 49000
...
2005 4/26/2005111000
thank you all in advance...
--
View this message in context:
http://www.nabble.com/problem-with-dates-in-zoo-package-tp22053103p22053103.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
