similar to: aggregate.zoo

hi everyone! hope you can help me here. i am a new R user. what i am trying to do is to find the maximum annual discharge from a daily record. i have a data.frame which includes date and the discharge. somewhat like this.. 10/1/1989 2410 10/2/1989 2460 10/3/1989 2890 ... ... ... 12/31/2005 5730 i have been browsing through the archives and fount out about the aggregate

I have an ordered series of 3 month t-bill rates (annual). I transform this to a daily series, however, the observations are constructed only from the dates on which the t-bills were issued, which is every week. So now I have ordered observations of the daily 'risk-free rate' for one day every week. I want to expand this zoo object to give a value for every day, and to do so, copy the

Hi, I'd like to make a time series at an annual frequency. > a<-xts(x=c(2,4,5), order.by=c("1991","1992","1993")) Error in xts(x = c(2, 4, 5), order.by = c("1991", "1992", "1993")) : order.by requires an appropriate time-based object > a<-xts(x=c(2,4,5), order.by=1991:1993) Error in xts(x = c(2, 4, 5), order.by =

I am trying to get monthly means for a daily data series using zoo(). I have found an odd problem, that seems to be caused by zoo()'s handling of leap years. Here's my R script with 2 methods (freq=365, 366) for aggregating the daily data to monthly series: library(zoo) J_link <- "http://www.ijis.iarc.uaf.edu/seaice/extent/plot.csv" JAXA_data <- read.table(J_link,

I have spent hours on R in Windows 7. Just installed 2 days ago so the R package should be current. Currently I am using the RGui (64-bit) for Windows. I can not read an Excel file into R from my computer. Have hours on this. Completely crazy!! I have the XLConnect package loaded. I think it is loaded because when I enter: > loadedNamespaces() [1] "base"

Hi, I'm removing non-unique time indices in a zoo time series by means of aggregate. The time series is bivariate, and the row to be kept only depends on the maximum of one of the two columns. Here's an example: x <- zoo(rbind( c(1,1), c(1.1, 0.9), c(1.1, 1.1), c(1,1) ), order.by=c(1,1,2,2)) The eventual aggregated result should be 1 1.1 0.9 2 1.1 1.1 that is, in

Hi Jakob, dat1<-read.table(text=" TIME, Value1, Value2 01.08.2011 02:30:00, 4.4, 4.7 01.09.2011 03:00:00, 4.2, 4.3 01.11.2011 01:00:00, 3.5, 4.3 01.12.2011 01:40:00, 3.4, 4.5 01.01.2012 02:00:00, 4.8, 5.3 01.02.2012 02:30:00, 4.9, 5.2 01.08.2012 02:30:00, 4.1, 4.7 01.12.2012 03:00:00, 4.7, 4.3 01.01.2013 01:00:00, 3, 4.3 01.01.2013 01:30:00, 3.8, 4.1 01.01.2013 02:00:00, 3.8,

Readers, I am trying to use the zoo package with an array of data: file1: hh:mm:ss 1 hh:mm:ss 2 hh:mm:ss 3 hh:mm:ss 4 file2: hh:mm:ss 11 55 hh:mm:ss 22 66 hh:mm:ss 33 77 hh:mm:ss 44 88 I wanted to merge these data set so I tried the following commands: library(chron) library(zoo) z1<-read.zoo("path/to/file1.csv",header=TRUE,sep=",",FUN=times)

Hi everybody, I have a small question about the function "na.locf" from the package "zoo". I saw in the help that this function is able to fill NA gaps with the last value before the NA gap (or with the next value). But it is possible to fill my NA gaps according to the last AND the next value at the same time? Actually, I want R to fill my gaps with the method of

I have a zoo object on daily data for 10 years. Now I want to create a list, wherein each member of that list is the monthly observations. For example, 1st member of list contains daily observation of 1st month, 2nd member contains daily observation of 2nd month etc. Then for a particular month, I want to divide all observations into 3 parts (arbitrary) and then want to calculate some statistics

Dear R People: The aggregate function works very well on regular time series. Is there a version for zoo or its that would take daily data and convert it to monthly, please? Thanks in advance, Sincerely, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodgess at gmail.com

Dear R People: I'm trying to convert a daily zoo object to a weekly zoo object: xdate <- seq(as.Date("2002-01-01"),as.Date("2010-07-10"),by="day") library(zoo) length(xdate) xt <- zoo(rnorm(3113),order=xdate) xdat2 <- seq(index(xt)[1],index(xt)[3113],by="week") xt.w <- aggregate(xt,by=xdat2,mean) Error: length(time(x)) ==

All: I have a SQlite database where I have stored some verification data by date & time (cycle Z/UTC), lead_time as well as type, duration, etc. I would like to analyze & plot the data as monthly averages. I have looked at a bunch of examples which use some combination of zoo and aggregate, but I have not been able to successfully apply bits and pieces from the examples I have found. Any

Dear Gabor, Thanks for your reply. however: > tail(DJd) ^DJI.Close 2010-04-01 10927.07 2010-04-05 10973.55 2010-04-06 10969.99 2010-04-07 10897.52 2010-04-08 10927.07 *2010-04-09 10997.35* > tail(ag) 2009-11-30 10344.84 2009-12-31 10428.05 2010-01-31 10067.33 2010-02-28 10325.26 2010-03-31 10856.63 *2010-04-30 10997.35 * It seems the script "makes up"

Dear all, I have several text files looking like this: 9063032 19700201 22:00 174.067 9063032 19700201 23:00 174.076 9063032 19700202 00:00 174.085 9063032 19700202 01:00 174.091 9063032 19700202 02:00 174.094 9063032 19700202 03:00 174.091 9063032 19700202 04:00 174.082 9063032 19700202 05:00 174.079 And I run this loop: for (j in 1:nr.of.files) { #Import: DF <-

Hi all, I am using the zoo package to plot time series. I have a problem with formatting the axes. my zoo object (z) looks like the following. c1 1992-01-10 21 1992-01-17 34 1992-01-24 33 1992-01-31 41 1992-02-07 39 1992-02-14 38 1992-02-21 37 1992-02-28 28 1992-03-06 33 1992-03-13 40

Can anyone show me how to add a log Pearson type III plot using the evdistq() command to an extreme value plot using the lmom package? Attached sample code below... Thanks in advance, Dave library(lmom) # annual maximum daily streamflows Mackenzie River mackenzieRiver = c(26600, 30300, 34000, 32000, 29200, 28300, 28600, 26400, 28300, 28800, 29000, 22100, 32900, 31800, 21600, 32100, 27000,

Dear Users, I have a monthly data for a number of years(1960-2007) for a number of stations and i wish to extract sesonal time-series for the months of March-May and October-November for very station. I have read this data with read.table in R with stations as columns and time (months) as rows. My attempt to aggregate with the zoo package using the function as.yearqtr failed since this sums

Hi R-experts, Thanks very much to Jim Holtman and Gabor on my previous question. I am having another problem with data manipulation in zoo. The following is data (Z) for first business day of every month in zoo format. I am trying to get mean of "open" for each year. I subset Z <- Z[,2] then > sapply(split(Z, format(index(Z), "%Y")),mean) I get error message:

Hello, I'm very new to R, and so my question is simple. I have data record with 80 years of daily temperatures in one long string. The dates are also recorded, in YYMMDD format. I'd like to learn an elegant simple way to pull out the annual averages. (Obviously, every 4th year has 366 days.) I know I can set up a formal loop to create annual records and then average. But R