similar to: aggregate.zoo on bivariate data

Hi Would anyone have any pointers on how to slice up a large zoo table. I have the following structure: - > str(ZOO_OBJ) ?zoo [1:632, 1:83] 30.4 30.4 30.4 30.4 30.3 ... ?- attr(*, "dimnames")=List of 2 ??..$ : NULL ??..$ : chr [1:83] "COL1" "COL2" "COL3" "COL4" ... ?- attr(*, "index")= POSIXct[1:632], format: "2009-05-01

Dear R-helpers, It seems to me that a character zoo cannot be coerced to a numeric zoo. Below is a minimal example. Can someone tell me what I have done wrong? > z<-zoo(1:4,order.by=1:4) > coredata(z)<-as.character(coredata(z)) > str(z) ‘zoo’ series from 1 to 4 Data: chr [1:4] "1" "2" "3" "4" Index: int [1:4] 1 2 3 4 >

Hello, I'm trying to use na.spline (package zoo) to fill some missing data in a time series. this works fine, however, if I apply the 'maxgap' argument, I always get the error: <------ Error in na.spline.vec(x., coredata(object.), xout = xout., ...) : attempt to apply non-function ------> I couldn't find a similar error for this case in the mailing lists and zoo vignette,

# Hi there, # I am trying to apply a function over a moving-window for a large number of multivariate time-series that are grouped in a nested set of factors. I have spent a few days searching for solutions with no luck, so any suggestions are much appreciated. # The data I have are for the abundance dynamics of multiple species observed in multiple fixed plots at multiple sites. (I total I

Just trying to create returns from prices, and do something like: returns.z = tail(prices.z,-1)/head(prices.z,-1) - 1 # should be equivalent to returns = exp(diff(log(prices.z))) - 1 Curiously, I get a zoo object back with zeros everywhere and also with the index having one fewer element than it should. Does anyone know how to pointwise divide zoo objects, and what exactly "/" is

(Sorry, sent the message before I finished it) Hello, everyone I have a long script that uses zoo objects. In this script I used simple moving averages and these I can very efficiently calculate with filter() functions. Now, I have to use special "exponential" moving averages, and the only way I could write the code was with a for-loop, which makes everything extremely slow. I don't

Hello, everyone I have a long script that uses zoo objects. In this script I used simple moving averages and these I can very efficiently calculate with filter() functions. Now, I have to use special "exponential" moving averages, and the only way I could write the code was with a for-loop, which makes everything extremely slow. I don't know how to optimize the code, but I need to

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

My problem is the following : I create 2 zoo objects and then I try to subset one of them using logic. indicesthatpass is a vector of trues and falses but when I send it into bckret, it returns an empty bckret. Obviously it has something to do with bckret being a zoo object and if I do the same subsctripting off of coredata(bckret), I'm confident it will work. But, I need to keep the minute

Dear all, I have an error message, when I try to convert a zoo object (called test) to ts (on R 2.4.0, Package zoo version 1.2-1, Windows XP) > test 1994-05-10 1994-06-09 1994-07-09 0.0024943889 0.0024881824 0.0006955831 > str(test) atomic [1:3] 0.002494 0.002488 0.000696 - attr(*, "index")=Class 'Date' num [1:3] 8895 8925 8955 > is.regular(test) [1] TRUE

Hi R-helpers I'm looking for a vectorised function which does missing value replacement as in last observation carried forward in the zoo package but instead of a locf, I would like the locf function to add +1 to each time a missing value occurred. See below for an example. > require(zoo) > x <- 5:15 > x[4:7] <- NA > coredata(na.locf(zoo(x))) [1] 5 6 7 7 7 7 7 12 13

Dear useRs, a new version of the zoo package for totally ordered observations is available from CRAN. It contains many new features, the most important of which are: support of regular time series, an improved merge method and functions for carrying out rolling analyses of time series. A more detailed list of changes and new features is given below. All features (both old and new) are also

Dear useRs, a new version of the zoo package for totally ordered observations is available from CRAN. It contains many new features, the most important of which are: support of regular time series, an improved merge method and functions for carrying out rolling analyses of time series. A more detailed list of changes and new features is given below. All features (both old and new) are also

Dear list-reader, by running the following script: library(zoo) sessionInfo() search() packageDescription("zoo") data(EuStockMarkets) dax <- as.zoo(EuStockMarkets[1:10, "DAX"]) daxr <- diff(log(dax)) identical(as.vector(qnorm(daxr)), qnorm(coredata(daxr))) qqnorm(coredata(daxr)) qqnorm(daxr) qqnorm() produces an error: > qqnorm(daxr) Fehler in if (xi == xj) 0L

Hi guys I have a reasonably basic question with zoo usage, but I havent been able to find a satisfactory workaround yet. Heres a simple example of what I'm talking about (the following data frame contains numeric columns that contains NAs): > head(ebs) time src tstamp code bid ask 1 2009-03-03 13:03:29.536 perf.Tib_listener 14980321164 EBS.REC.EURJPY=EBS.NaE 123.48 NA 2 2009-03-03

Hi, I have a financial (zoo) time series with prices and volumes (although I can get the coredata as a matrix). Due to the data-source some indices have multiple observations. I want to aggregate these according to a weighted average. 11:00:01 34 1000 11:00:01 35 500 11:00:01 35 1000 11:00:02 34 500 11:00:02 35 500 should become 11:00:01 34.6 2500 11:00:02 34.5 1000 I currently do this

Here's my code: http://pastebin.com/0yRxEVtm The important parts are uncommented and should be easy to find using the link above. For the following line of code, I plan on looking for a way to offset it up 7 rows so that the 15 minute timestamp would be considered the "median" of the subset being averaged to find the mean: avgCool = aggregate(intCool, trunc(time(intCool),

I'm doing: > alyL32007z <- zoo(alyL32007,alyL32007$time) > range(time(alyL32007z)) [1] "2007-01-01 00:00:00 UTC" "2007-12-31 23:30:00 UTC" But then, while the original variable is: > summary(alyL32007$NEE_st) Min. 1st Qu. Median Mean 3rd Qu. Max. NA's -15.340 -1.615 -0.054 -0.814 0.750 8.965 11124 the variable within the zoo object

I have a zoo object that contains 2 time series named "A-B" and "V1". When I create a third series "V2", the name of the "A-B" series is changed to "A.B". Although I could recreate the names for the 3 series I am wondering if there is a way of preventing the name change from happening ( ... maybe an equivalent of the keep.names=TRUE statement

Dear R community, I have the following two zoo objects: MONTHLY CPI > plot(z) > par("usr") [1] 1977.76333 2011.15333 70.39856 227.03744 > z=zooreg(cpius$Value,as.yearmon("1979-11"),frequency=12) > str(z) ?zooreg? series from Nov 1979 to Oct 2010 Data: num [1:372] 76.2 77 77.8 78.5 79.5 80.3 81.1 82 82 82.6 ... Index: Class 'yearmon' num [1:372]