similar to: Vector recycling and zoo

I have only just discovered the joys of the dotchart (since I am reading William Cleveland's -- Sean Carmody The Stubborn Mule http://www.stubbornmule.net http://twitter.com/seancarmody [[alternative HTML version deleted]]

With an ordinary plot, to customise the axis it is possible to suppress drawing the axis and then call Axis. I have been trying to change the location of the y-axis on a plot.table plot to the right hand side, but cannot even work out how to suppress drawing the labels. Here is a toy example of the sort of plot I am working with. Any suggestions as to how to have the axis on the right hand side

I have been plotting the same charts using png on a Windows machine and on a Mac OSX and the quality of the resulting images, particularly in relation to the fonts, look far superior in the plots produced on the Mac. Is there any way I can enhance the quality of the plots produced on the Windows machine? I have also tried using win.metafile on the Windows machine and the quality looks the same as

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 >

(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

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

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

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

hi all : the code pasted below runs but then, a dput on rollmeandifflogbidask gives me what is below the code. the structure of rollmeandifflogbidask is a zoo object but with a "frequency" so it's not the same structure as the original actual diff and this really causes things to blow up in later code. i'm sure gabor and achim know what to do but in the case that they are not

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 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

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

down vote favorite Hello I have a zoo series. It lasts 10 years and its frequency is 15min. I'd like to get a new zoo series (or vector) with the same number of elements, whith each element equal to the first element of the day. That's, The first element everyday is repeated throughout the wole day. This is not same as aggregate(originalseries,as.Date,head,1) because this gives a

Hello Where could I find examples on how to work with the time index in a timeseries or zoo series? Let say I've got this series DATA 1990-01-01 10:00:00 0.900 1990-01-01 10:01:00 0.910 1990-01-01 10:03:00 0.905 1990-01-01 10:04:00 0.905 1990-01-01 10:05:00 0.890 ....................... 2000-12-31 20:00:00 0.992 How do I make simple calculations such as ... ? Calculate the

A quick question x <- as.yearmon(2000 + seq(0, 23)/12) x [1] "Jan 2000" "Feb 2000" "Mar 2000" "Apr 2000" "May 2000" "Jun 2000" "Jul 2000" "Aug 2000" "Sep 2000" "Oct 2000" "Nov 2000" "Dec 2000" "Jan 2001" [14] "Feb 2001" "Mar 2001" "Apr

Suppose I have following dataset : > head(data1) Date Return 1 03/31/00 0.14230650 2 04/28/00 -0.03276228 3 05/31/00 -0.06527890 4 06/30/00 -0.04999873 5 07/31/00 -0.01447902 6 08/31/00 0.22265729 Now I convert it to zoo object : > data11 = zoo(data1[,2], as.Date(data1[,1], format="%m/%d/%y")) > head(data11) 2000-03-31 2000-04-28 2000-05-31

I'm stuck, but am sure it can be done I just don't understand how. I have data in an irregular timeseries. I want to be able to use stl to visualise the data (see seasonal parts etc), so I need to change to regular series of class ts (I think). I am using 2 zoo objects one is regular and the other is my irregular data. I am then merging to create the object I want but when I try to change

Hi I've been greatly enjoying the functionality the zoo package offers. However I've hit a snag with the following code > a <- zoo(matrix(1:10,5,2), 2001:2005) > a 2001 1 6 2002 2 7 2003 3 8 2004 4 9 2005 5 10 > a[I(2003), 2] 2003 8 > a[I(2003), 2] <- NA Error: subscript out of bounds > I've also tried > coredata(a[I(2003), 2])

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