similar to: How to use write.csv(x=zoo.object), and how to let the first column corret?

Hello all, I don' like to have ( ) around my datetime, when send the csv file to somebody. How to remove it? >library(chron) >mydate <- c("2009-05-03","2009-06-07") >mytime <- c("10:30:00","10:20:00") >mydatetime <- chron(dates=mydate,times=mytime,format=c("y-m-d","h:m:s")) >write.csv(mydatetime)  

Sorry, this is my first time to post. I have a big data set: first colume is date (ex: 2008-2-150, the second is time (10:30:00), and the following columes are variaty measurement data. Every 30 min, I have one data. I want to find an effecient way to calculate the hourly, daily, monthly and yearly average, and plot them, and eventually use these average data to do further analysis. Thanks!

Dear all, I'm trying to plot, in the same window, two different series, using barplot() for the first one and plot() for the second. What happens is that the second chart has a different axes origin, therefore the final plot is wrong. This piece of code shows the differences between the values of par()$usr: barplot(coredata(Z0), beside=TRUE, ylim=c(0,100)); par()$usr;

how to input multiple .txt files? A data folder has lots of .txt files from different customers. Want to read all these .txt files to different master files: such as: cust1.xx.txt, cust1.xxx.txt, cust1.xxxx.txt,.............. to master file: X.txt cust2.xx.txt, cust2.xxx.txt, cust2.xxxx.txt,.............. to master file: Y.txt Thanks! [[alternative HTML version deleted]]

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

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

Hi, I have an intraday timeseries of financial data (see below) which has gaps due to market opening and closing hours. I am trying to plot it, but the time gap is always visible in the plot. I tried converting data to xts, zoo, timeSeries and plotting it with different functions i.e. plot.xts, plot.zoo. The only way to make it work was with function 'chartSeries' in the quantmod package

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

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 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 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

I am a bit confused about the different approaches taken to recycling in plain data frames and zoo objects. When carrying out simple arithmetic, dataframe seem to recycle single arguments, zoo objects do not. Here is an example > x <- data.frame(a=1:5*2, b=1:5*3) > x a b 1 2 3 2 4 6 3 6 9 4 8 12 5 10 15 > x$a/x$a[1] [1] 1 2 3 4 5 > x <- zoo(x) > x$a/x$a[1] 1 1 >

Hi List, I have weekly sales observations for several products drawn via ODBC. Source data is available at https://www.dropbox.com/s/78vxae5ic8tnutf/asr.csv. This is retail sales data, so will contain seasonality and trend information. I expect to see 52 or 53 observations per year, each observation occuring on the same day of the week (Saturday). Ultimately I'm looking to feed these series

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

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