similar to: replace NA with 9999 in zoo object

# chron library(chron) fmt.chron <- function(x) { chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x)) } z1 <- read.zoo(SC3.csv, sep = ",", header = TRUE, FUN = fmt.chron) z2 <- read.zoo(SC2.csv, sep = ",", header = TRUE, FUN = fmt.chron) z3<-c(z1, z2) write.table(z3, sep="," , "SC.csv") How do you

DateTime RM61 11/30/2006 12:31 NA 11/30/2006 12:46 NA 11/30/2006 13:01 2646784125 11/30/2006 13:16 NA 11/30/2006 13:31 NA 11/30/2006 13:46 NA 11/30/2006 14:01 2666435177 11/30/2006 14:16 NA 11/30/2006 14:31 NA 11/30/2006 14:46 NA 11/30/2006 15:01 2653041914 11/30/2006 15:16 NA 11/30/2006 15:31 NA 11/30/2006 15:46 NA 11/30/2006 16:01 2693126189 11/30/2006 16:16 NA 11/30/2006 16:31 NA 11/30/2006

the comma seperated file is 37Mb, and I get the below message: it is zoo object read in this way: # chron > library(chron) > fmt.chron <- function(x) { + chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x)) + } > z1 <- read.zoo("all.csv", sep = ",", header = TRUE, FUN = fmt.chron) and then the plot is done with:

Another newbie question I got the 1 minute spine interpolation and 15 mean aggregation working with many thanks to Gabor Grothendieck using Zoo functions. I got a tip from Hasan Diwan to look at xts but it seemed I would make better progress using code from Gabor. Now I'm having trouble plotting this zoo object. I'm thinking I want a function to "split" the zoo object back to

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 <-

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)

DateTime,Temp,SpCond,DOConc,Depth,pH,ORP,Turbidity+,Chlorophyll,Battery,Cond,DO%,Salinity,TDS 01/13/2006 17:01,10.87,84,9.36,0.664,7.3,132,28.8,3.1,11.5,0.062,84.6,0.04,0.055 01/13/2006 17:16,10.9,84,9.36,0.66,7.31,133,28.7,2.9,11.5,0.062,84.7,0.04,0.055 01/13/2006 17:31,10.92,84,9.36,0.655,7.3,132,28.4,2.6,11.4,0.062,84.8,0.04,0.055 01/13/2006

I have a matrix with data that runs from 1/1/06 00:01:00-1/31/08 23:46:00. I have read in the data with this fmt.chron <- function(x) { chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x)) } x <- read.zoo(file.choose(), sep=",", header=T, FUN=fmt.chron) plotted with this plot(x[,(seq(3, by=9, length.out=12))],

Hello all, I'm trying to use zoo.read but can't figure out how to deal with the time format. (example below) would be nice if someone could help. best regards, Immanuel --------------------------- L <- "Date,Time,Open,High,Low,Close,Up,Down 05.02.2001,00:30,421.20,421.20,421.20,421.20,11,0 05.02.2001,01:30,421.20,421.40,421.20,421.40,7,0

I am was going to look at the as.yearmon function in the zoo package and write a as.day function to aggregate a time series of 96 observations per day into the mean for each day, but I don't know how to look at the code so that I can convert it into something I can use. On top of that I believe that it is probably an S3 method and I haven't quite gotten that far in my programming

R version 2.8.0 (2008-10-20) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] StreamMetabolism_0.01 chron_2.3-24 zoo_1.5-4 loaded

library(zoo) library(chron) t1 <- chron("1/1/2006", "00:00:00") t2 <- chron("1/31/2006", "23:45:00") deltat <- times("00:15:00") tt <- seq(t1, t2, by = times("00:15:00")) d <- sample(33:700, 2976, replace=TRUE) sin.zoo <- zoo(d,tt) #there are ninety six reading in a day d.max <- rollapply(sin.zoo, width=96, FUN=max)

I would like to merge zoo objects that are stored in a list into one big zoo object with one index for all of the observations. I have created the list (74 dataframes) with the code below, and have tried the do.call(merge, foo) in the call and the output is not what I expected. Any help would be greatly appreciated. Stephen Sefick ###################################################level logger

day<-structure(c(7.7, 7.7, 7.7, 7.7, 7.7, 7.71, 7.7, 7.71, 7.71, 7.7, 7.7, 7.7, 7.7, 7.69, 7.68, 7.68, 7.67, 7.67, 7.67, 7.66, 7.65, 7.65, 7.65, 7.64, 7.64, 7.63, 7.63, 7.63, 7.62, 7.62, 7.62, 7.62, 7.63, 7.63, 7.63, 7.63, 7.63, 7.64, 7.64, 7.65, 7.65, 7.65, 7.66, 7.66, 7.67, 7.67, 7.67, 7.68, 7.68, 7.69, 7.69, 7.69, 7.69, 7.7, 7.7, 7.7, 7.7, 7.7, 7.71, 7.7, 7.7, 7.71, 7.71, 7.7, 7.7, 7.7, 7.7,

#I would like to interpolate a straight line between 06/08/06 04:16:00 - 06/08/06 20:31:00 with values and then sum them. This is an estimate of ecosystem #respiration and I will be using this in a larger context(48 days of these diurnal curves), but for right now I am just trying to figure out how to do it for this one #day example. I have some other code for Ecosystem (stream) Metabolism that

NOTE: I will provide data if necessary, but I didn't want clutter everyones mailbox All: I have a time series with level and temperature data for 11 sites for each of three bases. I will have to do this more than once is what I am saying here. OK, The time series are zoo objects with index values in chron format. The problem is that the date and times should be at even 15 min intervals,

I would like to include this in a package. The S3 methods on R CMD check says * checking S3 generic/method consistency ... WARNING window: function(x, ...) window.chron: function(data, day1, hour1, day2, hour2, ...) See section 'Generic functions and methods' of the 'Writing R Extensions' manual. I have looked and can not figure it out. This function is for convience. What

Is there a easy way to create the time index for a zoo/xts object for every 100 milliseconds. eg. time Index would be: 10:00:00:100 10:00:00:200 10:00:00:300 10:00:00:400 I am looking to build an empty zoo/xts object with time index from 10am to 3pm, index jumps by 100ms each row. Thanks, Chris -- View this message in context:

I have two files with dates and prices in each. The number of rows in each of them will differ. How do I create a new file which contains data from both these files? Cbind and merge are not helpful. For cbind because the rows are not the same replication occurs. Also if I have similar data how do I write a vlookup kind of function? I am giving an example below: Say Price1 file contains the

Hi all I have an irregular zoo series, where the time index looks like the following: > head(time(l.zoo)) [1] "2009-06-15 01:44:20.802 GMT" "2009-06-15 01:44:20.812 GMT" "2009-06-15 01:44:20.837 GMT" "2009-06-15 01:44:20.848 GMT" "2009-06-15 06:00:01.320 GMT" [6] "2009-06-15 06:00:01.330 GMT" >