similar to: replacing elements of a zoo object

** Disclaimer: I'm looking for general suggestions ** I'm sorry, but can't send out the file I'm using, so there is no reproducible example. I'm using read.table and it's taking over 30 seconds to read a tiny file. The strange thing is that it takes roughly the same amount of time if the file is 100 times larger. After re-reviewing the data Import / Export manual I think

I have read ?zoo but am not sure how to relate the parameters (x, order.by, frequency, and style) to my data.frame. The structure of the data.frame is 'data.frame': 11169 obs. of 4 variables: $ stream : Factor w/ 37 levels "Burns","CIL",..: 1 1 1 1 1 1 1 1 1 1 ... $ sampdate: Date, format: "1987-07-23" "1987-09-17" ... $ param : Factor w/

I don't understand how this function can subset by i when i is missing.... ## My function: myfun = function(vec, i){ ret = vec[i] ret } ## My data: i = 10 vec = 1:100 ## Expected input and behavior: myfun(vec, i) ## Missing an argument, but error is not caught! ## How is subsetting even possible here??? myfun(vec) Is there a way to check for missing function arguments, *and*

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 >

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

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]

Hello I am new to R and I need to convert some dates (numeric format by matlab) to actual dates in R. For instance, Matlab -> 730456 -> >> datestr(730456) ans = 02-Dec-1999 R - > library(zoo) > as.Date(730456) [1] "3969-12-03" I don't not mind the output format but it needs to be right. Many thanks Ed

I would really like some help with understanding the panel function, in zoo. Thank you. R 15.1 and zoo 1.7-7. library(zoo) x = seq(0,3*pi,length.out=100) y = sin(x) zobj = zoo(y, x) ########################################################### ## EXAMPLE 1 - GLOBAL ARGUMENT ## This panel function works ## But, it relies on mycol, which is a global variable

When I try to convert the zoo object to a timeSeries object, which would allow me to utilize Rmetrics packages, I get an error message. > Data<-read.zoo("c:\\DOWUBSPRICING.txt,na.strings="NA",sep="\t",header=T) > is(Data) "zoo" > as.timeSeries.zoo(Data) Error in .local (.Object, . ) Is this happening because I am using daily data?

I have a zooreg object and I want to be able to generate a value for seasons and 11-day composites paste it onto my zoo data frame, along with year, month and days. Right now I have the following to work from: eg. dat.zoo.mdy <- with(month.day.year(time(dat.zoo)), cbind(dat.zoo, year, month, day, quarter = (month - 1) %/% 3 + 1, dow = as.numeric(format(time(dat.zoo), "%w")))) For

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 all, I would like to use the apply or a similar function belonging to this family, but applying for each column (or row) but let say for each q columns. For example I would like to apply a function FUN for the first q columns of matrix X then for q+1:2*q and so on. If I do apply (X, 2, FUN) it applies for each column and not for every q columns. Is that possible with any similar function?

I read in ?na.omit that it returns the object with incomplete cases removed. I interpret this to mean that any zoo object row where any column shows 'NA' will be removed from the data set. That's not what I need, since the 'NA' represents information in my context. However, what I would like to do is eliminate the rows where every column is 'NA'. When I aggregate

Dear All, While i am trying convert data frame object to zoo object I am getting numeric(0) error in performance analytics package. The source code i am using from this website to learn r in finance: https://faculty.washington.edu/ezivot/econ424/returnCalculations.r # create zoo objects from data.frame objects dates.sbux = as.yearmon(sbux.df$Date, format="%m/%d/%Y") dates.msft =

Hello, I would like to implement a "turn-of-the-month' trading strategy in R. Given a daily series of stock market return data as a zoo object, the strategy would go long (buy) four trading days before the end of the month, and sell the third trading day of the following month. How can I select these days, particularly the fourth day before and the third day after the turn of the

I thought the "apply" functions are faster than for loops, but my most recent test shows that apply actually takes a significantly longer than a for loop. Am I missing something? It doesn't matter much if I do column wise calculations rather than row wise ## Example of how apply is SLOWER than for loop: #rm(list=ls()) ## DEFINE VARIABLES mu=0.05 ; sigma=0.20 ; dt=.25 ; T=50 ;

Dear R users, I have 3 columns in a zoo object. I want to plot all of them in one screen but I want the first two to be in "lines" format and 3rd one to be barplot. Regards Vikram [[alternative HTML version deleted]]

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

I have date data as a numeric and hourly data in 0 to 2300 hours in a dataframe. d <- rep(20110101,24) h <- seq(from = 0, to = 2300, by = 100) df <- data.frame(LST_DATE = d, LST_TIME = h, data = rnorm(24, 0, 1)) S <- chron(dates. = as.character(df$LST_DATE), times. = paste(as.character(df$LST_TIME/100), ":0:0", sep = ""), format =

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 =