similar to: Using 'diff' on zoo vs zooreg classes (possible bug?)

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

Hi, I'm trying to make a ts object that has both NA values and a frequency other than 1 (so I can use stl). I've tried all permutations I can think of, but cannot get the desired (expected?) results. The values live in x and the corresponding semi-regular time stamps are in t: > library('zoo') > z = zoo(x, order.by=t, frequency=24) > zzr = as.zooreg(z, start=0) > zr

Hi all and thanks for your time in advance, I can't figure out why summary.lmrob complains when lmrob is used on a zooreg object. If the zooreg object is converted to vector before calling lmrob, no problems appear. Let me clarify this with an example: >library(robustbase) >library(zoo) >dad<-c(801.4625,527.2062,545.2250,608.2313,633.8875,575.9500,797.0500,706.4188,

Dear R users, I am currently working in subsetting a zooreg() object using either window or subset. I have a solution but it may be a bit cumbersome when I start working with actual data. Your inputs would be greatly appreciated. Example: I have a zooreg() object that starts in 1997 and ends in 2001. This object contains daily data for the 4 years

I create a zooreg object that runs from Jan-1-2002 0:00 to Jun-1-2005 0:00... regts.start = ISOdatetime(2002, 1, 1, hour=0, min=0, sec=0, tz="") regts.end = ISOdatetime(2005, 6, 1, hour=0, min=0, sec=0, tz="") regts.zoo <- zooreg( NA, regts.start, regts.end, deltat=3600 ) Upon inspection: > regts.zoo[1:3] 2002-01-01 00:00:00 2002-01-01 01:00:00 2002-01-01 02:00:00

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 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

I have *** attached *** an RData file containing an R object that is acting strangely. Try this in a fresh workspace. Do not load zoo or any other package. We load the object, zz2, from the attached RData file. It is just the number 1 with the class c("zooreg", "zoo"). Now create an S3 print routine that simply prints an X when given an object of class "zoo". If

Dear all, please consider my following workbook: library(zoo) lis1 <- vector('list', length = 2) lis2 <- vector('list', length = 2) lis1[[1]] <- zooreg(rnorm(20), start = as.Date("2010-01-01"), frequency = 1) lis1[[2]] <- zooreg(rnorm(20), start = as.yearmon("2010-01-01"), frequency = 12) lis2[[1]] <- matrix(1:40, 20) lis2[[2]] <-

No he podido crear una matriz con dos columnas una de fechas y la otra columna con números de casos de una enfermedad x library(zoo) zooreg(1:300, start = as.Date("2012-03-01")) cbind(zooreg(1:300, start = as.Date("2012-03-01")) , c(1:300)) #?como incluir un vector o una ts? -- Este mensaje le ha llegado mediante el servicio de correo electronico que

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]

Hi, I'm trying to plot the forecasts I generated using the Plot.Fore function of the BootPR package. But I got an error from zoo: My data: Time Series: Start = 1 End = 18 Frequency = 1 [1] 38731 38628 39117 92809 71984 31226 58613 72360 107956 92066 [11] 95208 99098 95848 120383 110717 105680 98469 101916 Script: y1<-ts(y1);

R-listers, I am using xts with a yearmon index, but am getting some inconsistent results with the date index when i drop observations (for example by using na.omit). The issue is illustrated in the example below. If I start with a monthly zooreg series starting in 2009, yearmon converts this to "Dec-2008". Not such a worry for my example, but strange. Having converted to xts, i drop

Please see that correlogram for a arbitrary time series : acf(zooreg(rnorm(39), start=as.yearmon("2008-01-01"), frequency=12)) What is the meaning of lag 0.2, 0.4, ........ in the plot? Those should not be integers? Or I am missing something? Thanks -- View this message in context: http://n4.nabble.com/Meaning-of-lag-0-2-0-4-tp1765093p1765093.html Sent from the R help mailing list

Estimados ?como se puede llevar el archivo que adjunto a vector? necesito hacer lo siguiente con el archivo a library(zoo) x <- c (1,3,1,1,1,1,2,1,1,2,1,1,2,1,1,1,2,1,2,2,1,1,1,3,1,1,2,2,2,2,1,1,1,2,1,1,1,1,1,2,2,5,1,1,1,1,2,1,1,1,2,4,1,2,1,3,1,1,1,1,3,1,2,1,1,3,1,3,3,1,2,1,2,2,2,3,1,2,1,1,1,1,2,1,1,1,1,1,1) names(x) <- format(index(zooreg(1:89, start =as.Date("2015-01-01"))),

Hi R Users! I would like to create an high frequency series but I am experiencing some difficulties. My series should start at 09.30 a.m. each day and end at 16.00 for, let's say, 2 years. I don't care on how many observations are for each day. It's ok also one observation each minute. In this case, I would have 390 observations each day. I have tried the following: start <-

Hi,   I observed that if I use window() function available with the 'zoo' package to extract a portion of my times series and if that time series data is stored in some 'zoo' object with only 1 column, then the resulting zoo object is becoming vector.   Here is my observation:   > library(zoo) > Dat <- matrix(1:3, nc = 1) > Dat      [,1] [1,]    1 [2,]    2 [3,]    3

Dear all, Please forgive me if there is a duplicate post; my previous mail perhaps didnt reach the list....... Let say I have following time series library(zoo) > dat1 <- zooreg(rnorm(10), start=as.Date("2010-01-01"), frequency=1) > dat1[c(3, 7,8)] = NA > dat1 2010-01-01 2010-01-02 2010-01-03 2010-01-04 2010-01-05 2010-01-06 2010-01-07 2010-01-08 2010-01-09

En el paquete R0 se despliega data(Germany.1918) y cuando aplicamos el comando is.vector, nos refleja TRUE. Mi duda es COMO GENERAR UNA DATA SIMILAR A: data(Germany.1918) para que pueda realizar las operaciones que deseo. Al tratar de hacerlo #combinar fechas y casos df <- data.frame( Fecha = format(index(zooreg(1:269, start = as.Date("2012-03-01"))),

Hi, Can you tell me what is the meaning for "tail, 1" in "aggregate"? I also want to get some similar graph, but the data is not time series data. Suppose here is my data one, I want a graph with x-axis is just the index(1:9). The graph plot all the variable A, B,C,D. So there should be 4 lines for each graph. For the A line, at each time point, the letter A should be on