similar to: zoo: bug with unique for yearmon

Hello, In zoo package, if I would like the time frame to be 1981M01 to 1982M12, then I code time_0<-as.yearmon("1981-01")+(0:23)/12 However, if the time frame of interest becomes 1981M01 to 2011M12, it is relatively hard to calculate the number of months. Is there any faster way to do it? Thanks, miao [[alternative HTML version deleted]]

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

It looks like a timezone issue, and it's causing confusion to me at least. My original data: gmt <- c("19880101 0000", "19880101 0100", "19880101 0300", "19880101 0400", "19880101 0500", "19880101 0600") These were converted to local dates/times with akst<-strptime(gmt,format="%Y%m%d %H%M")-(3600*9) # because I want

Dear useRs and developeRs, In my diploma thesis I work with a daily time series of glacier runoff data. I did already aggregate them to monthly means etc. Now i want to use just the summer values (I am indecisive by now what that means, but let's make it easy and use months like June). Is there a way to extract the data off this zoo into another zoo with frequency=1 ? Do you have

I have following dataset: > res [,1] [,2] [,3] [1,] 1946 4 1.27 [2,] 1946 5 1.27 [3,] 1946 6 1.27 [4,] 1946 7 1.27 [5,] 1946 8 1.52 [6,] 1946 9 1.52 [7,] 1946 10 1.52 [8,] 1946 11 1.52 [9,] 1946 12 1.62 [10,] 1947 1 1.62 [11,] 1947 2 1.62 [12,] 1947 3 1.62 [13,] 1947 4 1.87 [14,] 1947 5 1.87 [15,] 1947 6 1.87 Now I write following code

Is there any way to get the current memory used by R without running gc()? I'd like to include the memory usage in logging output, but calling gc() to get that information takes long enough to be noticeable (~ 6 s with ~ 20 GB of ram in use), which goes against the point of logging. Thanks, Johann

Hi all, I'm trying to make an integer-backed quarter (as in fraction of year) class, but I can't quite it to work. I want integer-backed so I don't have to worry about floating-point effects when doing math, and so that I can use it as in data.table. First of all, is there a good reference for this anywhere? All of the S4 tutorials that I've found have been too high-level, and

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 =

I am trying to get monthly means for a daily data series using zoo(). I have found an odd problem, that seems to be caused by zoo()'s handling of leap years. Here's my R script with 2 methods (freq=365, 366) for aggregating the daily data to monthly series: library(zoo) J_link <- "http://www.ijis.iarc.uaf.edu/seaice/extent/plot.csv" JAXA_data <- read.table(J_link,

Dear Sir, Thanks for your mail and help. I got this error while trying to run your code. sbux1.z <- read.csv.zoo(u, FUN = as.yearmon, format = fmt) Error in read.table(file = file, header = header, sep = sep, quote = quote, : 'file' must be a character string or connection Thanks and Regards, Upananda Pani On Tue, Sep 19, 2017 at 4:31 PM, Upananda Pani <upananda.pani at

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 Gabor's code works as expeceted without error. What is "u" in your case? Cheers Petr > -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Upananda > Pani > Sent: Wednesday, September 20, 2017 11:06 AM > To: Gabor Grothendieck <ggrothendieck at gmail.com> > Cc: r-help <r-help at r-project.org> >

Dear All, Thanks a lot for your help. Would you please let me know if i want to read a csv file as zoo object from my local file rather than directly from the website, how to do that? library(zoo) u <- "https://faculty.washington.edu/ezivot/econ424/sbuxPrices.csv" fmt <- "%m/%d/%Y" With sincere regards, Upananda Pani On Wed, Sep 20, 2017 at 3:22 PM, PIKAL Petr

Hi R gourmets, I am trying to convert an HTML table into an xts object. The table has six columns, with the data of interest in a single row with each cell containing a long, \n-delimited character string. Initially, I work with these strings as elements in a list. This is necessary because the strings in each cell do not translate into a regular matrix with equal-length columns. Once I fix

I would like to run an aggregation on a zoo object that has multiple series in it, with one of more series having NA values. The problem is that by default the aggregate function will produce an NA value in each aggregated period that contains an NA. For instance, if I run aggregate(x, as.yearmon(index(x)), mean) on the example object "x" which is printed below, I will just get a bunch

#the below code is the way that I would like the plot to look. I have tried to write a panel function: my.panel <- function(x, y, ..., pf = parent.frame()) { axis(side=1, at = seq(rng[1], rng[2], 1/12), labels = n, tcl = -0.3) } #but it does not work and I am at a loss and help would be appreciated. I will use this for multiple library(zoo) library(chron) #this is what I would like the

What is the current best package for manipulating HDF5 data files? I tried "hdf5" a long time ago, but I ran into memory problems. "h5r" is on CRAN now, and "rhdf5" is part of bioconductor. Ideally, I'd like to read simple vectors or tables, either the entire thing or a subset of rows. I don't need much writing support, but it would be nice. Compression is a

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]] <-

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 =

Given a long zoo matrix, the goal is to "sweep" out a statistic from the entire length of the sequences. longzoomatrix<-zoo(matrix(rnorm(720),ncol=6),as.yearmon(outer(1900,seq(0,length=120)/12,"+"))) cnames<-c(12345,23456,34567,45678,56789,67890) colnames(longzoomatrix)<-cnames longzoomatrix[1:24,] 12345 23456 34567 45678