similar to: Hourly Time Series

I have a setup where i am using samba to access my linux box through windows, to edit scripts and stuff. But say if a script as executable permissions for all when i open it in windows through samba, on saving it the prior permissions are overwritten by samba's default permission. Is there a way to tell samba to keep the file prior permission. Harshal Dharia

hi every one i dont know whether rsync excludes the open file while taking backup or not.? My problem is how to handle open file backup(the files which are open by any process ) with rsync .? Thanks Harshal SPsoft

Hey list, suppose I have several dates in a database-table, where these dates are marked as 'set' or 'not set'. If I do something like : SELECT ID FROM my_table WHERE client=clientID AND set=yes and this query results in several rows and thus several ID's like "2 5 7 11 13 14 17"... How can I iterate over these values ?? Something like : for each item in ARRAY {

Dear All, Assume I have a data.frame that contains also factors and I would like to get another data.frame containing the factors as numeric vectors, to apply functions like sapply(..., median) on them. I read the warning concerning as.numeric or unclass, but in my case this makes sense, because the factor levels are properly ordered. I can do it, if I write for each single column

Have I been sleeping in class? rw1071 from CRAN, windows XP incidencia is made by a call to tapply > class(incidencia) [1] "array" > incidencia <- unclass(incidencia) > class(incidencia) [1] "array" Kjetil Halvorsen

In a real example I was trying to remove the class from the result of table, just because it was to be used as a building block for other things and a simple integer vector seemed likely to be most efficient. I'm puzzled as to why unclass doesn't work. > zed <- table(1:5) > class(zed) [1] "table" > class(unclass(zed)) [1] "array" >

In a real example I was trying to remove the class from the result of table, just because it was to be used as a building block for other things and a simple integer vector seemed likely to be most efficient. I'm puzzled as to why unclass doesn't work. > zed <- table(1:5) > class(zed) [1] "table" > class(unclass(zed)) [1] "array" >

I have a time series of rainfall in a dataframe. I would like to be able to aggregate this using a sliding window approach- i.e. a new 24 hourly total is calculated for each hours rainfall. I'm struggling to understand how this might be achieved - currently I've tried looping a sum function to re-calculate a new total at every stage of the loop. for (inp[[9]] in

I have data from several sensors that recorded data at hourly intervals during seven months. I want to separate daily variation from the trend, and also be able to zoom in on only one month of data. I have not been able what functions to use, I can not figure out from the help for 'ts' how to use hourly data. I guess this is routine-work for a lot of people so I hope someone can

Is there a low-level function that returns the length of an object 'x' - the length that for instance .subset(x) and .subset2(x) see? An obvious candidate would be to use: .length <- function(x) length(unclass(x)) However, I'm concerned that calling unclass(x) may trigger an expensive copy internally in some cases. Is that concern unfounded? Thxs, Henrik

Full_Name: Bo Zhou Version: 2.6.1 (2007-11-26) OS: Windows XP Submission from: (NULL) (207.237.54.242) Hi, I found an arithmetic problem when I'm doing something with POSIXct The code to reproduce it is as follows (This is the recommended way of finding out time zone difference on R News 2004-1 Page 32 URL http://cran.r-project.org/doc/Rnews/Rnews_2004-1.pdf) a=Sys.time()

Hello, I have a dataframe containing various time series (not time series objects though!)with hourly time steps. I?d like to perform ccf for I need to know the correlation factors for different lags. Here is an example: x<-as.POSIXct(c("2008-12-25 16:00:00", "2008-12-25 17:00:00", "2008-12-25 18:00:00", "2008-12-25 19:00:00", "2008-12-25

The following command produces red axis line in a pairs plot: pairs(iris[1:4], main = "Anderson's Iris Data -- 3 species", pch = "+", col = c("red", "green3", "blue")[unclass(iris$Species)]) Trying to fool pairs in the following way produces the same plot as above: pairs(iris[1:4], main = "Anderson's Iris Data -- 3

Hello everyone, I am just a tyro in R and would like your kindly help for some problems which I've been struggling for a while but still in vain. I have a time-series file (with some missing value ) which looks like time[sec] , Factor1 , Factor2 00:00:00 01.01.2007 , 0.0000 , 0.176083 01:00:00 01.01.2007 , 0.0000 , 0.176417 [ ... ] 11:00:00 10.06.2007 , 0.0000 , 0.148250 12:00:00 10.06.2007

Hi, Encouraged by a tip from Simon Urbanek I tried to use the S3 machinery to write a faster version of the data.frame class. This quickly hits a snag: the "[.default"(x, i) for some reason cares about the dimensionality of x. In the end there is a full transcript of my R session. It includes the motivation for writing the class and the problems I have encountered. As a result I see

Dear list, I have to look examine hourly time - series and would like to plot variable section of them using plot.zoo. Hourly time series data which looks like this: YYYY MM DD HH P-uk P-kor P-SME EPOT EREA RO R1 R2 RGES S-SNO SI SSM SUZ SLZ 2003 1 1 1 0.385 0.456 0.021 0.000 0.000 0.000 0.013 0.223 0.235 0.01 0.38

Buenas tardes a todos los participantes del foro. Me dirijo a vosotros porque estoy atascado con una duda de programación respecto al data frame: > dd # Data frame de 5 variables, leído de un archivo txt id sexo nacim origen final 1 1 0 02/09/1955 01/04/1985 01/02/2014 2 2 1 29/10/1951 15/08/1996 01/05/2009 3 3 0 30/10/1942 02/08/2000 01/02/2014 4 4 1

R 3.5.0 Is it intended that the Date method of as.POSIXct does not respect the tz parameter? I suggest changing as.POSIXct.Date to this: function (x, tz = "", ...) .POSIXct(unclass(x) * 86400, tz = tz) Currently, the best workaround seems to be using the character method if one doesn't want the default timezone (which is often an annoying DST timezone). This came up on

The bottomline here is that one can always call a base method, inexpensively and without modifying the object, in, let's say, *formal* OOP languages. In R, this is not possible in general. It would be possible if there was always a foo.default, but primitives use internal dispatch. I was wondering whether it would be possible to provide a super(x, n) function which simply causes the

Hi All, I'm perplexed by the way the unclass function displays a factor whose labels have been swapped with the relevel function. I realize it won't affect any results and that the relevel did nothing useful in this particular case. I'm just doing it to learn ways to manipulate factors. The display of unclass leaves me feeling that the relevel had failed. I've checked three books