similar to: Error in length of vector ?

Hi, I'm trying to aggregate date values using the aggregate function. For example: aggregate(data,by=list(weekdays(LM),months(LM)),FUN=length) I would also like to aggregate by year but there seems to be no years() function. Should there be one? Is there any alternative choice? Also, a hours() function would be great. Any tip on this? Thanks in advance! S?rgio Nunes

Hi, Since "R" is a (very) generic name, I've been having some trouble searching the web for this topic. Due to this, I've just created a Google Custom Search Engine that includes several of the most relevant sites that have information on R. See it in action at: http://google.com/coop/cse?cx=018133866098353049407%3Aozv9awtetwy This is really a preliminary test. Feel free to

Hi, I have several files with data in this format: 20070102 20070102 20070106 20070201 ... The data is sorted and each line represents a date (YYYYMMDD). I would like to analyze this data using R. For instance, I would like to have a histogram by year, month or day. I've already made a simple Perl script that aggregates this data but I believe that R can be much more powerful and easy on

Hi, I'm looking for a function to measure the dispersion of a set of values ranging from 0 to 1. This function should be 0 if all the values are evenly spaced within the interval and it should be > 0 if values are clustered. The more clustered the values are, the higher should the function be. An example: [0; 0.2; 0.4; 0.6; 0.8; 1] - function should be ~ 0 [0; 0.1; 0.1; 0.15; 1] -

Hi, What's the best way to average dates? I though mean.POISXct would work fine but... > a [1] "2007-04-02 19:22:00 WEST" > b [1] "2007-03-17 16:23:00 WET" > class(a) [1] "POSIXt" "POSIXct" > class(b) [1] "POSIXt" "POSIXct" > mean(a,b) [1] "2007-04-02 19:22:00 WEST" > mean(b,a) [1] "2007-03-17

Hi, I'm trying to draw a 2D plot using multiple tints of red. The (simplified) setup is the following: || year | x | y || My idea is that each year is plotted with a different tint of red. Older year (lightest) -> Later year (darkest). I've managed to plot this with different scales of grays simply by doing: palette(gray(length(years):0/length(years))) before the plot and for each

Just for the record, here are my steps for producing a date based histogram. Data is stored in a file where each line only has a date - 2007/02/16 >d<-readLines("filename.dat") >d<-as.Date(d, format="%Y/%m/%d") >pdf(yearly.pdf) >hist(d, "years") >dev.off() Instead of "years" you can also use "days", "weeks",

Hi Everyone, I am trying to solve this problem, but couldn''t able to get the solution. I hope i will get some help here. I am using these three find methods to get different types of devices. I dont have any relationships between any models. windows_devices = BaseManagedEntity.find(:all, :select => "b.BaseManagedEntityInternalId, c.NetworkName, c.IPAddress,

Hi, I'm pretty new to R and I've been having some problems filtering data in matrices. I have the following initial dataset: || year | name | varA || I have multiple values for "varA" for the same "year" and the same "name". Having this as the input I would like to obtain the following: || year | name | {varA mean} || Where I only have one line for each

Hi, It seems that hist() has a buggy behavior when breaking over "days". The bug can be reproduced in a few steps: > d=data.frame(date=c("2009-01-01", "2009-01-02", "2009-01-02")) > d$date=as.Date(d$date) > d$date [1] "2009-01-01" "2009-01-02" "2009-01-02" > h=hist(d$date, "days") > h$count [1] 3

Hi, I'm pretty new to R, I've just installed version 2.1.1 on Windows. I have a simple (it seems) doubt - how do I change the Console default Language ? It's set to Portuguese but I would like to view it in English. Thanks in advance, S??rgio Nunes

The length() of a POSIXlt object is given as 9 regardless of the actual length. For example: > make.date.time function (year=c(2006,2006),month=c(8,8),day=2:5,hour=13,minute=45) {# convert year, etc., into POSIXlt object # d=as.character(make.date(year,month,day)) t=paste(hour,minute,sep=":") as.POSIXlt(paste(d,t)) } > t=make.date.time() > t [1] "2006-08-02 13:45:00"

Check the generated OVF for -o rhv and -o vdsm outputs. Variable UUIDs and date/times are filtered out. Make sure the the important UUIDs (disk, volume, VM) are where we think they should be. Signed-off-by: Tomáš Golembiovský <tgolembi@redhat.com> --- v2v/test-v2v-o-rhv.ovf.expected | 92 ++++++++++++++++++++++++++++++++ v2v/test-v2v-o-rhv.sh | 20 +++++++

In the following, can anyone tell me why length(eee) returns 9? I was expecting 15398, and when I try to add this vector to a data frame with that many rows, it fails complaining that the vector is of length 9. In what I thought was an identical situation with a related dataset, the same code worked as expected. > length(fff) [1] 15398 > str(fff) int [1:15398] 20010102 20010102

Arrays of POSIXlt dates always return a length of 9. This is correct (they're really lists of vectors of seconds, hours, and so forth), but other methods disguise them as flat vectors, giving superficially surprising behaviour: strings <- paste('2009-1-', 1:31, sep='') dates <- strptime(strings, format="%Y-%m-%d") print(dates) # [1]

I have a vector of POSIX times/dates (called "times") that I want to plot. But I'm having trouble because R reports the length of the vector incorrectly. Can someone help me figure out what's going on here? This code.... print(times) print(class(times)) print(length(times)) Produces this... . . . [355] "2006-03-06 18:34:00" "2006-03-06 18:32:00"

Full_Name: Petr Simecek Version: 2.5.1, 2.6.1 OS: Windows XP Submission from: (NULL) (195.113.231.2) Several times I have experienced that a length of a POSIXt vector has not been computed right. Example: tv<-structure(list(sec = c(50, 0, 55, 12, 2, 0, 37, NA, 17, 3, 31 ), min = c(1L, 10L, 11L, 15L, 16L, 18L, 18L, NA, 20L, 22L, 22L ), hour = c(12L, 12L, 12L, 12L, 12L, 12L, 12L, NA, 12L,

On 28/05/2018 13:23, kfx wrote: > On 28/05/2018 13:04, Timo Sirainen wrote: >> On 28 May 2018, at 13.28, kfx <kadafax at gmail.com >> <mailto:kadafax at gmail.com>> wrote: >>> >>> >>>> Especially what is in the "fts" header vs. "next uid" header? Does >>>> the UID in "fts" header keep changing

On Thu, 2019-12-05 at 17:15 +0000, Rowland penny via samba wrote: > On 05/12/2019 17:00, S?rgio Basto wrote: > > On Thu, 2019-12-05 at 10:15 +0000, Rowland penny via samba wrote: > > > On 05/12/2019 06:16, S?rgio Basto wrote: > > > > Sorry , I spoke too soon getent passwd "a new user to this > > > > server" > > > > doesn't work .

I have OpenSSL forgenrate the CA root file in my server and work fine. My question is, ?howto i say to Samba (configuration) for work with CA certificates? . I dont find information about this. Thanks. Saludos. --- Miguel El mar., 10 nov. 2020 a las 15:22, S?rgio Basto (<sergio at serjux.com>) escribi?: > On Tue, 2020-11-10 at 14:48 -0300, Miguel Angel Coa M. via samba wrote: >