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Hi, I am having trouble trying to figure out why if_else is behaving the way it is, it may be my code or the way the data is structured. Below is a snapshot of a database am working on and it represents a longitudinal survey of study participants in a trial with weekly follow up. The variable "survey_start" represents the start of the study-defined one year follow up (which we called

Technically, there is a round() for 'Date' objects, but it doesn't seem very useful, because it basically just fall back to the default round() method, which only takes the 'digits' argument. Here's an example: > date <- Sys.Date() > class(date) [1] "Date" We see that there are only two round() methods in addition to the implicit built-in one; >

Proper calendar dates in R are great for plotting and calculating. However for the non-wonks among us, they can be very frustrating. I have recently discussed the pains that people in my lab have had with dates in R. Especially the frustration of bringing date data into R from Excel, which we have to do a lot. Please find below a simple analgesic for R date importation that I discovered over

This is a workaround, and could be the basis for a round.Date improvement: date <- Sys.Date() as.Date(round(as.POSIXct(date), "years")) as.Date(round(as.POSIXct(Sys.Date() + 180), "years")) Duncan Murdoch On 08/02/2024 12:23 p.m., Henrik Bengtsson wrote: > Technically, there is a round() for 'Date' objects, but it doesn't > seem very useful,

Dear R users, I have a column with dates (character) in a data frame: 12-Jan-01 11-Jan-01 10-Jan-01 9-Jan-01 8-Jan-01 5-Jan-01 and I need to convert them to (Julian) dates so that I can sort the whole data frame by date. I thought it would be very simple, but after checking the documentation and the list I still don't have something that works. 1. as.Date returns the error below. What am I

Thanks it worked for me. I wanted to plot days since planting on x-axis 1 and years on x-axis 3. LAI_simulation$Date <- as.Date( LAI_simulation$Date, '%Y/%m/%d') LAI_simulation$Date <- as.integer(LAI_simulation$Date - as.Date("2009-10-07")) plot(LAI~Date,data=LAI_simulation,xlab="Days since Oct, 7,

I am trying to calculate the number of days between 2 dates using Time.now - object.date where object.date is a date pulled from a database. I want to display the age of the entry in the database in days. I keep getting an error about not being able to convert a date into a float. What am I doing wrong. If I use object.date - object.date it seems to work but the answer is useless (0). I

?s 14:36 de 08/02/2024, Olivier Benz via R-devel escreveu: >> On 8 Feb 2024, at 15:15, Martin Maechler <maechler at stat.math.ethz.ch> wrote: >> >>>>>>> Ji?? Moravec >>>>>>> on Wed, 7 Feb 2024 10:23:15 +1300 writes: >> >>> This is my first time working with dates, so if the answer is "Duh, work >>> with

Hi R users, Wonder if somebody could give me help on how to reshape this type of data: ----------------------------------------------------------------------------------------------------------------------- Date:10.09.19 Time:21:39:05 Lat:N62.37.18 Long:E018.07.32 0000-0010 | 28| 28 0010-0020| 302| 302 0020-0030| 42| 42 0030-0040| 2| 2 0040-0050| 1| 1 0060-0070| 1| 1

I.e., I see this when I run `git fetch`: ``` $ git fetch -v llvm.org From http://llvm.org/git/llvm = [up to date] master -> llvm.org/master = [up to date] release_1 -> llvm.org/release_1 = [up to date] release_16 -> llvm.org/release_16 = [up to date] release_20 -> llvm.org/release_20 = [up to date] release_21 -> llvm.org/release_21 = [up to date]

I don't fully understand the logic you are trying to implement, but something along the lines of foo <- cut(trialData$date, breaks = as.Date(c("2007-01-01", "2008-05-01", "2009-04-01", "2010-05-01",

Hi, I am trying to install R 2.9.0 on AIX 5.3. During the 'make' run I get the following error: Error: Invalid DESCRIPTION file Invalid Priority field. Packages with priorities 'base' or 'recommended' or 'defunct-base' must already be known to R. See the information on DESCRIPTION files in section 'Creating R packages' of the

What''s the best way to handle a requirement that shows up as a sub-requirement requirement in other features? For example let''s say users can enter dates in various forms throughout my application. There is one set of global rules specifying the formats in which dates may be entered, and how they are interpreted. I put that in one feature. In various other features, like

That many ifelse statements is obviously rather a pain. Would you not have got what you want with ... paste("survey", year, sep="_") ? If that is not what you're looking for (eg because 'year' is the observation year and not the study start year), perhaps something that picks the minimum year for a subject or other relevant group might work? For example

Colleagues I want to download the package "date" from CRAN <http://cran.r- project.org/> to get the R-scripts (for version 1.2.0) *** date/INDEX *** as.date Coerce Data to Dates date.ddmmmyy Format a Julian date date.mdy Convert from Julian Dates to Month, Day, and Year date.mmddyy Format a Julian date date.mmddyyyy Format a Julian date date.object Date Objects mdy.date

Hello, I gather variants of this question have been asked previously. I have done some reading but only became more confused, as I suspect what I am trying to do is more basic than other applications. The following code readily calculates the difference in days between two dates: newdays <- ISOdate(2005, 5,12) - ISOdate(2006, 12, 22) However, I wanted to be able to deduct the dates in

Hi, Given a date, how do I get the last date of that month? I have data in the form YYYYMM, that I've read as a date using > x$Date <- as.Date(ISOdate(substr(x$YearEnd,1,4),substr(x$YearEnd,5,6),1)) But this gives the first day of the month. To get the last day of the month, I tried > as.Date(as.yearmon(x$Date,frac=0)) But I don't get the last day of the month here. (Tried

Hi Kevin, I ran the code on the full data set and was able to reproduce the problem that you are facing. My guess is that you have an error in your intuition and/or logic, and that this relates to the use of the subscript [1]. Specifically, on the full dataset, the condition trialData$date[trialData$survey_start == "Y" & trialData$year == 2013 & trialData$site ==

i know it''s probably really simple, how do i work out someone''s age if i have their d.o.b. stored as a date in my db. cheers -- Posted via http://www.ruby-forum.com/.

I am a relatively new user of R and have a question about using the read.table command for reading in .csv files that contain dates. My ultimate goal is to be able to read several different files into data frames and then do a merge along the "Date" column of each data frame. It seems I should be able to specify the name of the column that contains dates and then automatically convert