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Home/ Questions/Q 6860619
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T02:25:51+00:00 2026-05-27T02:25:51+00:00

I have started using data.table. Indeed it is very fast and quite nice syntax.

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I have started using data.table. Indeed it is very fast and quite nice syntax. I am having trouble with dates. I like to use lubridate. In many of my data sets I have dates or dates and times and have used lubridate to manipulate them. Lubridate stores the instant as a POSIX class. I have seen answers here that create new variables for instance just to get the year eg. 2005. I do not like that. There are times that I will be analyzing by year and other times by quarter and other times by month and other times by durations. I would like to do something simple such as this

mydatatable[,length(medical.record.number),by=year(date.of.service)]

that should give me the number of patient encounters in a given year. The by function is not working.

Error in names(byval) = as.character(bysuborig) : 
  'names' attribute [2] must be the same length as the vector [1]

Can you please point me to vignettes where data.tables is used with dates and where manipulations and categorizations of those dates are done on the fly.

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  1. Editorial Team
    Editorial Team
    2026-05-27T02:25:52+00:00Added an answer on May 27, 2026 at 2:25 am

    This uses one of the examples in the help(IDateTime) page. It shows that you canc hange to syntax for the by=argument to a character value in the form ” = ” or (after @Matthew Dowle’s comment below) you can try to use the functional form that you were using (although I have not been able to get it to work myself. I did get the preferred form: by=list(wday=wday(idate)) to work.) Note that the key creation assumes an IDateTime class since there is no idate or itime variable. Those are attributes of the class

    datetime <- seq(as.POSIXct("2001-01-01"), as.POSIXct("2001-01-03"), by = "5 hour")    
    (af <- data.table(IDateTime(datetime), a = rep(1:2, 5), key = "a,idate,itime"))
    
     af[, length(a), by = "wday = wday(idate)"]
             wday V1
    [1,]    2  4
    [2,]    3  5
    [3,]    4  1
    
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