Re: aggregate function does not work: arguments must have same length

Às 05:08 de 28/02/2026, Sorkin, John escreveu:
> PLEASE READ THIS ENTIRE MESSAGE. It explains my problem and contains a reproducible example of the problem.
> 
> I am trying to run an aggregate function. The function produces the following error message:
> ERROR MESSAGE:
> Error in aggregate.data.frame(as.data.frame(x), ...) :
>    arguments must have same length
> 
> Everything beyond this line can be pasted into an R session and run:
> 
> ####Start of code
> #I am trying to write a code that will compute the number of hours for which we have data for each day.
> #Here are my data:
> 
>    inJan<-
>      structure(list(NewTime = structure(c(1597363200, 1597366800,
>       1597370400, 1597374000, 1597377600, 1597381200, 1597384800, 1597388400,
>       1597392000, 1597395600, 1597399200, 1597402800, 1597406400, 1597410000,
>       1597413600, 1597417200, 1597420800, 1597424400, 1597428000, 1597431600,
>       1597435200, 1597438800, 1597442400, 1597446000, 1597449600, 1597453200,
>       1597456800, 1597460400, 1597464000, 1597467600, 1597471200, 1597474800,
>       1597478400, 1597482000, 1597485600, 1597489200, 1597492800, 1597496400,
>       1597500000, 1597503600, 1597507200, 1597510800, 1597514400, 1597518000,
>       1597521600, 1597525200, 1597528800, 1597532400, 1597536000, 1597539600
>      ), class = c("POSIXct", "POSIXt"), tzone = "UTC"), Days = structure(c(18488,
>       18488, 18488, 18488, 18488, 18488, 18488, 18488, 18488, 18488,
>       18488, 18488, 18488, 18488, 18488, 18488, 18488, 18488, 18488,
>       18488, 18488, 18488, 18488, 18488, 18489, 18489, 18489, 18489,
>       18489, 18489, 18489, 18489, 18489, 18489, 18489, 18489, 18489,
>       18489, 18489, 18489, 18489, 18489, 18489, 18489, 18489, 18489,
>       18489, 18489, 18490, 18490), class = "Date")), class = "data.frame", row.names = c(NA,-50L))
> inJan
> names(inJan)
> 
> #I have written a function which finds the number of hours between a last time last(x) and a first time first(x).
> 
> # Number of follow-up hours in each day
> timediff <- function(x) {
> difftime(last(x),first(x),units="hours")
>   }
> 
> #The function I wrote is used in an aggregate function which
> # finds the minimum and maximum time in each day, and then
> # calls timediff to find the number of hours in each day.
> zoop <- aggregate(inJan[,"NewTime"],as.list(inJan[,"Days"]),timediff)
> 
> #When I run the aggregate function I receive and error:
> #Error in aggregate.data.frame(as.data.frame(x), ...) :
> #arguments must have same length
> 
> #HOWEVER I believe inJan[,"NewTime"] and inJan[,"Days"] have the same length
> 
> llength(inJan[,"NewTime"])
> #[1] 50
> length(inJan[,"Days"])
> #[1] 50
> 
> #Can someone tell my why the aggregate function thinks the arguments are a different length?
> #### End of code
> 
> Thank you,
> John
> 
> John David Sorkin M.D., Ph.D.
> Professor of Medicine, University of Maryland School of Medicine;
> Associate Director for Biostatistics and Informatics, Baltimore VA Medical Center Geriatrics Research, Education, and Clinical Center;
> Former PI Biostatistics and Informatics Core, University of Maryland School of Medicine Claude D. Pepper Older Americans Independence Center;
> Senior Statistician University of Maryland Center for Vascular Research;
> 
> Division of Gerontology, Geriatrics and Palliative Medicine,
> 10 North Greene Street
> GRECC (BT/18/GR)
> Baltimore, MD 21201-1524
> Cell phone 443-418-5382
> 
> 
> 
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Hello,

The problem is not in the input vectors' lengths, it's the output length.

When aggregating, I find it better to use the formula interface.
It avoids the list() call and eventual errors due to as.list() being 
mistaken for it.

Also, when using non base packages, please include a call to library() 
or similar. I had to revise the function for it to work.


# Number of follow-up hours in each day
timediff <- function(x) {
   difftime(dplyr::last(x), dplyr::first(x), units = "hours")
}

aggregate(NewTime ~ Days, inJan, timediff)
#>         Days  NewTime
#> 1 2020-08-14 23 hours
#> 2 2020-08-15 23 hours
#> 3 2020-08-16  1 hours


Alternative versions:

# This is what Peter Daalgard proposed (list, not as.list)
aggregate(inJan[,"NewTime"], list(inJan[,"Days"]), timediff)

# another one
aggregate(inJan[["NewTime"]], list(inJan[["Days"]]), timediff)


# not relevant, but if you ever want just a vector, here it is.
tapply(inJan[,"NewTime"], inJan[,"Days"], timediff)
#> 2020-08-14 2020-08-15 2020-08-16
#>         23         23          1


Hope this helps,

Rui Barradas