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Editorial Team
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Asked: 2026-06-07T06:04:47+00:00

> f = function(x) as.Date(as.character(x), format=’%Y%m%d’) > f(20110606) [1] 2011-06-06 > sapply(20110606, f) [1]

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> f =  function(x) as.Date(as.character(x), format='%Y%m%d')
> f(20110606)
[1] "2011-06-06"
> sapply(20110606, f)
[1] 15131

Why 2 returned values are not the same. I need to apply this function to a long vector of dates, but I’m not getting dates with sapply()!

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1 Answer

  1. Editorial Team
    > lapply(20110606, f)
[[1]]
[1] "2011-06-06"

> unlist(lapply(20110606, f))
[1] 15131

    sapply unlists lapply and in doing so unclasses the date

    > unclass(lapply(20110606, f)[[1]])
[1] 15131
> class(lapply(20110606, f)[[1]])
[1] "Date"

    as @Joshua Ulrich noted there is no need to use apply type functions however for interest

     d <- 20110606 + 0:10
 do.call("c",lapply(d, f))

    would be one possible way to “unlist” the dates

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