Got it, thank you!
On Tue, 10 Aug 2021, 00:12 David Winsemius, <dwinsem...@comcast.net> wrote:
>
> On 8/9/21 12:22 PM, Luigi Marongiu wrote:
> > Thank you! it worked fine! The only pitfall is that `NA` became
> > `<NA>`. This is essentially the same thing anyway...
>
>
> It's not "essentially the same thing". It IS the same thing. The print
> function displays those '<>' characters flanking NA's when the class is
> factor. Type this at your console:
>
>
> factor(NA)
>
>
> --
>
> David
>
> >
> > On Mon, Aug 9, 2021 at 5:18 PM Ivan Krylov <krylov.r...@gmail.com>
> wrote:
> >> Thanks for providing a reproducible example!
> >>
> >> On Mon, 9 Aug 2021 15:33:53 +0200
> >> Luigi Marongiu <marongiu.lu...@gmail.com> wrote:
> >>
> >>> df[df[['vect[2]']] == 2, 'vect[2]'] <- "No"
> >> Please don't quote R expressions that you want to evaluate. 'vect[2]'
> >> is just a string, like 'hello world' or 'I want to create a new column
> >> named "vect[2]" instead of accessing the second one'.
> >>
> >>> Error in `[<-.data.frame`(`*tmp*`, df[[vect[2]]] == 2, vect[2], value
> >>> = "No") : missing values are not allowed in subscripted assignments
> >>> of data frames
> >> Since df[[2]] containts NAs, comparisons with it also contain NAs. While
> >> it's possible to subset data.frames with NAs (the rows corresponding to
> >> the NAs are returned filled with NAs of corresponding types),
> >> assignment to undefined rows is not allowed. A simple way to remove the
> >> NAs and only leave the cases where df[[vect[2]]] == 2 is TRUE would be
> >> to use which(). Compare:
> >>
> >> df[df[[vect[2]]] == 2,]
> >> df[which(df[[vect[2]]] == 2),]
> >>
> >> --
> >> Best regards,
> >> Ivan
> >
> >
>
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