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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