[R] substitute column data frame based on name stored in variable in r

[David Winsemius]

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.r00t using gmail.com> wrote:
>> Thanks for providing a reproducible example!
>>
>> On Mon, 9 Aug 2021 15:33:53 +0200
>> Luigi Marongiu <marongiu.luigi using 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
>
>