On Jun 28, 2012, at 09:42 , Rui Barradas wrote:
> Hello,
>
> Another way is to use index vectors:
>
>
> v1.factor <- c("S","S","D","D","D",NA)
> v2.factor <- c("D","D","S","S","S","S")
>
> td2 <- test.data <- data.frame(v1.factor,v2.factor)
>
> for (i in 1:nrow(test.data) ) {
>
> [... etc ...]
>
> } #End FOR
>
> # Create index vectors
> na1 <- is.na(v1.factor)
> na2 <- is.na(v2.factor)
>
> # Create 'newvar' with default value
> td2$newvar <- NA
> # Now, set values if condition is met.
> td2$newvar[!na1 & !na2] <- as.character(td2$v1.factor[!na1 &!na2])
>
> all.equal(test.data, td2)
> [1] TRUE
>
Shouldn't this rather be something like
new <- td2$v1.factor
i <- is.na(new)
new[i] <- td2$v2.factor[i]
td2$newvar <- new
??
(Caveat: things could go wrong if the levels differ.)
>
> I find this way better when doing a multiple if/else based on combinations of
> a small number of conditions. It's also very readable.
>
> Hope this helps,
>
> Rui Barradas
>
> Em 28-06-2012 08:00, Miguel Manese escreveu:
>> Hi James,
>>
>> On Thu, Jun 28, 2012 at 12:33 AM, James Holland <holland.ag...@gmail.com>
>> wrote:
>>> I need to look through a dataset with two factor variables, and depending
>>> on certain criteria, create a new variable containing the data from one of
>>> those other variables.
>>>
>>> The problem is, R keeps making my new variable an integer and saving the
>>> data as a 1 or 2 (I believe the levels of the factor).
>>>
>>> I've tried using as.factor in the IF output statement, but that doesn't
>>> seem to work.
>>>
>>> Any help is appreciated.
>>>
>>>
>>>
>>> #Sample code
>>>
>>> rm(list=ls())
>>>
>>>
>>> v1.factor <- c("S","S","D","D","D",NA)
>>> v2.factor <- c("D","D","S","S","S","S")
>>>
>>> test.data <- data.frame(v1.factor,v2.factor)
>>
>> The vectorized way to do that would be
>>
>> # v1.factor if present, else v2.factor
>> test.data$newvar <- ifelse(!is.na(v1.factor), v1.factor, v2.factor)
>>
>> I suggest you work with the character levels first then convert it
>> into a factor, e.g. if v1.factor & v2.factor are already factors, do:
>>
>> test.data$newvar <- as.factor(ifelse(!is.na(v1.factor),
>> as.character(v1.factor), as.character(v2.factor)))
>>
>>
>>
>> Regards,
>>
>> Jon
>>
>> ______________________________________________
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>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> ______________________________________________
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd....@cbs.dk Priv: pda...@gmail.com
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