Hi,
Using ddply,
ddply(df, .(ID), mutate, nrows=length(rel.head), test = nrows==2 &
all(rel.head %in% c(1,3)))
HTH,
baptiste
On 5 October 2011 06:02, Dennis Murphy <djmu...@gmail.com> wrote:
> Hi:
>
> Here's another way to do it with the plyr package, also not terribly
> elegant. It assumes that rel.head is a factor in your original data
> frame:
>> str(df)
> 'data.frame': 11 obs. of 2 variables:
> $ ID : Factor w/ 6 levels "17100","17101",..: 1 1 2 3 4 4 5 5 5 6 ...
> $ rel.head: Factor w/ 3 levels "1","2","3": 1 3 1 1 1 2 1 2 3 1 ...
>
> If this is not the case in your data, then you need to modify the
> function f below accordingly. (This is why use of dput() is preferred
> when sending example data to R-help, BTW.)
>
> library('plyr')
> f <- function(d) {
> tvec <- factor(c(1, 3), levels = 1:3) # target vector
> if(nrow(d) != 2L) {d$dummy <- rep(0, nrow(d)); return(d)}
> # If the first if statement is FALSE, then the following code is run:
> d$dummy <- ifelse(!identical(d[, 2], tvec), 0, 1)
> d
> }
>
> ddply(df, .(ID), f)
>
> ID rel.head dummy
> 1 17100 1 1
> 2 17100 3 1
> 3 17101 1 0
> 4 17102 1 0
> 5 17103 1 0
> 6 17103 2 0
> 7 17104 1 0
> 8 17104 2 0
> 9 17104 3 0
> 10 17105 1 1
> 11 17105 3 1
>
> HTH,
> Dennis
>
> On Tue, Oct 4, 2011 at 8:44 AM, <gra...@stat.columbia.edu> wrote:
>> Hi all,
>>
>> I have a dataset of individuals where the variable ID corresponds to the
>> identification of the household where the individual lives. rel.head stands
>> for the relationship with the household head. so rel.head=1 is the household
>> head, rel.head=2 is the spouse, rel.head=3 is the children.
>>
>> Here is an example to see how it looks like:
>>
>> df<-data.frame(ID=c("17100", "17100", "17101", "17102", "17103", "17103",
>> "17104", "17104", "17104", "17105", "17105"),
>> rel.head=c("1","3","1","1","1", "2", "1", "2", "3", "1", "3"))
>>
>>
>> I want to add a dummy variable that is equal to 1 when these conditions
>> held simultaneously :
>>
>> a) the number of rows with same ID is equal to 2
>> b) the variable rel.head=1 and rel.head=3
>>
>>
>> So my ideal output is:
>>
>> ID rel.head added.dummy
>> 1 17100 1 1
>> 2 17100 3 1
>> 3 17101 1 0
>> 4 17102 1 0
>> 5 17103 1 0
>> 6 17103 2 0
>> 7 17104 1 0
>> 8 17104 2 0
>> 9 17104 3 0
>> 10 17105 1 1
>> 11 17105 3 1
>>
>> Is there a simple way to do that?
>> Can somebody help?
>>
>> Thanks in advance,
>> Grazia
>>
>> ______________________________________________
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> 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
> https://stat.ethz.ch/mailman/listinfo/r-help
> 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
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
