Hannah,
If Gen is a factor you can simply build the new factor "on top" of it:
dataframe$Gen<- factor( c('Wynda' , 'A_2' , 'B_1' , 'Wynda' , 'Wynda'
, 'OP1_5')[Gen] )
Just make sure the replacement labels are in the same order as levels(Gen).
Cheers
On Tue, Feb 28, 2012 at 8:39 PM, hannahmaohuang
<hannahmaohu...@gmail.com> wrote:
> Dear All,
> I've been searching relevant topics about replacing values, none seemed to
> be applicable to me...
>
> I have a file with many many varieties, and want to replace some of them
> into different names.
> I tried various of ways, still don't know how to do that most efficiently..
> Here is part of the example data:
>
>
> Gen Rep
> A_1 1
> A_1 2
> A_2 1
> A_2 2
> B_1 1
> B_1 2
> B_3 1
> B_3 2
> OP1_1 1
> OP1_1 2
> OP1_5 1
> OP1_5 2
>
> For example, I want to replace A_1, B_3, OP1_1 into different name
> "Wynda"
>
> So that the expected file should become:
>
> Gen Rep
> Wynda 1
> Wynda 2
> A_2 1
> A_2 2
> B_1 1
> B_1 2
> Wynda 1
> Wynda 2
> Wynda 1
> Wynda 2
> OP1_5 1
> OP1_5 2
>
>
> I have created a link file, which contains two rows, translating which Gen
> correlating to which Name. Not sure if this file helps or not, example as
> below:
>
> Column1(Gen) Column2(Name)
> A_1 Wynda
> A_2 A_2
> B_1 B_1
> B_3 Wynda
> OP1_1 Wynda
> OP1_5 OP1_5
>
>
> Though I can replace one by one in excel, since there are too many files and
> too many reps, it'll be very time-consuming also easy to make mistakes.
>
> Please give me any guidance or help in terms of finish this with R.
>
> Thanks so much !
>
> Sincerely
> Hannah
>
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/How-to-replace-the-values-in-a-column-tp4430448p4430448.html
> Sent from the R help mailing list archive at Nabble.com.
>
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