s on numeric, character and factor.
Jason
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Trevor Davies
Sent: Friday, July 12, 2013 2:57 PM
To: r-help@r-project.org
Subject: Re: [R] replace multiple values in vector at once
I always think t
David is right, but it's trivial if x is a factor (which is the
default when you create character columns in a data frame).
(Note also how to use rep() properly -- read the docs: ?rep)
x <- factor(rep(LETTERS[1:3],e=3))
x
[1] A A A B B B C C C
Levels: A B C
levels(x) <- 1:3
x
[1] 1 1 1 2 2 2 3 3
Hi,
library(car)
recode(x,"'x'=1;'y'=2;'z'=3")
#[1] 1 1 1 2 2 2 3 3 3
#or
as.numeric(factor(x))
#[1] 1 1 1 2 2 2 3 3 3
A.K.
- Original Message -
From: Trevor Davies
To: "r-help@r-project.org"
Cc:
Sent: Friday, July 12, 2013 5:56 PM
Yes, I caught my error once I posted it - I was fiddling with match prior
to hammering down with merge but your solution is much better. Thank you.
On Fri, Jul 12, 2013 at 3:05 PM, David Winsemius wrote:
>
> On Jul 12, 2013, at 2:56 PM, Trevor Davies wrote:
>
> > I always think that replying to
On Jul 12, 2013, at 2:56 PM, Trevor Davies wrote:
> I always think that replying to your own r-help feels silly but it's good
> to close these things out.
>
> here's my hack solution:
>
> x1<-merge(data.frame(A=x),data.frame(A=c('x','y','z'),B=c(1,2,2)),by='A')[,2]
That fairly tortured compare
I always think that replying to your own r-help feels silly but it's good
to close these things out.
here's my hack solution:
x1<-merge(data.frame(A=x),data.frame(A=c('x','y','z'),B=c(1,2,2)),by='A')[,2]
Well that works and should for my more complex situation. If anyone has
something a little
I'm trying to find a function that can replace multiple instances of values
or characters in a vector in a one step operation. As an example, the
vector:
x <- c(rep('x',3),rep('y',3),rep('z',3))
> x
[1] "x" "x" "x" "y" "y" "y" "z" "z" "z"
I would simply like to replace all of the x's with 1's,
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