R help - Jun 2005 - data transformation

i have data fame da:
> da
    x y
1   1 a
2   2 a
3   3 a
4   4 a
5   5 a
6   6 b
7   7 b
8   8 b
9   9 b
10 10 b
> str(da)
`data.frame':   10 obs. of  2 variables:
 $ x: num  1 2 3 4 5 6 7 8 9 10
 $ y: Factor w/ 2 levels "a","b": 1 1 1 1 1 2 2 2 2 2

and i want to generate new variable da$z,when
y=="a",da$z=da$x-mean(x[y=="a"])  ,when  
y=="b",da$z=da$x-mean(x[y=="b"]).

this data frame is simple and i can do it by hand,if the y has many levels and i
have x1,x2....
can i do it quickly?

Try:
> da$z <- da$x - ave(da$x, da$y)
> da
    x y  z
1   1 a -2
2   2 a -1
3   3 a  0
4   4 a  1
5   5 a  2
6   6 b -2
7   7 b -1
8   8 b  0
9   9 b  1
10 10 b  2

Andy
> From: ronggui
> 
> i have data fame da:
> > da
>     x y
> 1   1 a
> 2   2 a
> 3   3 a
> 4   4 a
> 5   5 a
> 6   6 b
> 7   7 b
> 8   8 b
> 9   9 b
> 10 10 b
> > str(da)
> `data.frame':   10 obs. of  2 variables:
>  $ x: num  1 2 3 4 5 6 7 8 9 10
>  $ y: Factor w/ 2 levels "a","b": 1 1 1 1 1 2 2 2 2 2
> 
> and i want to generate new variable da$z,when 
> y=="a",da$z=da$x-mean(x[y=="a"])  ,when   
> y=="b",da$z=da$x-mean(x[y=="b"]).
> 
> this data frame is simple and i can do it by hand,if the y 
> has many levels and i have x1,x2....
> can i do it quickly?
> 
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>

On 6/4/05, ronggui <0034058 at fudan.edu.cn>
wrote:
> i have data fame da:
> > da
>    x y
> 1   1 a
> 2   2 a
> 3   3 a
> 4   4 a
> 5   5 a
> 6   6 b
> 7   7 b
> 8   8 b
> 9   9 b
> 10 10 b
> > str(da)
> `data.frame':   10 obs. of  2 variables:
>  $ x: num  1 2 3 4 5 6 7 8 9 10
>  $ y: Factor w/ 2 levels "a","b": 1 1 1 1 1 2 2 2 2 2
> 
> and i want to generate new variable da$z,when
y=="a",da$z=da$x-mean(x[y=="a"])  ,when  
y=="b",da$z=da$x-mean(x[y=="b"]).
> 
> this data frame is simple and i can do it by hand,if the y has many levels
and i have x1,x2....
> can i do it quickly?
> 
Try this:

da$z <- resid(lm( x ~ y, da ))