R help - Oct 2012 - Efficient method: Equality of all elements of two vectors

n=10
x1<-(1:n)/n
y1<-rnorm(n,x1^2,1)
m=20
x2<-(1:m)/m

The value of y2 will be rnorm (m, x2^2,1) if none of the elements of x2 is same
as x1, but for every same elements in x1 and x2, the value of y2 will be same as
y1. I know the following is correct, but for large vectors, this won't work
efficiently as this is taking longer time. Can you please let me know how to
write the following more efficiently.
-------------------
y2<-rep(0,m)
for(i in 1:m)
{ 
  for (j in 1:n)
  {
    if(x2[i]==x1[j]
    {
       y2[i]=y1[j]
    }
    else
    {
    y2[i]=rnorm(1,x2[i]^2,1);
  }
}
	[[alternative HTML version deleted]]

Hi,

On Sun, Oct 14, 2012 at 11:23 AM, Shant Ch <sha1one at yahoo.com>
wrote:
> n=10
> x1<-(1:n)/n
> y1<-rnorm(n,x1^2,1)
> m=20
> x2<-(1:m)/m
>
> The value of y2 will be rnorm (m, x2^2,1) if none of the elements of x2 is
same as x1, but for every same elements in x1 and x2, the value of y2 will be
same as y1. I know the following is correct, but for large vectors, this
won't work efficiently as this is taking longer time. Can you please let me
know how to write the following more efficiently.
Sounds like
y2 <- rnorm (m, x2^2,1)
y2 <- ifelse(x2==x1, y1, y2)

Best,
Ista
> -------------------
> y2<-rep(0,m)
> for(i in 1:m)
> {
>   for (j in 1:n)
>   {
>     if(x2[i]==x1[j]
>     {
>        y2[i]=y1[j]
>     }
>     else
>     {
>     y2[i]=rnorm(1,x2[i]^2,1);
>   }
> }
>         [[alternative HTML version deleted]]
>
>
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