

On 20-07-2010, at 20:30, litao.ext [via R] wrote:

> yes, the matrix is symmetric. 
> 
> But when I apply normalize upon rows, I expect to get the following result: 
> > apply( a, 1, normalize ) 
>      [,1] [,2] [,3] [,4] [,5] 
> [1,] 0.00 0.25 0.50 0.75 1.00 
> [2,] 0.00 0.25 0.50 0.75 1.00 
> [3,] 0.00 0.25 0.50 0.75 1.00 
> [4,] 0.00 0.25 0.50 0.75 1.00 
> [5,] 0.00 0.25 0.50 0.75 1.00 
> 
> Actually we can get asymmetric matrix like: 
> > a = c(1:5) %*% t(1:5) 
> > a 
>      [,1] [,2] [,3] [,4] [,5] 
> [1,]    5    4    3    2    1 
> [2,]   10    8    6    4    2 
> [3,]   15   12    9    6    3 
> [4,]   20   16   12    8    4 
> [5,]   25   20   15   10    5 
> 
> > apply(a, 2, normalize) 
>      [,1] [,2] [,3] [,4] [,5] 
> [1,] 0.00 0.00 0.00 0.00 0.00 
> [2,] 0.25 0.25 0.25 0.25 0.25 
> [3,] 0.50 0.50 0.50 0.50 0.50 
> [4,] 0.75 0.75 0.75 0.75 0.75 
> [5,] 1.00 1.00 1.00 1.00 1.00 
> > 
> > apply(a, 1, normalize) 
>      [,1] [,2] [,3] [,4] [,5] 
> [1,] 1.00 1.00 1.00 1.00 1.00 
> [2,] 0.75 0.75 0.75 0.75 0.75 
> [3,] 0.50 0.50 0.50 0.50 0.50 
> [4,] 0.25 0.25 0.25 0.25 0.25 
> [5,] 0.00 0.00 0.00 0.00 0.00 
> 
> the latter result is very strange. 

Try ?apply

You will see  that apply concatenates the result for each row in a single vector.
This results in the matrix you show.
You need to transpose the result of apply(a,1,..) if you want to get what you want.

/Berend


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