R devel - Jun 2001 - v matrix of svd(X) loses dimensions if nrow(X)==1 (PR#963)

Dear R-developers

I'm not very sure whether this is really a bug and not a feature:
> is.matrix(svd(matrix(1:12,nrow=1))$v)  
[1] FALSE

In all other cases the $v component is a matrix. Also, the $u component
always seems to be a matrix as indicated in the doc.

My R-version:
> version
         _                
platform i686-pc-linux-gnu
arch     i686             
os       linux-gnu        
system   i686, linux-gnu  
status                    
major    1                
minor    2.3              
year     2001             
month    04               
day      26               
language R  

Yours
Marcel Wolbers

PS: 
- On S-Plus Version 5.1 Release 1 for Linux 2.0.31 : 1999 
  > is.matrix(svd(matrix(1:12,nrow=1))$v)
  [1] T
- Thanks a lot for the great piece of software you've written

-- 
Marcel Wolbers, LEO C14, Seminar for Statistics, ETH Zurich
CH-8092 Zurich, Switzerland; phone: (01) 632 2252	

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Marcel> Dear R-developers
    Marcel> I'm not very sure whether this is really a bug and not a
feature:

    Marcel> is.matrix(svd(matrix(1:12,nrow=1))$v)  
    Marcel> [1] FALSE

    Marcel> In all other cases the $v component is a matrix. Also, the $u
    Marcel> component always seems to be a matrix as indicated in the doc.

It's the famous `programmer bug' of not using  `` , drop = FALSE
''
when matrix subscripting in functions....

in the 2nd-to-last line of svd() :

    if(nv && nv < p) z$v <- z$v[, 1:nv, drop = FALSE]
				      ^^^^^^^^^^^^^^
				       add this.

{merci, Marcel!}


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