[Rd] v matrix of svd(X) loses dimensions if nrow(X)==1 (PR#963)

Martin Maechler Martin Maechler <maechler@stat.math.ethz.ch>
Fri, 1 Jun 2001 14:39:10 +0200

    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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