[Rd] Inverting a square... (PR#13762)

[rvaradhan at jhmi.edu]

Yes=2C Peter=2E  I did look at it=2C but not carefully enought to catch =
that=2E

Thanks=2C
Ravi=2E

=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=
=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=
=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F

Ravi Varadhan=2C Ph=2ED=2E
Assistant Professor=2C
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University

Ph=2E (410) 502-2619
email=3A rvaradhan=40jhmi=2Eedu


----- Original Message -----
From=3A Peter Dalgaard =3CP=2EDalgaard=40biostat=2Eku=2Edk=3E
Date=3A Thursday=2C June 18=2C 2009 9=3A15 am
Subject=3A Re=3A =5BRd=5D Inverting a square=2E=2E=2E (PR=2313762)
To=3A rvaradhan=40jhmi=2Eedu
Cc=3A R-bugs=40r-project=2Eorg


=3E Refiling this=2E  The actual fix was slightly more complicated=2E Wi=
ll soon
=3E  be committed to R-Patched (aka 2=2E9=2E1 beta)=2E
=3E  =

=3E  	-p
=3E  =

=3E  rvaradhan=40jhmi=2Eedu wrote=3A
=3E  =3E Full=5FName=3A Ravi Varadhan
=3E  =3E Version=3A 2=2E8=2E1
=3E  =3E OS=3A Windows
=3E  =3E Submission from=3A (NULL) (162=2E129=2E251=2E19)
=3E  =3E =

=3E  =3E =

=3E  =3E Inverting a matrix with solve()=2C but using LAPACK=3DTRUE=2C g=
ives erroneous
=3E  =3E results=3A
=3E  =

=3E  Thanks=2C but there seems to be a much easier fix=2E
=3E  =

=3E  Inside coef=2Eqr=2C we have
=3E  =

=3E  coef=5Bqr=24pivot=2C =5D =3C-
=3E  =2ECall(=22qr=5Fcoef=5Freal=22=2C qr=2C y=2C PACKAGE =3D =22base=22=
)=5Bseq=5Flen(p)=5D
=3E  =

=3E  which should be =5Bseq=5Flen(p)=2C=5D
=3E  =

=3E  (otherwise=2C in the matrix case=2C the RHS will recycle only the 1=
st p
=3E  elements=2C i=2Ee=2E=2C the 1st column)=2E
=3E  =

=3E  =3E =

=3E  =3E Here is an example=3A
=3E  =3E =

=3E  =3E      hilbert =3C- function(n) =7B i =3C- 1=3An=3B 1 / outer(i -=
 1=2C i=2C =22+=22) =7D
=3E  =3E 	h5 =3C- hilbert(5)
=3E  =3E 	hinv1 =3C- solve(qr(h5))
=3E  =3E 	hinv2 =3C- solve(qr(h5=2C LAPACK=3DTRUE))	=

=3E  =3E 	all=2Eequal(hinv1=2C hinv2)  =23 They are not equal
=3E  =3E =

=3E  =3E Here is a function that I wrote to correct this problem=3A
=3E  =3E =

=3E  =3E 	solve=2Elapack =3C- function(A=2C LAPACK=3DTRUE=2C tol=3D1=2Ee=
-07) =7B
=3E  =3E 	=23 A function to invert a matrix using =22LAPACK=22 or =22LIN=
PACK=22
=3E  =3E         if (nrow(A) !=3D ncol(A)) stop(=22Matrix muxt be square=
=22)
=3E  =3E         qrA =3C- qr(A=2C LAPACK=3DLAPACK=2C tol=3Dtol)
=3E  =3E         if (LAPACK) =7B
=3E  =3E 	apply(diag(1=2C ncol(A))=2C 2=2C function(x) solve(qrA=2C x)) =

=3E  =3E         =7D else  solve(qrA)
=3E  =3E 	=7D
=3E  =3E =

=3E  =3E         hinv3 =3C- solve=2Elapack(h5)
=3E  =3E 	all=2Eequal(hinv1=2C hinv3)  =23 Now=2C they are equal
=3E  =3E =

=3E  =3E =5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=
=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=
=5F
=3E  =3E R-devel=40r-project=2Eorg mailing list
=3E  =3E =

=3E  =

=3E  =

=3E  -- =

=3E     O=5F=5F  ---- Peter Dalgaard             =D8ster Farimagsgade 5=2C=
 Entr=2EB
=3E    c/ /=27=5F --- Dept=2E of Biostatistics     PO Box 2099=2C 1014 C=
ph=2E K
=3E   (*) =5C(*) -- University of Copenhagen   Denmark      Ph=3A  (+45)=
 35327918
=3E  =7E=7E=7E=7E=7E=7E=7E=7E=7E=7E - (p=2Edalgaard=40biostat=2Eku=2Edk)=
              FAX=3A (+45) 35327907
=3E  =

=3E