Re: CRIMCOORD transformation in QUEST

2002-02-27 Thread Paul Thompson
That is either a sloppiness in writing or reliance on the relationship between eigen decomposition and SVD. SSM - square symmetric matrix AM - arbitrary matrix In ED, SSM = Q E Q' In SVD, AM = P D Q' SSM = AM' AM = Q D P' P D Q' = Q D D Q' = Q E Q', if E = D D I haven't checked

CRIMCOORD transformation in QUEST

2002-02-26 Thread David Chang
Hi, thank you for reading this message. I have the following problems in getting the "correct" CRIMCOORD transformation of categorical variables in QUEST decision tree algorithm. Your help will be greatly appreciated. Q1: In Loh & Shih's paper (Split Selection Models for Classification Trees, Sta