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From |
[email protected] (Vince Wiggins, StataCorp) |

To |
[email protected] |

Subject |
Re: RE: st: Stata 11 Announcement (statalist-digest V4 #3467) |

Date |
Sat, 27 Jun 2009 11:07:02 -0500 |

About the new factor-variables features in Stata 11 Allan Reese <[email protected]> writes, > The # notation makes sense, but it may be subtle for many users and > leaves the possibility to confuse fitting variables as continuous or > discrete. If I understand Bill's example sex#group is parsed as > i.sex#i.group simply because # was used. Hence do you *have* to use > the new c. in i.sex#c.age ... The short answer is yes, the c. is required to designate a variable in an interaction as continuous. The short reason is that most people will create more interactions with factor variables than with continuous variables. Some people will do the opposite and wish that we had jumped the other way. Those people may be able to console themselves with the fact that it requires very little typing to designate a set of variables as continuous. To create a regression of mpg on foreign and several continuous covariates, where we want to estimate the effect of each covariate separately for foreign and domestic cars, we might be tempted to type, . regress mpg i.foreign price weight trunk turn headroom foreign#c.price foreign#c.weight foreign#c.trunk foreign#c.turn foreign#c.headroom Or, if we knew a bit more about factor-variable notation, . regress mpg foreign##(c.price c.weight c.trunk c.turn c.headroom) The latter takes advantage of the factorial operator ## and parenthesis binding. Because the new factor variable syntax is an algebra, however, we don't even need to repeat the c.'s. We can also type, . regress mpg foreign##c.(price weight trunk turn headroom) -- Vince [email protected] * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

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