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st: RE: AW: AW: egen cut(runiform()) inconsistent across versions 10 and 11?


From   "Nick Cox" <n.j.cox@durham.ac.uk>
To   <statalist@hsphsun2.harvard.edu>
Subject   st: RE: AW: AW: egen cut(runiform()) inconsistent across versions 10 and 11?
Date   Tue, 25 Aug 2009 18:10:22 +0100

On a different note, the presumption seems to be that absolutely all glitches are documented on being fixed. This just isn't true, and you shouldn't want it to be true. For example, typos in documentation and output are typically fixed without an echo in -whatsnew-. I do see why there is an expectation that whatever was changed here might have been documented, but the principle is not absolute. 

What's especially bizarre about -egen, cut()- is that the syntax requires a varname, but the test that one was supplied is idiosyncratic. It appears to have been introduced when the function was re-written on adoption as part of official Stata. 

Nick 
n.j.cox@durham.ac.uk 

Martin Weiss

You can trace this back to the question whether 

******
sum (runiform()), meanonly
******

is legal or not (line 9 of _gcut.ado), which it seems to be in 10.1, but not 11. So the change might not be entered in the -whatsnew- under -egen, cut()- , but something else... 

Also note that both -version-s use the same 


C:\Program Files\Stata11\ado\base\_\_gcut.ado
*! version 2.0.8  01oct2004

so it is unlikely to be specific to _gcut...

László Sándor

I used the command -egen treat = cut(runiform()), group(2)- with
success on a copy of Stata 10 locally, with executables and ado files
updated to last week's builds. When I give the same command to Stata
11 on a UNIX cluster, it reports the error "invalid, variable name
expected". This is the case even if I add "version 10:" to the command
(-version 10: egen treat = cut(runiform()), group(2)-).

Why is this the case? I didn't find the change documented. And
shouldn't -version- solve this in any case?

(Of course, I can work around this, but still, what's wrong with
efficient and elegant code? And why only recently?)

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