Hello, I have the following piece of code in my script: loop t=(jsample+2)..tend loop foreach i ylist $i[t]=$ihat[t] endloop endloop where jsample and tend are scalars and ylist is a list with a few time series, say "hires" and "quits". I had generated hireshat and quitshat with the forecasts for the end of the sample. The idea of the loop is to included the forecasts into the original time series "hires" and "quits". It all worked fine when I last run the code (about 3 months ago). Now, I get an error message saying that $i[t] = $ihat[t] is a syntax error. Has anything changed in how the script understands $i[t] ?? Thanks! Paulo

On Mon, 18 May 2015, Paulo Grahl wrote: > Hello, > I have the following piece of code in my script: > > loop t=(jsample+2)..tend > loop foreach i ylist > $i[t]=$ihat[t] > endloop > endloop > > where jsample and tend are scalars and ylist is a list with a few time > series, say "hires" and "quits". > I had generated hireshat and quitshat with the forecasts for the end of the > sample. The idea of the loop is to included the forecasts into the original > time series "hires" and "quits". > It all worked fine when I last run the code (about 3 months ago). Now, I > get an error message saying that > $i[t] = $ihat[t] is a syntax error. > > Has anything changed in how the script understands $i[t] ?? Yes, but inadvertantly. Sorry we didn't catch that before the 1.10.1 release. It's now fixed in CVS and the snapshots for Windows and Mac. But there's a simple workaround for use with the 1.10.1 release: preface the $i expression with "genr": genr $i[t] = whatever