[GRASS-dev] Re: [GRASS-user] Get current location projection in a Python Script

[Nikos Alexandris]

Nikos Alexandris wrote:
> > > > >>     # faster/ easier way: use of the "grass.region()" function
> > > > >>     rows = int(grass.region()['rows'])
> > > > >>     cols = int(grass.region()['cols'])
> > > > > 
> > > > > In 7.0, these fields are already integers; r40555 should be
> > > > > backported.
> > > > 
> > > > AFAIK it has been backported some time ago.
> > > > 
> > > > http://trac.osgeo.org/grass/browser/grass/branches/releasebranch_6_4/
> > > > lib/ python/core.py#L503
> > > 
> > > Okay; change that to "I should 'svn update' the other branches more
> > > often", then ;)
> > 
> > This means it should be like:
> > 
> > rows = grass.region()['rows']
> > cols = grass.region()['cols']
> > 
> > Right?

Glynn:
> Right; although you should store the result of grass.region() to avoid
> redundant g.region commands, i.e.:
> 
> 	rgn = grass.region()
> 	rows = rgn['rows']
> 	cols = rgn['cols']

Thank you Glynn, Nikos