[GRASSLIST:4533] Re: r.mapcalc and environment variables - BINGO

[Daniel Victoria]

Thanks for all the help...
If anyone is interested I'm just about finishing a
bash/perl script to filter out clouds in AVHRR imagery
based on a paper called "Improved Cloud Detection for
Daytime AVHRR Scenes over land"

Again, many thanks

Daniel
--- Glynn Clements <glynn.clements at virgin.net> wrote:
> 
> Daniel Victoria wrote:
> 
> > Thanks for all the help, now I can output from a
> bash
> > script to r.mapcalc. But now I bumped into a new
> > problem.
> > In my formula line I have two integer constants
> that
> > are calculated during the bash script but in the
> > formula I need the result of the division on a
> > floating point form. Since I don't think there is
> a
> > way to do floating point calculations in bash I
> just
> > put on my formula line something like ($m1/$m2)
> and
> > I'd get (13/18).
> > 
> > The problem now is this. When I run r.mapcalc like
> > this
> > r.mapcalc output=blabla(13/18)blabla
> > I get  different result from
> > r.mapcalc output=blabla(0.722223)blabla
> > 
> > Is this an illigal thing in r.mapcalc (13/18)??
> > How should I go about this???
> 
> r.mapcalc will treat "13/18" as integer division,
> with a result of
> zero. You need to explicitly convert one of the
> values to
> floating-point, e.g.:
> 
> 	r.mapcalc "output=blabla($m1.0/$m2)blabla"
> or:
> 	r.mapcalc "output=blabla(float($m1)/$m2)blabla"
> 
> The first will expand to:
> 
> 	r.mapcalc "output=blabla(13.0/18)blabla"
> 
> resulting in the numerator being treated as a
> floating-point value.
> 
> The second will expand to:
> 
> 	r.mapcalc "output=blabla(float(13)/18)blabla"
> 
> resulting in the integer value being converted to a
> floating-point
> value by r.mapcalc's float() function.
> 
> -- 
> Glynn Clements <glynn.clements at virgin.net>


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