[GRASS-user] Getting statistical values (r.univar) via python

[Johannes Radinger]

Hi Madi,

thank you, that's just what I thought of...
anyway do you know does r.univar consider MASK if it is existing?

/johannes

-------- Original-Nachricht --------
> Datum: Fri, 25 May 2012 13:38:05 +0200
> Von: Margherita Di Leo <diregola at gmail.com>
> An: Johannes Radinger <JRadinger at gmx.at>
> CC: grass user list <grass-user at lists.osgeo.org>
> Betreff: Re: [GRASS-user] Getting statistical values (r.univar) via python

> Hi Johannes,
> 
> On Fri, May 25, 2012 at 1:12 PM, Johannes Radinger <JRadinger at gmx.at>
> wrote:
> 
> > Hi,
> >
> > I am trying to get statistical values for a raster map in a python
> script.
> > What I am doing so far is using r.univar (e.g. for the nc-dataset):
> >
> > grass.read_command("r.univar", map ="elevation")
> >
> > but how to get e.g the mean value as float? I thought of two
> > options:
> >
> > 1) either indexing like
> > float(grass.read_command("r.univar", map ="elevation")[x:y])
> >
> > 2) or splitting the string into lines and then at the ":"
> >
> >
> You can use the following:
> 
> import sys
> import os
> import grass.script as grass
> 
> univar = grass.read_command('r.univar', map='elevation')
> 
> print univar #so that you can see how the output looks like
> 
> The output is:
> total null and non-null cells: 2025000
> total null cells: 0
> 
> Of the non-null cells:
> ----------------------
> n: 2025000
> minimum: 55.5788
> maximum: 156.33
> range: 100.751
> mean: 110.375
> mean of absolute values: 110.375
> standard deviation: 20.3153
> variance: 412.712
> variation coefficient: 18.4057 %
> sum: 223510266.5581016541
> 
> Now you want to split every line of the output with .split('\n'), you are
> interested in the line [10], then you split against split(':') and take
> the
> argument [1].
> 
> mean = float(univar.split('\n')[10].split(':')[1])
> 
> Hope this helps,
> 
> madi
> -- 
> Ing. Margherita Di Leo, Ph.D.

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