Hello Martin,
On Mon, 31 Mar 2014 23:08:52 +1100, Martin Landa
wrote:
Hi,
2014-03-31 6:56 GMT+02:00 Terry Duell :
Hello All,
Running grass-6.4.3-5.fc20.x86_64 (Fedora 20)
If I go to "help" (GRASS GIS Layer Manager window) and choose "GRASS
Help"
or "GUI help', both fail, as follows...
I answered your question at
http://gis.stackexchange.com/questions/90784/why-my-python-script-of-r-watershed-has-no-output/91399#91399
and added an example with full module call:
gcore.run_command('r.watershed', elevation='elev_lid792_1m',
drainage='elev_lid_drainage', half='elev_lid_half_basin'
On Mon, Mar 31, 2014 at 1:10 PM, Heinz Chen wrote:
> Sorry for asking questions to just helping me debug my python script, but
> I really can't find where to be edited, and here's my script to generate
> new mapset,
>
> import osimport sys
> #set up GRASS environment variables
> sys.path.append(
Sorry for asking questions to just helping me debug my python script, but I
really can't find where to be edited, and here's my script to generate new
mapset,
import osimport sys
#set up GRASS environment variables
sys.path.append(os.path.join(os.environ['GISBASE'], 'etc',
'python'))import grass.s
As title, I want to use r.watershed module, and here's my python script in
pyscripter,
import osimport sys
#set up GRASS environment variables
sys.path.append(os.path.join(os.environ['GISBASE'], 'etc',
'python'))import grass.script as gimport grass.script.setup as gsetup
gisbase = os.environ['GISB
Hi Veronica,
you can use t.rast.series for this kind of task. Use a SQL expression
to select only specific months of a year. Please try this code
(untested, may contain errors):
# January averages
t.rast.series input=monthly_aggregates \
output=jan_average method=average \
where="start_tim
Hi all,
I have this MODIS time series of 8-day composite of chlorophyll
concentration from 2003 to 2013. I've already aggregated it by month
(average), so from 506 images I now have 132 (12 x year, 11 years)
but what I need doing now is to aggregate months over the years, what
i want are 12 summa
Thank you for your answer. I read on the pages you mention :
"If the point falls exactly upon a grid line, the exact result will be
determined by the direction of any rounding error. One consequence of this is
that downsampling by a factor which is an even integer will always sample
exactly on
On 31/03/14 15:33, BLANDENIER Lucien wrote:
Hello,
I'm wondering how does r.mapcalc compute the value of a new pixcel if the
resolution have been reduced before (i.e. the pixcel is larger). Is it an
average ?
I did a simple test with a 10*10 pixcel raster. I reduced the resolution to 5*5 and
Hello,
I'm wondering how does r.mapcalc compute the value of a new pixcel if the
resolution have been reduced before (i.e. the pixcel is larger). Is it an
average ?
I did a simple test with a 10*10 pixcel raster. I reduced the resolution to 5*5
and made a new raster with r.mapcalc "newrast=old
Hello,
Am 31.03.2014 12:38, schrieb Markus Neteler:
> Yes, I think so: you should just be aware that a "lot" of increase
> needs to be supported by information, otherwise it is pure cosmetics.
> see
> http://grasswiki.osgeo.org/wiki/Interpolation#Resampling_of_raster_maps_to_finer_resolution
Tha
Hi,
2014-03-31 6:56 GMT+02:00 Terry Duell :
> Hello All,
> Running grass-6.4.3-5.fc20.x86_64 (Fedora 20)
> If I go to "help" (GRASS GIS Layer Manager window) and choose "GRASS Help"
> or "GUI help', both fail, as follows...
>
> (Mon Mar 31 15:26:07 2014)
> g.manual entry=wxGUI
> ERROR: No HTML man
On Mon, Mar 31, 2014 at 11:16 AM, Christian Willmes
wrote:
> Hello,
>
> I want to interpolate a regular 1° x 1° temperature (°K) data vector
> point grid to a higher resolution, say a regular 0.1° x 0.1° raster grid.
>
> I assume that the temperature data is continous and the data between the
> ce
Hello,
I want to interpolate a regular 1° x 1° temperature (°K) data vector
point grid to a higher resolution, say a regular 0.1° x 0.1° raster grid.
I assume that the temperature data is continous and the data between the
cells should be interpolated as if they were on a straight gradient
line/p
Hi and thanks for your hints and sorry for answering that late,
i didn't run into the memory-problem with a value of 10% until now, but
that way i can't guarantee that it works on even higher pixel numbers.
The problem is caused by the exp() function, maybe it isn't even wrong, but
now i can filte
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