Re: [Matplotlib-users] basemap and omerc
Evan Mason wrote: Hi, I am having some problems using the oblique mercator projection in basemap. I want to define a rectangular orthogonal grid, rotated clockwise by about 13 degrees. I want to define grid cells of size, say, about 20x20 km. The script I have so far is below. The problem is that at some point (the makegrid step) I lose the rotation, as seen in the plot. I'd appreciate any help with this, thanks, Evan from matplotlib.toolkits.basemap import Basemap M = Basemap(projection = 'omerc', \ resolution = None, \ llcrnrlon = -43.7, \ llcrnrlat = 14.7,\ urcrnrlon = -4.0,\ urcrnrlat = 41.9,\ lat_2 = 11.0,\ lat_1 = 45.5,\ lon_2 = -27.8, \ lon_1 = -19.9) dl = 2. nx = int((M.xmax - M.xmin) / dl) + 1 ny = int((M.ymax - M.ymin) / dl) + 1 lonr, latr = M.makegrid(nx, ny) plot(lonr, latr, 'c.') show() Evan: I have to admit, I'm not too familiar with the Oblique Mercator projection. What exactly should it look like? If I plot M = Basemap(projection = 'omerc', \ resolution = 'l', \ llcrnrlon = -43.7, \ llcrnrlat = 14.7,\ urcrnrlon = -4.0,\ urcrnrlat = 41.9,\ lat_2 = 11.0,\ lat_1 = 45.5,\ lon_2 = -27.8, \ lon_1 = -19.9) M.drawcoastlines() M.drawparallels(arange(10,51,10)) M.drawmeridians(arange(-50,1,10)) M.show() I see a reasonable looking map, but then I don't really know exactly what to expect. It seems that there are two ways to specify oblique mercator in proj4 1) by specifying 2 points (lon1,lat1), (lon2,lat2) along the central line 2) by specifying a central point and an azimuth that passes through the central point. Basemap uses (1), but every example on the web I've seen uses (2). It could be there are bugs in (1), and (2) would produce more reasonable results in your case. If you can give me an example of what your map *should* look like, it would help a lot. -Jeff -- Jeffrey S. Whitaker Phone : (303)497-6313 NOAA/OAR/CDC R/PSD1FAX : (303)497-6449 325 BroadwayBoulder, CO, USA 80305-3328 - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2008. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] healpix and basemap
jlu wrote: Has anyone had any luck plotting a Healpix (sky pixelization used in astronomy) map using matplotlib basemap... or any other python plotting package for that matter? Cheers, Jessica Jessica: I don't know anything about Healpix, but if you are more specific about what the problem is I may be able to help. For instance, how are the locations of the pixels in your image defined? -Jeff -- Jeffrey S. Whitaker Phone : (303)497-6313 NOAA/OAR/CDC R/PSD1FAX : (303)497-6449 325 BroadwayBoulder, CO, USA 80305-3328 - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2008. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] basemap and omerc
On Feb 12, 2008, at 8:39 PM, Evan Mason wrote: dl = 2. nx = int((M.xmax - M.xmin) / dl) + 1 ny = int((M.ymax - M.ymin) / dl) + 1 lonr, latr = M.makegrid(nx, ny) plot(lonr, latr, 'c.') show() I think you might be looking for M.plot() rather than plot()?? plot() will just overwrite the basemap in the figure (although that might be possible to fix with some axis-magic, but that is far beyond my knowledge of matplotlib). Cheers Tommy - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2008. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] basemap and omerc
Evan Mason wrote:
Thanks for the replies. The map you produced, Jeff, looks as it
should. However, I am trying to make an ocean model grid, and so I
require two 2d arrays of lon and lat, at my desired grid spacing.
This is why I try the steps:
dl = 2.
nx = int((M.xmax - M.xmin) / dl) + 1
ny = int((M.ymax - M.ymin) / dl) + 1
lonr, latr = M.makegrid(nx, ny) - it seems to be here that it loses
'memory' of omerc projection that I specified, and maybe there is a
bug here?
Evan: Why do you say it 'loses' memory of the projection? The values
look fine to me - they are just equally spaced points in map projection
coordinates that cover the map projection region. Take a look at
M = Basemap(projection = 'omerc', \
resolution = 'l', \
llcrnrlon = -43.7, \
llcrnrlat = 14.7,\
urcrnrlon = -4.0,\
urcrnrlat = 41.9,\
lat_2 = 11.0,\
lat_1 = 45.5,\
lon_2 = -27.8, \
lon_1 = -19.9)
dl = 20.
nx = int((M.xmax - M.xmin) / dl) + 1
ny = int((M.ymax - M.ymin) / dl) + 1
lonr, latr,x,y= M.makegrid(nx, ny, returnxy=True)
M.drawcoastlines()
M.scatter(x.flatten(), y.flatten(),5,marker='o')
M.drawparallels(arange(10,51,10))
M.drawmeridians(arange(-50,1,10))
show()
If you have matlab, the following lines do what I am looking for:
incx = 0.00310/2;
incy = 0.00306/2;
Xstr = -0.275;
Xend = 0.275;
Ystr = 0.17;
Yend = 0.8;
X = [Xstr:incx:Xend];
Y = [Ystr:incy:Yend];
[XX,YY]= meshgrid(X,Y);
[Lonr,Latr] = m_xy2ll(XX,YY);
m_proj('Oblique Mercator','lon',[-23.75 -28.25],'lat',[45.5
11],'direction','vertical');
plot(Lonr, Latr, 'c.')
Sorry, I don't have matlab - but it looks at first glance like it ought
to be doing the same thing.
-Jeff
-Evan
On Feb 13, 2008 5:14 AM, Jeff Whitaker [EMAIL PROTECTED]
mailto:[EMAIL PROTECTED] wrote:
Evan Mason wrote:
Hi, I am having some problems using the oblique mercator
projection in
basemap. I want to define a rectangular orthogonal grid, rotated
clockwise by about 13 degrees. I want to define grid cells of size,
say, about 20x20 km. The script I have so far is below. The
problem
is that at some point (the makegrid step) I lose the rotation,
as seen
in the plot.
I'd appreciate any help with this, thanks, Evan
from matplotlib.toolkits.basemap import Basemap
M = Basemap(projection = 'omerc', \
resolution = None, \
llcrnrlon = -43.7, \
llcrnrlat = 14.7,\
urcrnrlon = -4.0,\
urcrnrlat = 41.9,\
lat_2 = 11.0,\
lat_1 = 45.5,\
lon_2 = -27.8, \
lon_1 = -19.9)
dl = 2.
nx = int((M.xmax - M.xmin) / dl) + 1
ny = int((M.ymax - M.ymin) / dl) + 1
lonr, latr = M.makegrid(nx, ny)
plot(lonr, latr, 'c.')
show()
Evan: I have to admit, I'm not too familiar with the Oblique Mercator
projection. What exactly should it look like?
If I plot
M = Basemap(projection = 'omerc', \
resolution = 'l', \
llcrnrlon = -43.7, \
llcrnrlat = 14.7,\
urcrnrlon = -4.0,\
urcrnrlat = 41.9,\
lat_2 = 11.0,\
lat_1 = 45.5,\
lon_2 = -27.8, \
lon_1 = -19.9)
M.drawcoastlines()
M.drawparallels(arange(10,51,10))
M.drawmeridians(arange(-50,1,10))
M.show()
I see a reasonable looking map, but then I don't really know exactly
what to expect.
It seems that there are two ways to specify oblique mercator in proj4
1) by specifying 2 points (lon1,lat1), (lon2,lat2) along the
central line
2) by specifying a central point and an azimuth that passes
through the
central point.
Basemap uses (1), but every example on the web I've seen uses (2). It
could be there are bugs in (1), and (2) would produce more reasonable
results in your case. If you can give me an example of what your map
*should* look like, it would help a lot.
-Jeff
--
Jeffrey S. Whitaker Phone : (303)497-6313
NOAA/OAR/CDC R/PSD1FAX : (303)497-6449
325 BroadwayBoulder, CO, USA 80305-3328
--
Jeffrey S. Whitaker Phone : (303)497-6313
Meteorologist FAX: (303)497-6449
NOAA/OAR/PSD R/PSD1Email : [EMAIL PROTECTED]
325 BroadwayOffice : Skaggs Research Cntr 1D-124
Re: [Matplotlib-users] Random colourmap?
José Gómez-Dans wrote: Hi, Is there a simple way to come up with a random colourmap? I need to plot discrete values, and I would like that the colours do not show a trend, to easily distinguish them? It might help if you described the application. I am guessing that you have a small number of values (5? 10? 15?) and that using direct indexing with a ListedColormap will be the simplest way to get what you want. I have recently made this easier in svn, as illustrated by the second panel in examples/colorbar_only.py, but this is not yet available in a released version of mpl. A simpler use of a ListedColormap is given in examples/contourf_demo.py, in the last plot; this is the same in recent mpl releases as in svn. I read the cookbook entry on Doing your own colormap, but can't seem to bring my mind into it! Are you referring to this http://www.scipy.org/Cookbook/Matplotlib/ColormapTransformations ? Eric Cheers! Jose - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2008. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] pyinstall and matplotlib
Well, looks like nobody has an answer to this question. How'bout py2exe or other ways of creating exe files out of matplotlib projects? Has anybody been able to do that? --- [EMAIL PROTECTED] wrote: Has anybody been able to create an exe of their python applications involving matplotlib using pyinstall (ver 1.3)? I am getting a: RuntimeError: Could not find the matplotlib data files when I attempt to run the exe created. In searching the web, it appears this is an issue when others tried to use py2exe as well. Unfortunately, the few hits I saw doesn't include enough details to inspire me as to what I should be doing in my pyinstall .spec file. Does anybody has an example or information about this? Thanks, -- John Henry -- John Henry - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2008. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] basemap and omerc
Thanks for the replies. The map you produced, Jeff, looks as it should.
However, I am trying to make an ocean model grid, and so I require two 2d
arrays of lon and lat, at my desired grid spacing. This is why I try the
steps:
dl = 2.
nx = int((M.xmax - M.xmin) / dl) + 1
ny = int((M.ymax - M.ymin) / dl) + 1
lonr, latr = M.makegrid(nx, ny) - it seems to be here that it loses
'memory' of omerc projection that I specified, and maybe there is a bug
here?
If you have matlab, the following lines do what I am looking for:
incx = 0.00310/2;
incy = 0.00306/2;
Xstr = -0.275;
Xend = 0.275;
Ystr = 0.17;
Yend = 0.8;
X = [Xstr:incx:Xend];
Y = [Ystr:incy:Yend];
[XX,YY]= meshgrid(X,Y);
[Lonr,Latr] = m_xy2ll(XX,YY);
m_proj('Oblique Mercator','lon',[-23.75
-28.25],'lat',[45.511],'direction','vertical');
plot(Lonr, Latr, 'c.')
-Evan
On Feb 13, 2008 5:14 AM, Jeff Whitaker [EMAIL PROTECTED] wrote:
Evan Mason wrote:
Hi, I am having some problems using the oblique mercator projection in
basemap. I want to define a rectangular orthogonal grid, rotated
clockwise by about 13 degrees. I want to define grid cells of size,
say, about 20x20 km. The script I have so far is below. The problem
is that at some point (the makegrid step) I lose the rotation, as seen
in the plot.
I'd appreciate any help with this, thanks, Evan
from matplotlib.toolkits.basemap import Basemap
M = Basemap(projection = 'omerc', \
resolution = None, \
llcrnrlon = -43.7, \
llcrnrlat = 14.7,\
urcrnrlon = -4.0,\
urcrnrlat = 41.9,\
lat_2 = 11.0,\
lat_1 = 45.5,\
lon_2 = -27.8, \
lon_1 = -19.9)
dl = 2.
nx = int((M.xmax - M.xmin) / dl) + 1
ny = int((M.ymax - M.ymin) / dl) + 1
lonr, latr = M.makegrid(nx, ny)
plot(lonr, latr, 'c.')
show()
Evan: I have to admit, I'm not too familiar with the Oblique Mercator
projection. What exactly should it look like?
If I plot
M = Basemap(projection = 'omerc', \
resolution = 'l', \
llcrnrlon = -43.7, \
llcrnrlat = 14.7,\
urcrnrlon = -4.0,\
urcrnrlat = 41.9,\
lat_2 = 11.0,\
lat_1 = 45.5,\
lon_2 = -27.8, \
lon_1 = -19.9)
M.drawcoastlines()
M.drawparallels(arange(10,51,10))
M.drawmeridians(arange(-50,1,10))
M.show()
I see a reasonable looking map, but then I don't really know exactly
what to expect.
It seems that there are two ways to specify oblique mercator in proj4
1) by specifying 2 points (lon1,lat1), (lon2,lat2) along the central line
2) by specifying a central point and an azimuth that passes through the
central point.
Basemap uses (1), but every example on the web I've seen uses (2). It
could be there are bugs in (1), and (2) would produce more reasonable
results in your case. If you can give me an example of what your map
*should* look like, it would help a lot.
-Jeff
--
Jeffrey S. Whitaker Phone : (303)497-6313
NOAA/OAR/CDC R/PSD1FAX : (303)497-6449
325 BroadwayBoulder, CO, USA 80305-3328
-
This SF.net email is sponsored by: Microsoft
Defy all challenges. Microsoft(R) Visual Studio 2008.
http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/___
Matplotlib-users mailing list
Matplotlib-users@lists.sourceforge.net
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[Matplotlib-users] Random colourmap?
Hi, Is there a simple way to come up with a random colourmap? I need to plot discrete values, and I would like that the colours do not show a trend, to easily distinguish them? I read the cookbook entry on Doing your own colormap, but can't seem to bring my mind into it! Cheers! Jose - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2008. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/ ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
[Matplotlib-users] day-night terminator and/or solar zenith angle code ?
Hello all, Could someone offer suggestions for (preferably) python code to compute: * day/night terminator position * solar zenith angle I'm developing a basemap application that is required to hide contour data in the night regions of the earth. The projections will be Mercator and polar stereographic; presumably, it will suffice to have Mercator-based terminator map only. Thanks for any suggestions. -- jv - This SF.net email is sponsored by: Microsoft Defy all challenges. Microsoft(R) Visual Studio 2008. http://clk.atdmt.com/MRT/go/vse012070mrt/direct/01/___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] basemap and omerc
By losing the memory I mean that the grid is no longer rotated; that the
rotation I introduced through lat1, lon1, lat2, lon2 is lost. If you look
at the latitude of the two bottom corners you see that they are the same,
they should be different - for the matlab script they are different. In
other words, I want my great circle not to be the equator or a meridian,
instead I want it to be between lat1, lon1, lat2, lon2. See for example:
http://erg.usgs.gov/isb/pubs/MapProjections/projections.html#mercator
At present, basemap seems to be reverting to a standard mercator projection.
-Evan
On Feb 13, 2008 10:48 AM, Jeff Whitaker [EMAIL PROTECTED] wrote:
Evan Mason wrote:
Thanks for the replies. The map you produced, Jeff, looks as it
should. However, I am trying to make an ocean model grid, and so I
require two 2d arrays of lon and lat, at my desired grid spacing.
This is why I try the steps:
dl = 2.
nx = int((M.xmax - M.xmin) / dl) + 1
ny = int((M.ymax - M.ymin) / dl) + 1
lonr, latr = M.makegrid(nx, ny) - it seems to be here that it loses
'memory' of omerc projection that I specified, and maybe there is a
bug here?
Evan: Why do you say it 'loses' memory of the projection? The values
look fine to me - they are just equally spaced points in map projection
coordinates that cover the map projection region. Take a look at
M = Basemap(projection = 'omerc', \
resolution = 'l', \
llcrnrlon = -43.7, \
llcrnrlat = 14.7,\
urcrnrlon = -4.0,\
urcrnrlat = 41.9,\
lat_2 = 11.0,\
lat_1 = 45.5,\
lon_2 = -27.8, \
lon_1 = -19.9)
dl = 20.
nx = int((M.xmax - M.xmin) / dl) + 1
ny = int((M.ymax - M.ymin) / dl) + 1
lonr, latr,x,y= M.makegrid(nx, ny, returnxy=True)
M.drawcoastlines()
M.scatter(x.flatten(), y.flatten(),5,marker='o')
M.drawparallels(arange(10,51,10))
M.drawmeridians(arange(-50,1,10))
show()
If you have matlab, the following lines do what I am looking for:
incx = 0.00310/2;
incy = 0.00306/2;
Xstr = -0.275;
Xend = 0.275;
Ystr = 0.17;
Yend = 0.8;
X = [Xstr:incx:Xend];
Y = [Ystr:incy:Yend];
[XX,YY]= meshgrid(X,Y);
[Lonr,Latr] = m_xy2ll(XX,YY);
m_proj('Oblique Mercator','lon',[-23.75 -28.25],'lat',[45.5
11],'direction','vertical');
plot(Lonr, Latr, 'c.')
Sorry, I don't have matlab - but it looks at first glance like it ought
to be doing the same thing.
-Jeff
-Evan
On Feb 13, 2008 5:14 AM, Jeff Whitaker [EMAIL PROTECTED]
mailto:[EMAIL PROTECTED] wrote:
Evan Mason wrote:
Hi, I am having some problems using the oblique mercator
projection in
basemap. I want to define a rectangular orthogonal grid, rotated
clockwise by about 13 degrees. I want to define grid cells of
size,
say, about 20x20 km. The script I have so far is below. The
problem
is that at some point (the makegrid step) I lose the rotation,
as seen
in the plot.
I'd appreciate any help with this, thanks, Evan
from matplotlib.toolkits.basemap import Basemap
M = Basemap(projection = 'omerc', \
resolution = None, \
llcrnrlon = -43.7, \
llcrnrlat = 14.7,\
urcrnrlon = -4.0,\
urcrnrlat = 41.9,\
lat_2 = 11.0,\
lat_1 = 45.5,\
lon_2 = -27.8, \
lon_1 = -19.9)
dl = 2.
nx = int((M.xmax - M.xmin) / dl) + 1
ny = int((M.ymax - M.ymin) / dl) + 1
lonr, latr = M.makegrid(nx, ny)
plot(lonr, latr, 'c.')
show()
Evan: I have to admit, I'm not too familiar with the Oblique
Mercator
projection. What exactly should it look like?
If I plot
M = Basemap(projection = 'omerc', \
resolution = 'l', \
llcrnrlon = -43.7, \
llcrnrlat = 14.7,\
urcrnrlon = -4.0,\
urcrnrlat = 41.9,\
lat_2 = 11.0,\
lat_1 = 45.5,\
lon_2 = -27.8, \
lon_1 = -19.9)
M.drawcoastlines()
M.drawparallels(arange(10,51,10))
M.drawmeridians(arange(-50,1,10))
M.show()
I see a reasonable looking map, but then I don't really know exactly
what to expect.
It seems that there are two ways to specify oblique mercator in
proj4
1) by specifying 2 points (lon1,lat1), (lon2,lat2) along the
central line
2) by specifying a
Re: [Matplotlib-users] basemap and omerc
Evan Mason wrote: Hi Jeff Here are the corners: lon_corners = N.array([-4.09300764,-35.76003475,-43.72330207, -12.05627497]) lat_corners = N.array([41.90278813, 49.2136974, 14.7209971, 7.41008784]) The reason for the differences is that the matlab script is very fiddly, lots of trial and error to get the grid in the right place. The attraction of using basemap is it allows me to specify the corners, so that I have it right first time, that's the idea anyway. That would be great if you could turn off that rotation, maybe a keyword True/False Thanks, Evan Evan: I've changed Basemap in svn so you can set 'no_rot=True' when creating a Basemap instance for the 'omerc' projection to get what you want. If you don't feel like upgrading (since that requires upgrading matplotlib to svn head at the same time), this will work in the version you have: from matplotlib.toolkits.basemap import Basemap, pyproj from pylab import * p = pyproj.Proj(lon_2=-27.8,lon_1=-19.9,no_rot=True,proj='omerc',\ lat_2=11.0,lat_1=45.5) xmax,ymax = p(-4.093,41.9027) # upper right corner xmin,ymin = p(-43.723,14.721) # lower left corner x = linspace(xmin,xmax,35) y = linspace(ymin,ymax,35) x, y = meshgrid(x,y) lonr,latr = p(x,y, inverse=True) m = Basemap(llcrnrlon=-60,llcrnrlat=5,\ urcrnrlon=15,urcrnrlat=60,resolution='i') m.drawcoastlines() m.scatter(lonr.flatten(),latr.flatten(),5,marker='o') m.drawmeridians(arange(-60,21,10),labels=[0,0,0,1]) m.drawparallels(arange(0,61,10),labels=[1,0,0,0]) show() Let me know if this fixes it for you. -Jeff On Feb 13, 2008 12:56 PM, Jeff Whitaker [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] wrote: Evan Mason wrote: Hi Jeff By losing the memory I mean that the grid is no longer rotated; that the rotation I introduced through lat1, lon1, lat2, lon2 is lost. If you look at the latitude of the two bottom corners you see that they are the same, they should be different. In other words, I want my great circle not to be the equator or a meridian, instead I want it to be between lat1, lon1, lat2, lon2. See for example: http://erg.usgs.gov/isb/pubs/MapProjections/projections.html#mercator Attached is a png from the matlab script. Here you can see the rotation that I am looking for. The latitude of the two bottom corners is different, unlike what happens presently with my basemap script. Thanks, Evan Evan: OK, I was confused by your use of the term 'losing the memory'. Basemap didn't lose the rotation, it never had it in the first place. It looks like matlab and Basemap define the projection regions differently. They both are right, but are showing you different regions of the same projection. The difference is that proj4 (and therefore Basemap) automatically rotates the y axis to lie along true north. I think I know how to modify Basemap to display the region you want, by turning off that rotation. Can you send me the lat/lon values of the 4 corners of the region that matlab produces? -Jeff P.S. I don't know if this is relevant or not, but you appear to be giving matlab different points to define the center of the projection than you did in Basemap (the lons you gave matlab are -23.75,-28.25, the lons you give in Basemap are -27.8 and 19.9). On Feb 13, 2008 10:48 AM, Jeff Whitaker [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] wrote: Evan Mason wrote: Thanks for the replies. The map you produced, Jeff, looks as it should. However, I am trying to make an ocean model grid, and so I require two 2d arrays of lon and lat, at my desired grid spacing. This is why I try the steps: dl = 2. nx = int((M.xmax - M.xmin) / dl) + 1 ny = int((M.ymax - M.ymin) / dl) + 1 lonr, latr = M.makegrid(nx, ny) - it seems to be here that it loses 'memory' of omerc projection that I specified, and maybe there is a bug here? Evan: Why do you say it 'loses' memory of the projection? The values look fine to me - they are just equally spaced points in map projection coordinates that cover the map projection region. Take a look at M = Basemap(projection = 'omerc', \ resolution = 'l', \ llcrnrlon = -43.7, \ llcrnrlat = 14.7,\ urcrnrlon = -4.0,\ urcrnrlat = 41.9,\ lat_2 = 11.0,\
