[Matplotlib-users] setting the default markerfacecolor

2011-04-07 Thread Matt Funk
Hi,
i am farily new to matplotlib so my question might be fairly basic. I would 
like to be able to set certain default values at the beginning of my script. 
The way i did this with the other values is via changing the value stored in 
rcparams. So something like:
import matplotlib.pyplot as mpl
mpl.rcParams['lines.markersize'] = 20

But i would like to set the markerfacecolor in such a way but it is not 
included in rcParams. I would really like to avoid setting it in each 
individual plot call. 

Is there a way to change the default at the start of the script?

thanks
matt

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Re: [Matplotlib-users] setting the default markerfacecolor

2011-04-08 Thread Matt Funk
Hi Paul,

thanks for the response. Emmhh, i am not sure what's going on since what you 
are saying matches what's listed at 
http://matplotlib.sourceforge.net/users/customizing.html. The problem is that 
when i try it I get a key error, correspondingly, when i print rcParams i 
cannot find axes.color_cycle key:

So printing rcParams gives:
print mpl.rcParams
{'figure.subplot.right': 0.90002, 'mathtext.cal': 'cursive', 
'font.fantasy': ['Comic Sans MS', 'Chicago', 'Charcoal', 'ImpactWestern', 
'fantasy'], 'xtick.minor.pad': 4, 'tk.pythoninspect': False, 'image.aspect': 
'equal', 'font.cursive': ['Apple Chancery', 'Textile', 'Zapf Chancery', 
'Sand', 'cursive'], 'figure.subplot.hspace': 0.20001, 
'xtick.direction': 'out', 'axes.facecolor': 'w', 'mathtext.fontset': 'cm', 
'ytick.direction': 'out', 'svg.image_inline': True, 'lines.markersize': 10.0, 
'figure.dpi': 80, 'text.usetex': True, 'text.fontangle': 'normal', 
'patch.edgecolor': 'k', 'legend.labelspacing': 0.5, 'ps.useafm': False, 
'mathtext.bf': 'serif:bold', 'lines.solid_joinstyle': 'round', 
'font.monospace': ['Bitstream Vera Sans Mono', 'DejaVu Sans Mono', 'Andale 
Mono', 'Nimbus Mono L', 'Courier New', 'Courier', 'Fixed', 'Terminal', 
'monospace'], 'xtick.minor.size': 2, 'axes.formatter.limits': [-7, 7], 
'figure.subplot.wspace': 0.20001, 'savefig.edgecolor': 'w', 
'text.fontvariant': 'normal', 'image.cmap': 'jet', 'axes.edgecolor': 'k', 
'tk.window_focus': False, 'image.origin': 'upper', 'text.fontsize': 'medium', 
'font.serif': ['Bitstream Vera Serif', 'DejaVu Serif', 'New Century 
Schoolbook', 'Century Schoolbook L', 'Utopia', 'ITC Bookman', 'Bookman', 
'Nimbus Roman No9 L', 'Times New Roman', 'Times', 'Palatino', 'Charter', 
'serif'], 'savefig.facecolor': 'w', 'maskedarray': 'obsolete', 
'ytick.minor.size': 2, 'numerix': 'obsolete', 'font.stretch': 'normal', 
'text.dvipnghack': None, 'ytick.color': 'k', 'lines.linestyle': '-', 
'xtick.color': 'k', 'xtick.major.pad': 4, 'text.fontweight': 'normal', 
'patch.facecolor': 'b', 'figure.figsize': [8.0, 6.0], 'axes.linewidth': 1.0, 
'legend.handletextpad': 0.80004, 'mathtext.fallback_to_cm': True, 
'lines.linewidth': 1.0, 'savefig.dpi': 100, 'verbose.fileo': 'sys.stdout', 
'svg.image_noscale': False, 'docstring.hardcopy': False, 'font.size': 24.0, 
'ps.fonttype': 3, 'path.simplify': True, 'polaraxes.grid': True, 'toolbar': 
'toolbar2', 'pdf.compression': 6, 'grid.linewidth': 0.5, 'figure.autolayout': 
False, 'figure.facecolor': '0.75', 'ps.usedistiller': False, 'legend.isaxes': 
True, 'figure.edgecolor': 'w', 'mathtext.tt': 'monospace', 
'contour.negative_linestyle': 'dashed', 'image.interpolation': 'bilinear', 
'lines.markeredgewidth': 1.5, 'axes3d.grid': True, 'lines.marker': 'None', 
'legend.shadow': False, 'axes.titlesize': 24.0, 'backend': 'TkAgg', 
'xtick.major.size': 4, 'legend.fontsize': 24.0, 'lines.solid_capstyle': 
'projecting', 'mathtext.it': 'serif:italic', 'font.variant': 'normal', 
'xtick.labelsize': 'medium', 'axes.unicode_minus': True, 'ps.distiller.res': 
6000, 'pdf.fonttype': 3, 'patch.linewidth': 1.0, 'pdf.inheritcolor': False, 
'lines.dash_capstyle': 'butt', 'lines.color': 'b', 'text.latex.preview': 
False, 'figure.subplot.top': 0.90002, 'pdf.use14corefonts': False, 
'legend.markerscale': 1.0, 

[Matplotlib-users] it is possible to use basemap to create regular spaced lat/lon grids?

2011-09-06 Thread Matt Funk
Hi,
i want to interpolate irregular spaced satellite data onto a regular
spaced grid. The regular spaced grid should have cell sizes of 1km^2. Is
it possible to use basemap to create such a grid. It looked like it
includes some facilities like that, but i am not sure if they are meant
to be used by end user or more like internal fcns (the makegrid fcn for
example).

Any advice would be appreciated.

thanks
matt

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Re: [Matplotlib-users] it is possible to use basemap to create regular spaced lat/lon grids?

2011-09-06 Thread Matt Funk
Hi Aman,
thanks for your code. I am testing it right now, but i think this might
what i need.
Not sure if you know this: what is the difference between:
1) scipy.interpolate.griddata
2) matplotlib.mlab.griddata

For 2) you have specify the interpolation method and i think the calling
convention is different. Is one a wrapper for the other?

thanks
matt


On 9/6/2011 12:36 PM, Aman Thakral wrote:
> Hi Matt,
>
> Something like this?:
>
> def create_map(ax, llcrnrlon,llcrnrlat,urcrnrlon,urcrnrlat):
> m =
> Basemap(llcrnrlon=llcrnrlon,llcrnrlat=llcrnrlat,urcrnrlon=urcrnrlon,urcrnrlat=urcrnrlat,resolution='i',projection='cyl',lon_0=(urcrnrlon+llcrnrlon)/2,lat_0=(urcrnrlat+llcrnrlat)/2)
> m.drawcoastlines()
> m.drawmapboundary()
> m.drawstates(linewidth=3)
> m.fillcontinents(color='lightgrey',lake_color='white')
> m.drawcountries(linewidth=3)
> return m
>
>
> def plotMapData(ax,data):
>
> lats = []
> lons = []
> val = []
>   
> for k,v in data.iteritems():
> lats.append(float(k[0]))
> lons.append(float(k[1]))
> val.append(float(v))
>   
> value = np.array(val)
> lat = np.array(lats)
> lon = np.array(lons)
>
> llcrnlon = lon.min()-0.5
> llcrnlat = lat.min()-0.5
> urcrnlon = lon.max()+0.5
> urcrnlat = lat.max()+0.5
>
> xi = np.linspace(llcrnlon,urcrnlon,1000)
> yi = np.linspace(llcrnlat,urcrnlat,1000)
> zi = griddata(lon,lat,value,xi,yi)
>
> cmap = cm.jet
> m = create_map(ax,llcrnlon,llcrnlat,urcrnlon,urcrnlat)
> cs = ax.contour(xi,yi,zi,15,linewidth=0.5,cmap=cmap,alpha=0.5)   
> ax.contourf(xi,yi,zi,15,cmap=cmap,zorder=1000,alpha=0.5)
>
> colorscale = cm.ScalarMappable()
> colorscale.set_array(value)
> colorscale.set_cmap(cmap)
>
> colors = colorscale.to_rgba(value)
> ax.scatter(lon,lat,c=colors,zorder=1000,cmap=cmap,s=10)
> colorbar(colorscale, shrink=0.50, ax=ax,extend='both')
>
>
> On Tue, Sep 6, 2011 at 1:28 PM, Matt Funk  <mailto:matze...@gmail.com>> wrote:
>
> Hi,
> i want to interpolate irregular spaced satellite data onto a regular
> spaced grid. The regular spaced grid should have cell sizes of
> 1km^2. Is
> it possible to use basemap to create such a grid. It looked like it
> includes some facilities like that, but i am not sure if they are
> meant
> to be used by end user or more like internal fcns (the makegrid
> fcn for
> example).
>
> Any advice would be appreciated.
>
> thanks
> matt
>
> 
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> Finally, a world-class log management solution at an even better
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>


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Re: [Matplotlib-users] it is possible to use basemap to create regular spaced lat/lon grids?

2011-09-08 Thread Matt Funk
Hi,
sorry that it has taken me so long to reply. Anyway, i could be wrong,
but i don't think that the code:
xi = np.linspace(llcrnlon,urcrnlon,1000)
yi = np.linspace(llcrnlat,urcrnlat,1000)

will produce a grid which gives the lat/lon coordinates with 1km
spacing. The reason being is that the distance between 2 lons (say
-117.731659 and -91.303642) is different depending on where you are in
terms of the latitude (i.e. the extreme examples are of course the north
pole vs the equator). So the above gives a regular grid in terms of
degrees but not in terms of distance.
Anyway, but the example was still helpful in terms of getting me started
with the griddata issue. In my experience the mlab.griddate fcn did not
work as well as the scipy.griddata (but that could be a user error as
well ... ). Not sure why though. It might be the size of my source data
and the destination grid. I had to upgrade to the 64-bit python to be
able to access enough memory.

thanks
matt



On 9/6/2011 12:36 PM, Aman Thakral wrote:
> Hi Matt,
>
> Something like this?:
>
> def create_map(ax, llcrnrlon,llcrnrlat,urcrnrlon,urcrnrlat):
> m =
> Basemap(llcrnrlon=llcrnrlon,llcrnrlat=llcrnrlat,urcrnrlon=urcrnrlon,urcrnrlat=urcrnrlat,resolution='i',projection='cyl',lon_0=(urcrnrlon+llcrnrlon)/2,lat_0=(urcrnrlat+llcrnrlat)/2)
> m.drawcoastlines()
> m.drawmapboundary()
> m.drawstates(linewidth=3)
> m.fillcontinents(color='lightgrey',lake_color='white')
> m.drawcountries(linewidth=3)
> return m
>
>
> def plotMapData(ax,data):
>
> lats = []
> lons = []
> val = []
>   
> for k,v in data.iteritems():
> lats.append(float(k[0]))
> lons.append(float(k[1]))
> val.append(float(v))
>   
> value = np.array(val)
> lat = np.array(lats)
> lon = np.array(lons)
>
> llcrnlon = lon.min()-0.5
> llcrnlat = lat.min()-0.5
> urcrnlon = lon.max()+0.5
> urcrnlat = lat.max()+0.5
>
> xi = np.linspace(llcrnlon,urcrnlon,1000)
> yi = np.linspace(llcrnlat,urcrnlat,1000)
> zi = griddata(lon,lat,value,xi,yi)
>
> cmap = cm.jet
> m = create_map(ax,llcrnlon,llcrnlat,urcrnlon,urcrnlat)
> cs = ax.contour(xi,yi,zi,15,linewidth=0.5,cmap=cmap,alpha=0.5)   
> ax.contourf(xi,yi,zi,15,cmap=cmap,zorder=1000,alpha=0.5)
>
> colorscale = cm.ScalarMappable()
> colorscale.set_array(value)
> colorscale.set_cmap(cmap)
>
> colors = colorscale.to_rgba(value)
> ax.scatter(lon,lat,c=colors,zorder=1000,cmap=cmap,s=10)
> colorbar(colorscale, shrink=0.50, ax=ax,extend='both')
>
>
> On Tue, Sep 6, 2011 at 1:28 PM, Matt Funk  <mailto:matze...@gmail.com>> wrote:
>
> Hi,
> i want to interpolate irregular spaced satellite data onto a regular
> spaced grid. The regular spaced grid should have cell sizes of
> 1km^2. Is
> it possible to use basemap to create such a grid. It looked like it
> includes some facilities like that, but i am not sure if they are
> meant
> to be used by end user or more like internal fcns (the makegrid
> fcn for
> example).
>
> Any advice would be appreciated.
>
> thanks
> matt
>
> 
> --
> Special Offer -- Download ArcSight Logger for FREE!
> Finally, a world-class log management solution at an even better
> price-free! And you'll get a free "Love Thy Logs" t-shirt when you
> download Logger. Secure your free ArcSight Logger TODAY!
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>


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Re: [Matplotlib-users] it is possible to use basemap to create regular spaced lat/lon grids?

2011-09-09 Thread Matt Funk
On 9/9/2011 6:42 AM, Scott Sinclair wrote:
> On 8 September 2011 19:20, Matt Funk  wrote:
>> Hi,
>> sorry that it has taken me so long to reply. Anyway, i could be wrong, but i
>> don't think that the code:
>> xi = np.linspace(llcrnlon,urcrnlon,1000)
>> yi = np.linspace(llcrnlat,urcrnlat,1000)
>>
>> will produce a grid which gives the lat/lon coordinates with 1km spacing.
>> The reason being is that the distance between 2 lons (say -117.731659 and
>> -91.303642) is different depending on where you are in terms of the latitude
>> (i.e. the extreme examples are of course the north pole vs the equator). So
>> the above gives a regular grid in terms of degrees but not in terms of
>> distance.
> Yes, that's correct. You'll need to project your original data
> locations into a cartesian co-ordinate system before interpolating
> their values onto a regular grid in that co-ordinate system using
> griddata et al.
>
> You might like to use pyproj (included with the basemap toolkit) to
> help you project from lat/lon to your chosen co-ordinate system..
I have been using gdal for many of my geographic needs. Is there an
advantage/disadvantage using pyproj vs capabilities found in gdal (from
what i understand both are based on PROJ.4)? Can you comment on this?
Also, i was thinking of projecting things to UTM for interpolation
purposes. Is there any apparent reason this is a bad idea vs a different
projected coordinate system?

matt

>
> Cheers,
> Scott
>
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