On Mar 28, 2013, at 14:36 PM, Eric Firing wrote:
> The problem is that it would be entirely redundant, given the vmin and vmax
> kwargs that have been there for a long time.
Fair enough - I just often play with symmetric axis limits (i.e. [-1.,1.]) and
so its helpful to be able to specify as
On 2013/03/28 11:04 AM, Jody Klymak wrote:
>
> On Mar 28, 2013, at 11:51 AM, Eric Firing wrote:
>
>> Pay attention only to the listed keyword arguments above that table.
>
> Sounds good
>
> OTOH a clim argument for pcolormesh would be nice, as I often end up calling
> "clim" after I've made
On Mar 28, 2013, at 11:51 AM, Eric Firing wrote:
> Pay attention only to the listed keyword arguments above that table.
Sounds good
OTOH a clim argument for pcolormesh would be nice, as I often end up calling
"clim" after I've made a contour overtop of my pcolor, and then call clim,
whi
On 2013/03/28 8:23 AM, Jody Klymak wrote:
> Hi Eric,
>
> The docs seem to indicate "clim" is an acceptable kwarg, hence my confusion...
Jody,
You are right, that chunk of the docs is completely fouled up with
respect to kwargs. Thanks for pointing it out. The whole table of
supposed QuadMesh
Hi Eric,
The docs seem to indicate "clim" is an acceptable kwarg, hence my confusion...
http://matplotlib.org/api/pyplot_api.html?highlight=pcolormesh#matplotlib.pyplot.pcolormesh
Thanks, Jody
On Mar 28, 2013, at 11:12 AM, Eric Firing wrote:
> On 2013/03/28 7:56 AM, Jody Klymak wrote:
>>
>
On 2013/03/28 7:56 AM, Jody Klymak wrote:
>
> Hi all,
>
> In 1.2.0:
>
> pcolormesh(x,z,U,rasterized='True',cmap=cm.RdBu_r,clim=(-1.,1.))
Jody,
There is no clim kwarg, only a clim pyplot function. You can do this,
though:
pcolormesh(..., vmin=-1, vmax=1)
Eric
> #clim((-1.,1.))
>
> Doesn't se
Hi all,
In 1.2.0:
pcolormesh(x,z,U,rasterized='True',cmap=cm.RdBu_r,clim=(-1.,1.))
#clim((-1.,1.))
Doesn't seem to work, where as
pcolormesh(x,z,U,rasterized='True',cmap=cm.RdBu_r,clim=(-1.,1.))
clim((-1.,1.))
does work. Is this a bug or am I misunderstanding "clim" in the context of
pcolo
On 2013-03-28, at 7:29 AM, KURT PETERS wrote:
> If I'm using this SGP4 library: https://pypi.python.org/pypi/sgp4/
> which provides x,y,z of a satellite with respect to the center of
> the Earth, what do you think would be the best way to calculate the
> distance from the satellite to a lat/l
There is no documentation supplied for the first call. Should I file
an issue for this on github?
On Wed, Mar 27, 2013 at 7:10 AM, Benjamin Root wrote:
>
>
> On Wed, Mar 27, 2013 at 1:03 AM, Gökhan Sever wrote:
>>
>> Hello,
>>
>> Aren't these two log scaling calls supposed to be performing the s
Hi,
I'm cross-posting a question I asked on SO:
http://stackoverflow.com/q/15681890/152439
How can I vertically align text in matplotlib? My situation is as follows:
I'm using `ImageGrid`, to plot a row of five images:
grid = ImageGrid(fig,
rect=(0.06, 0., 0.885, 1.
matplotlib does not support the `\begin{array}` construct. You can see
what is supported here:
http://matplotlib.org/users/mathtext.html
If you need something like that in Sphinx, there are a number of other
math plugins here:
http://sphinx-doc.org/ext/math.html
Mike
On 03/28/2013 02:45 A
If I'm using this SGP4 library: https://pypi.python.org/pypi/sgp4/which
provides x,y,z of a satellite with respect to the center of the Earth, what do
you think would be the best way to calculate the distance from the satellite to
a lat/long point on the Earth using MATPLOT library basemaps? I s
12 matches
Mail list logo
