Andreas,
Perhaps you would be better off making your own colormap considering that
your data is not symmetric around zero. You could do something like the
following:
--------------------------------------
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.colors as plc
data = np.random.randn(12,72)
data = data*5. - 5.
zero = -1*data.min()/(data.max() - data.min())
cdict = {'red': [(0.0, 1.0, 1.0),
(zero, 1.0, 1.0),
(1.0, 0.0, 0.0)],
'green': [(0.0, 0.0, 0.0),
(zero, 1.0, 1.0),
(1.0, 0.0, 0.0)],
'blue': [(0.0, 0.0, 0.0),
(zero, 1.0, 1.0),
(1.0, 1.0, 0.0)],
}
cmap = plc.LinearSegmentedColormap('cmap', cdict, N=1000)
mappable = plt.cm.ScalarMappable(cmap=cmap)
mappable.set_array(data)
fig = plt.figure()
plt.pcolormesh(data, cmap=cmap)
fig.colorbar(mappable)
plt.show()
--------------------------------------
Of course, the zero calculation assumes that zero actually exists in your
data set. That can be fixed with a simple if...else statement if you want
this to be more robust.
You can get rid of a few lines from this code if you don't want to set a
ScalarMappable object. However, I was doing some stuff with line
collections where I wanted some colors to have an alpha associated with
them, and I found that I was losing that info with the simpler pyplot
interface for colorbar generation. So I left in the slightly more complex
code for reference. (This might be changed in newer versions of mpl, I just
haven't needed to rework my code to check.)
Ryan
On Fri, Oct 12, 2012 at 4:17 AM, Andreas Hilboll <li...@hilboll.de> wrote:
> Hi,
>
> I have some data I want to plot using pcolormesh. It's 2d climatological
> data, see the attached plot. My data is in a range from -7 to +0.6. I want
> to be 0.0 to be clearly visible, while at the same time, the color range
> should show the full dynamic of the values. I played with the bwr and
> seismic color maps, centering on zero, so that white is 0.0. However, I'm
> not too happy with the dynamic range I get in the negative.
>
> Any suggestions are very welcome ...
>
> Cheers, Andreas.
>
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