how to display data with a "fixed" color scale ?

Jim_Vickroy · June 3, 2010, 8:00pm

I want to generate a 2-d
figure with a (fixed) color scale that does not vary with the range of
the data being plotted.

How do I do this? Attempts to specify vimin and vmax appear to be
ignored.

The following example:

#

import numpy

data = numpy.zeros(shape=(240,240),dtype=int)

data[ 0: 80] = -1

data[ 80:160] = 0

data[160:] = 1

import matplotlib.pyplot as plot

figure = plot.figure()

ax = figure.add_subplot(111)

cax = ax.imshow(data, interpolation=‘bilinear’)

ax.set_title(‘test data with fixed colorbar’)

colorbar = figure.colorbar(cax, ticks=[-1, 0, 1])

colorbar.ax.set_yticklabels([’-1’, ‘0’, ‘1’])

plot.show()

#

produces a figure with 3
color bands (blue,green,red) and matching color bar with labels
(-1,0,1) as expected.

if the data[160:]=1 specification is deleted, in the above code, the
resulting figure has 2 color bands (blue,red) and the associated color
bar is identical to the original, but the labels are (-1,0).

What I want, in this second case, is a blue-green figure and a color
bar with labels identical to the original example.

– jv

_Jae-Joon_Lee1 · June 3, 2010, 8:10pm

Hmm,
vmin and vmax should work.

cax = ax.imshow(data, interpolation='bilinear', vmin=-1, vmax=1)

If these are still ignored, what the following line prints?

print cax.get_clim()

Regards,

-JJ

···

On Thu, Jun 3, 2010 at 4:00 PM, Jim Vickroy <Jim.Vickroy@...259...> wrote:

How do I do this? Attempts to specify vimin and vmax appear to be ignored.

Jim_Vickroy · June 3, 2010, 8:22pm

Jae-Joon Lee wrote:

How do I do this?  Attempts to specify vimin and vmax appear to be ignored.


Hmm,
vmin and vmax should work.
cax = ax.imshow(data, interpolation='bilinear', vmin=-1, vmax=1)
If these are still ignored, what the following line prints?
print cax.get_clim()
Regards,
-JJ

You are absolutely correct!
My apologies for not testing this prior to posting.

I’m really specifying vmin and vmax as parameters for pcolormesh and
the labels, for the associated color bar, vary (despite being
explicitly specified in figure.colorbar(…,ticks=…)) with the range
of the data being plotted.

I over-simplified, so I will try to produce a better version of the
actual code demonstrating the problem.

Thank-you for your quick reply.

– jv

···

<Jim.Vickroy@…259…>

Eric_Firing1 · June 3, 2010, 8:25pm

I want to generate a 2-d figure with a (fixed) color scale that does
not vary with the range of the data being plotted.

How do I do this? Attempts to specify vimin and vmax appear to be ignored.

The following example:

#<code>
import numpy
data = numpy.zeros(shape=(240,240),dtype=int)
data[ 0: 80] = -1
data[ 80:160] = 0
data[160:] = 1

import matplotlib.pyplot as plot
figure = plot.figure()
ax = figure.add_subplot(111)
cax = ax.imshow(data, interpolation='bilinear')
ax.set_title('test data with fixed colorbar')

Adding to what JJ said, note that setting the ticks on the colorbar has no effect on the norm used in color mapping. The vmin and vmax kwargs to imshow get passed to the norm, so they do set the mapping range.

colorbar = figure.colorbar(cax, ticks=[-1, 0, 1])
colorbar.ax.set_yticklabels(['-1', '0', '1'])

Please avoid setting the ticklabels directly--it is almost always unnecessary, and it is too easy to shoot yourself in the foot. If the default tick label formatting is inadequate, you can use the format kwarg in colorbar.

From the docstring:

         *ticks* [ None | list of ticks | Locator object ]
                       If None, ticks are determined automatically from the
                       input.
         *format* [ None | format string | Formatter object ]
                       If None, the
                       :class:`~matplotlib.ticker.ScalarFormatter` is used.
                       If a format string is given, e.g. '%.3f', that is
                       used. An alternative
                       :class:`~matplotlib.ticker.Formatter` object may be
                       given instead.

Eric

···

On 06/03/2010 10:00 AM, Jim Vickroy wrote:

plot.show()
#</code>

produces a figure with 3 color bands (blue,green,red) and matching color
bar with labels (-1,0,1) as expected.

if the data[160:]=1 specification is deleted, in the above code, the
resulting figure has 2 color bands (blue,red) and the associated color
bar is identical to the original, but the labels are (-1,0).

What I want, in this second case, is a blue-green figure and a color bar
with labels identical to the original example.

-- jv

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Jim_Vickroy · June 3, 2010, 8:43pm

Eric Firing wrote:


I want to generate a 2-d figure with a (fixed) color scale that does
not vary with the range of the data being plotted.
How do I do this? Attempts to specify vimin and vmax appear to be ignored.
The following example:
#<code>
import numpy
data = numpy.zeros(shape=(240,240),dtype=int)
data[ 0: 80] = -1
data[ 80:160] = 0
data[160:] = 1
import matplotlib.pyplot as plot
figure = plot.figure()
ax = figure.add_subplot(111)
cax = ax.imshow(data, interpolation='bilinear')
ax.set_title('test data with fixed colorbar')

Adding to what JJ said, note that setting the ticks on the colorbar has no effect on the norm used in color mapping. The vmin and vmax kwargs to imshow get passed to the norm, so they do set the mapping range.

colorbar = figure.colorbar(cax, ticks=[-1, 0, 1])
colorbar.ax.set_yticklabels(['-1', '0', '1'])


 Please avoid setting the ticklabels directly--it is almost always unnecessary, and it is too easy to shoot yourself in the foot. If the default tick label formatting is inadequate, you can use the format kwarg in colorbar.
From the docstring:
*ticks* [ None | list of ticks | Locator object ]
If None, ticks are determined automatically from the
input.
*format* [ None | format string | Formatter object ]
If None, the
:class:`~matplotlib.ticker.ScalarFormatter` is used.
If a format string is given, e.g. '%.3f', that is
used. An alternative
:class:`~matplotlib.ticker.Formatter` object may be
given instead.
Eric

Thanks for this advice.

In my case, the data being plotted is in the range 0-255, but the
color-bar labels are to be in the range 1-4095. So I have the
following code snippet:

colorbar = figure.colorbar(image, cax, orientation=‘vertical’,
ticks=(0, 64, 128, 192, 254))

colorbar.ax.set_yticklabels((‘1’,‘8’,‘64’,‘512’,‘4095’)) # colorbar
labels (which are to be in units of DN/sec on a log10 scale)

Is there a better way to do this?

– jv

···

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Eric_Firing1 · June 3, 2010, 9:17pm

Eric Firing wrote:

I want to generate a 2-d figure with a (fixed) color scale that does
not vary with the range of the data being plotted.

How do I do this? Attempts to specify vimin and vmax appear to be ignored.

The following example:

#<code>
import numpy
data = numpy.zeros(shape=(240,240),dtype=int)
data[ 0: 80] = -1
data[ 80:160] = 0
data[160:] = 1

import matplotlib.pyplot as plot
figure = plot.figure()
ax = figure.add_subplot(111)
cax = ax.imshow(data, interpolation='bilinear')
ax.set_title('test data with fixed colorbar')

Adding to what JJ said, note that setting the ticks on the colorbar has
no effect on the norm used in color mapping. The vmin and vmax kwargs
to imshow get passed to the norm, so they do set the mapping range.

colorbar = figure.colorbar(cax, ticks=[-1, 0, 1])
colorbar.ax.set_yticklabels(['-1', '0', '1'])

Please avoid setting the ticklabels directly--it is almost always
unnecessary, and it is too easy to shoot yourself in the foot. If the
default tick label formatting is inadequate, you can use the format
kwarg in colorbar.

  From the docstring:

          *ticks* [ None | list of ticks | Locator object ]
                        If None, ticks are determined automatically from the
                        input.
          *format* [ None | format string | Formatter object ]
                        If None, the
                        :class:`~matplotlib.ticker.ScalarFormatter` is used.
                        If a format string is given, e.g. '%.3f', that is
                        used. An alternative
                        :class:`~matplotlib.ticker.Formatter` object may be
                        given instead.

Eric

Thanks for this advice.

In my case, the data being plotted is in the range 0-255, but the
color-bar labels are to be in the range 1-4095. So I have the following
code snippet:

colorbar = figure.colorbar(image, cax, orientation='vertical', ticks=(0,
64, 128, 192, 254))
colorbar.ax.set_yticklabels(('1','8','64','512','4095')) # colorbar
labels (which are to be in units of DN/sec on a log10 scale)

Is there a better way to do this?

The advantage of using a custom formatter is that it formats actual tick values, so if you decide to use a different set of tick locations, you don't have to remember to change the labels. A formatter for a complicated case such as the above could use a dictionary, which would at least generate a KeyError if you changed a tick without adding the new location to the dictionary, or, better, it could calculate the label numbers. Suppose you have a function to do the translation:

def to_DNpersec(x):
dn = ... whatever function of x
return dn

import matplotlib as mpl

class DNpersecFormatter(mpl.ticker.Formatter):
     def __call__(self, val, pos=None):
         dn = to_DNpersec(val)
         return "%d" % round(dn)

...
colorbar = figure.colorbar(image, cax, orientation='vertical',
ticks=(0, 64, 128, 192, 254),
format=DNpersecFormatter())

Eric

···

On 06/03/2010 10:43 AM, Jim Vickroy wrote:

On 06/03/2010 10:00 AM, Jim Vickroy wrote:

-- jv

plot.show()
#</code>

produces a figure with 3 color bands (blue,green,red) and matching color
bar with labels (-1,0,1) as expected.

if the data[160:]=1 specification is deleted, in the above code, the
resulting figure has 2 color bands (blue,red) and the associated color
bar is identical to the original, but the labels are (-1,0).

What I want, in this second case, is a blue-green figure and a color bar
with labels identical to the original example.

-- jv

Jim_Vickroy · June 3, 2010, 9:33pm

OK, upon a more careful review of the code, I made a simple(-minded)
error. Specifying vmin and vmax do work (as everyone already knew).

Thanks to Jae-Joon and Eric for their quick replies. and valuable
suggestions.

– jv

Jim Vickroy wrote:

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