Getting Colorbar instance of scatter plot in pandas/matplotlib

Question

How do I get the internally created colorbar instance of a plot created by pandas.DataFrame.plot?

Here is an example for generating a colored scatter plot:

import matplotlib.pyplot as plt
import pandas as pd
import numpy as np
import itertools as it

# [ (0,0), (0,1), ..., (9,9) ]
xy_positions = list( it.product( range(10), range(10) ) )

df = pd.DataFrame( xy_positions, columns=['x','y'] )

# draw 100 floats
df['score'] = np.random.random( 100 )

ax = df.plot( kind='scatter',
              x='x',
              y='y',
              c='score',
              s=500)
ax.set_xlim( [-0.5,9.5] )
ax.set_ylim( [-0.5,9.5] )

plt.show()

which gives me a figure like this:

How do I get the colorbar instance in order to manipulate it, for instance for changing the label or setting the ticks?

CT Zhu · Accepted Answer · 2015-10-20 16:36:35Z

24

pandas does not return the axis for the colorbar, therefore we have to locate it:

1st, let's get the figure instance: i.e., use plt.gcf()

In [61]:

import matplotlib.pyplot as plt
import pandas as pd
import numpy as np
import itertools as it

# [ (0,0), (0,1), ..., (9,9) ]
xy_positions = list( it.product( range(10), range(10) ) )

df = pd.DataFrame( xy_positions, columns=['x','y'] )

# draw 100 floats
df['score'] = np.random.random( 100 )

ax = df.plot( kind='scatter',
              x='x',
              y='y',
              c='score',
              s=500)
ax.set_xlim( [-0.5,9.5] )
ax.set_ylim( [-0.5,9.5] )

f = plt.gcf()

2, how many axes does this figure have?

In [62]:

f.get_axes()
Out[62]:
[<matplotlib.axes._subplots.AxesSubplot at 0x120a4d450>,
 <matplotlib.axes._subplots.AxesSubplot at 0x120ad0050>]

3, The first axes (that is, the first one created), contains the plot

In [63]:

ax
Out[63]:
<matplotlib.axes._subplots.AxesSubplot at 0x120a4d450>

4, Therefore, the second axis is the colorbar axes

In [64]:

cax = f.get_axes()[1]
#and we can modify it, i.e.:
cax.set_ylabel('test')

answered Oct 20, 2015 at 16:36

CT Zhu

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4 Comments

desiato Over a year ago

Ok, that gives me the axes where probably the colorbar is located. But how can I be sure that this is the right axes? I know in my setting that the second one in the axes list is my colorbar.

CT Zhu Over a year ago

I don't think there is a easy way to identify colorbar axis when you have a complex figure containing a lot more axis (rather than just 2 in this case). You can identify colorbar axes by checking if the axis contains a matplotlib.collections.QuadMesh as one of its children (colorbar axis will have it), which is the color gradient. However, this is a unsafe hack as matplotlib.collections.QuadMesh may also appear in other types of plot. So I think it is better to figure out the order of creation of the axes and pick the colorbar axis(axes) that way.

Mark Andersen Over a year ago

Would be nice for the answer also to specify how to set the ticks. At this time, calling cax.set_yticks(i) will generate a warning rather than set the ticks, so the answer does not cover the specific challenges called out in the question.

Halyn Betchkal Over a year ago

After trying to wrestle with the colorbar for years this is easily one of the most satisfying and straightforward approaches I've ever encountered.

matt_s · Accepted Answer · 2015-10-20 16:33:39Z

2

It's not quite the same but you could just plot using matplotlib:

import matplotlib.pyplot as plt
import pandas as pd
import numpy as np
import itertools as it

# [ (0,0), (0,1), ..., (9,9) ]
xy_positions = list( it.product( range(10), range(10) ) )

df = pd.DataFrame( xy_positions, columns=['x','y'] )

# draw 100 floats
df['score'] = np.random.random( 100 )

fig = plt.figure()
ax = fig.add_subplot(111)

s = ax.scatter(df.x, df.y, c=df.score, s=500)
cb = plt.colorbar(s)
cb.set_label('desired_label')

ax.set_xlim( [-0.5,9.5] )
ax.set_ylim( [-0.5,9.5] )

plt.show()

answered Oct 20, 2015 at 16:33

matt_s

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3 Comments

desiato Over a year ago

yes, that is also possible but is loses the beauty of using the DataFrame directly.

matt_s Over a year ago

Maybe... but is ax = df.plot( kind='scatter', x='x', y='y', c='score', s=500) really more direct than s = ax.scatter(df.x, df.y, c=df.score, s=500)? I guess I prefer this to hunting around for the correct axes ;-)

desiato Over a year ago

Probably, using matplotlib directly has more advantages, especially if you want to use and modify the axes later.

JohanC · Accepted Answer · 2024-02-24 03:02:05Z

To answer the original question: you can set colorbar=False and generate it separately. It needs a "mappable", i.e. a matplotlib object which contains colormap information. Here, it are the scatter dots stored in ax.collections[0]. With the return value of cbar = plt.colorbar(...) you get access to the colorbar and its ax.

import matplotlib.pyplot as plt
from matplotlib.ticker import MultipleLocator
import pandas as pd
import numpy as np

# [ (0,0), (0,1), ..., (9,9) ]
df = pd.DataFrame({'x': np.repeat(range(10), 10), 'y': np.tile(range(10), 10)})

# draw 100 floats
df['score'] = np.random.random(100)

ax = df.plot(kind='scatter',
             x='x',
             y='y',
             c='score',
             s=500,
             cmap='RdYlGn',
             vmin=0,
             vmax=1,
             colorbar=False)
ax.set_xlim([-0.5, 9.5])
ax.set_ylim([-0.5, 9.5])
ax.xaxis.set_major_locator(MultipleLocator(1))
ax.yaxis.set_major_locator(MultipleLocator(1))
cbar = plt.colorbar(ax.collections[0], ax=ax)
cbar.set_ticks([0, 0.5, 1])
cbar.set_ticklabels(['low', 'medium', 'high'])
cbar.ax.set_title('Score', ha='left')

plt.tight_layout()
plt.show()

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