matplotlib colorbar for scatter

PythonColorsMatplotlib

Python Problem Overview


I'm working with data that has the data has 3 plotting parameters: x,y,c. How do you create a custom color value for a scatter plot?

Extending this example I'm trying to do:

import matplotlib
import matplotlib.pyplot as plt
cm = matplotlib.cm.get_cmap('RdYlBu')
colors=[cm(1.*i/20) for i in range(20)]
xy = range(20)
plt.subplot(111)
colorlist=[colors[x/2] for x in xy] #actually some other non-linear relationship
plt.scatter(xy, xy, c=colorlist, s=35, vmin=0, vmax=20)
plt.colorbar()
plt.show()

but the result is TypeError: You must first set_array for mappable

Python Solutions


Solution 1 - Python

From the matplotlib docs on scatter 1: > cmap is only used if c is an array of floats

So colorlist needs to be a list of floats rather than a list of tuples as you have it now. plt.colorbar() wants a mappable object, like the CircleCollection that plt.scatter() returns. vmin and vmax can then control the limits of your colorbar. Things outside vmin/vmax get the colors of the endpoints.

How does this work for you?

import matplotlib.pyplot as plt
cm = plt.cm.get_cmap('RdYlBu')
xy = range(20)
z = xy
sc = plt.scatter(xy, xy, c=z, vmin=0, vmax=20, s=35, cmap=cm)
plt.colorbar(sc)
plt.show()

Image Example

Solution 2 - Python

Here is the OOP way of adding a colorbar:

fig, ax = plt.subplots()
im = ax.scatter(x, y, c=c)
fig.colorbar(im, ax=ax)

Solution 3 - Python

If you're looking to scatter by two variables and color by the third, Altair can be a great choice.

Creating the dataset
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd

df = pd.DataFrame(40*np.random.randn(10, 3), columns=['A', 'B','C'])

Altair plot

from altair import *
Chart(df).mark_circle().encode(x='A',y='B', color='C').configure_cell(width=200, height=150)
Plot

enter image description here

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionAdam GreenhallView Question on Stackoverflow
Solution 1 - PythonmattView Answer on Stackoverflow
Solution 2 - PythonneuriteView Answer on Stackoverflow
Solution 3 - PythonNipun BatraView Answer on Stackoverflow