Matplotlib Scatterplot / Color as function of third variable / Symmetric Log Colorscale

Question

I have a scatter plot with the points shaded according to a third variable. I want to use a symmetric logarithmic scale for my colormap as described in the api: SymLogNorm

Unfortunately I get the following error:

TypeError: array cannot be safely cast to required type

Here a mini example. I'm using matplotlib 1.3.0.

# loading modules
import matplotlib as mpl
import matplotlib.pyplot as plt

# defining variables
x=[0,1,2,3]
y=[0,1,2,3]
c=[-1000,-100,100,1000]

# making scatterplot

plt.scatter(x, y, c=c, norm=mpl.colors.SymLogNorm(linthresh=10))

Without the symmetric logarithmic colormap the plot works fine.

plt.scatter(x, y, c=c)

see here

Thank you very much for your help.

What version of matplotlib are you using? Your "broken" example works for me with 1.3.0. — Benjamin Bannier
– Benjamin Bannier, Commented Oct 8, 2013 at 9:55
I'm using 1.3.0. So I don't understand why it is not working. — Chris
– Chris, Commented Oct 8, 2013 at 11:41

Community · Accepted Answer · 2020-06-20 09:12:55Z

2

The documentation for SymLogNorm is not particularly clear, as a result I am not confident everything I say in this answer is correct. It seems the vmin and vmax arguments should be used to determine the range of data your consider e.g:

# loading modules
import matplotlib as mpl
import matplotlib.pyplot as plt

# defining variables
x=[0,1,2,3]
y=[0,1,2,3]
c=[-1000,-100,100,1000]

# making scatterplot
plt.scatter(x, y, c=c, s=100, norm=mpl.colors.SymLogNorm(linthresh=10, vmin=-1e3, vmax=1e3))
plt.colorbar(ticks=c)

enter image description here

The colorbar ticks are then not going to know that it is log scaled but I think this is the effect you were aiming for.

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Oct 8, 2013 at 12:41

Greg

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1 Comment

Chris Over a year ago

Thanks a lot for the help! Actually the arguments vmin and vmax are needed to make SymLogNorm work. It should be written somewhere in the documentation.

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