CODEWITHSUNDEEP
pythonmatplotlibplotseaborn
Dani Carrillo
Dani Carrillo

Reputation: 153

Barplot colored according a colormap?

First of all, I'm pretty new to colors in Matplotlib or Seaborn. My purpose is to create a barplot with bars coloured according to a custom palette. Something like this, but with my custom palette (see below, a palette with red, orange, green and blue):

barplot

I have created my custom sequential palette using the LinearSegmentedColormap method, but I'm not able to use it in a simple plt.barplot(). Sure it's not difficult, but I can't see the way. I created the palette using the function below, got from this thread: Create own colormap using matplotlib and plot color scale

def make_colormap(seq):
"""Return a LinearSegmentedColormap
seq: a sequence of floats and RGB-tuples. The floats should be increasing
and in the interval (0,1).
"""
seq = [(None,) * 3, 0.0] + list(seq) + [1.0, (None,) * 3]
cdict = {'red': [], 'green': [], 'blue': []}
for i, item in enumerate(seq):
    if isinstance(item, float):
        r1, g1, b1 = seq[i - 1]
        r2, g2, b2 = seq[i + 1]
        cdict['red'].append([item, r1, r2])
        cdict['green'].append([item, g1, g2])
        cdict['blue'].append([item, b1, b2])

return mcolors.LinearSegmentedColormap('CustomMap', cdict)

#main#
c = mcolors.ColorConverter().to_rgb
rvb = make_colormap(
[c('red'), 0.125, c('red'), c('orange'), 0.25, c('orange'),c('green'),0.5, c('green'),0.7, c('green'), c('blue'), 0.75, c('blue')])

N = 1000
array_dg = np.random.uniform(0, 10, size=(N, 2))
colors = np.random.uniform(0, 5, size=(N,))
plt.scatter(array_dg[:, 0], array_dg[:, 1], c=colors, cmap=rvb)
plt.colorbar()
plt.show()

That returns this plot:

plot palette ok

As far as I can understand, I can't use a colormap (object type from LinearSegmentedColormap()? ) for barplots, but colormap is the unique way I have achieved a custom sequential palette.

In summary, I want to apply the colormap of the second plot (the scatterplot) to the first plot (the barplot). For now I can't do it because the barplot() function has not an argument that accepts a LinearSegmentedColormap object type.

I'm probably making it harder than it really is, so I would appreciate any cleaner or more correct way.

Upvotes: 15

Views: 46536

Answers (2)

ImportanceOfBeingErnest
ImportanceOfBeingErnest

Reputation: 339230

To obtain a barplot with the bars colored according to a colormap you can use the color argument of bar(x,y, color=colors), where colors is a list of length number of bars, containing all the colors. I.e. the ith entry in that list is the color for the ith bar.
In order to create this list from the colormap, you need to call the colormap with the respective value.

enter image description here

import matplotlib.pyplot as plt
import matplotlib.colors as mcolors
import numpy as np

clist = [(0, "red"), (0.125, "red"), (0.25, "orange"), (0.5, "green"), 
         (0.7, "green"), (0.75, "blue"), (1, "blue")]
rvb = mcolors.LinearSegmentedColormap.from_list("", clist)

N = 60
x = np.arange(N).astype(float)
y = np.random.uniform(0, 5, size=(N,))

plt.bar(x,y, color=rvb(x/N))
plt.show()

Upvotes: 14

Emanuel
Emanuel

Reputation: 432

Seaborn barplot is great for this, example:

ax = sns.barplot("size", y="total_bill", data=tips, palette="Blues_d")

Upvotes: 10

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