Creating a dark, reversed color palette in Seaborn

Question

I am creating a figure that contains several plots using a sequential palette like so:

import matplotlib.pyplot as plt
import seaborn as sns
import math

figure = plt.figure(1)
x = range(1, 200)
n_plots = 10

with sns.color_palette('Blues_d', n_colors=n_plots):
    for offset in range(n_plots):
        plt.plot(x, [offset + math.sin(float(i) / 10) for i in range(len(x))])

figure.show()

However, I would like to reverse the color palette. The tutorial states that I can add '_r' to a palette name to reverse it and '_d' to make it "dark". But I do not appear to be able to do these together: '_r_d', '_d_r', '_rd' and '_dr' all produce errors. How can I create a dark, reversed palette?

Answer 1

I'm answering my own question to post the details and explanation of the solution I used, because mwaskom's suggestion required a tweak. Using

with reversed(sns.color_palette('Blues_d', n_colors=n_plots)):

throws AttributeError: __exit__, I believe because the with statement requires an object with __enter__ and __exit__ methods, which the reversed iterator doesn't satisfy. If I use sns.set_palette(reversed(palette)) instead of a with statement, the number of colors in the plot is ignored (the default of 6 is used - I have no idea why) even though the color scheme is obeyed. To solve this, I use list.reverse() method:

figure = plt.figure(1)
x = range(1, 200)
n_plots = 10
palette = sns.color_palette("Blues_d", n_colors=n_plots)
palette.reverse()

with palette:
    for offset in range(n_plots):
        plt.plot(x, [offset + math.sin(float(i) / 10) for i in range(len(x))])

figure.show()

Edit: I discovered that the reason the n_colors argument was ignored in the call to set_palette was because the n_colors argument must also be specified in that call. Another solution is therefore:

figure = plt.figure(1)
x = range(1, 200)
n_plots = 10

sns.set_palette(reversed(sns.color_palette("Blues_d", n_plots)), n_plots)

for offset in range(n_plots):
    plt.plot(x, [offset + math.sin(float(i) / 10) for i in range(len(x))])

figure.show()