CODEWITHSUNDEEP
pythonseabornxticks
dododips
dododips

Reputation: 91

How can I mark xticks at the center of bins for a seaborn distplot?

I want to plot a distplot using seaborn with xticks at the mid point of the bins. I am using the below code:

sns.distplot(df['MILEAGE'], kde=False, bins=20)

This is what my displot currently looks like

Upvotes: 4

Views: 7307

Answers (2)

JohanC
JohanC

Reputation: 80299

To get the midpoints of the bars, you can extract the generated rectangle patches and add half their width to their x position. Setting these midpoints as xticks will label the bars.

import matplotlib.pyplot as plt
import seaborn as sns
import numpy as np

ax = sns.distplot(np.random.randn(1000).cumsum(), kde=False, bins=20)
mids = [rect.get_x() + rect.get_width() / 2 for rect in ax.patches]
ax.set_xticks(mids)
plt.show()

example plot

If the tick labels would overlap too much, you could rotate them and/or adapt their fontsize:

ax.tick_params(axis='x', rotation=90, labelsize=8)

If you need the bin edges instead of their centers:

edges = [rect.get_x() for rect in ax.patches] + [ax.patches[-1].get_x() + ax.patches[-1].get_width()]

Upvotes: 4

flurble
flurble

Reputation: 1106

In your specific case you could just get the ticks of the figure and add 25 to them to shift them into the middle of the bars. You could also completely reset them.

ticks = plt.gca().get_xticks()
ticks += 25
plt.xticks(ticks)

Alternatively, if you plot a histogram with matplotlib, you can get the bins directly:

x = np.random.rand(100)

# Matplotlib only
counts, bins, patches = plt.hist(x)
plt.xticks(bins + 25)

Upvotes: 0

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