Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
493 views
in Technique[技术] by (71.8m points)

python - Matplotlib: plotting transparent histogram with non transparent edge

I am plotting a histogram, and I have three datasets which I want to plot together, each one with different colours and linetype (dashed, dotted, etc). I am also giving some transparency, in order to see the overlapping bars.

The point is that I would like the edge of each bar not to become transparent as the inner part does. Here is an example:

import matplotlib.pyplot as plt
import numpy as np

x = np.random.random(20)
y =np.random.random(20)
z= np.random.random(20)

fig = plt.figure()
ax = fig.add_subplot(111)
ax.hist(x, bins=np.arange(0, 1, 0.1), ls='dashed', alpha = 0.5, lw=3, color= 'b')
ax.hist(y, bins=np.arange(0, 1, 0.1), ls='dotted', alpha = 0.5, lw=3, color= 'r')
ax.hist(z, bins=np.arange(0, 1, 0.1), alpha = 0.5, lw=3, color= 'k')
ax.set_xlim(-0.5, 1.5)
ax.set_ylim(0, 7)
plt.show()

enter image description here

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

plt.hist accepts additional keyword arguments that are passed to the constructor for matplotlib.patches.Patch. In particular you can pass an fc= argument which lets you set the patch facecolor using an (R, G, B, A) tuple when you create the histograms. Changing the alpha value of the facecolor does not affect the transparency of the edges:

ax.hist(x, bins=np.arange(0, 1, 0.1), ls='dashed', lw=3, fc=(0, 0, 1, 0.5))
ax.hist(y, bins=np.arange(0, 1, 0.1), ls='dotted', lw=3, fc=(1, 0, 0, 0.5))
ax.hist(z, bins=np.arange(0, 1, 0.1), lw=3, fc=(0, 0, 0, 0.5))

enter image description here


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...