如何在Matplot散点图中为每个气泡设置set_gid()? [英] How to set_gid() for each bubble in matplot scatter chart?

问题描述

我用下面的代码画了一个散点图:

I draw a scatter chart by below codes:

import matplotlib.pyplot as plt

x = [2,4,6]
y = [1,3,7]
r = [650,890,320]
clr = ['r','b','g']
bubble_id = ['C0','C1','C2']

H0 = plt.scatter(x,y,s=r,c=clr)

然后我想将 'set_gid()' 分别设为 'C0', 'C1' ,'C2' 三个气泡.怎么做 ?由于 H0 是单个对象 ，我不知道如何分解 H0 并找到三个泡泡儿子"的H0.感谢您的提示.

Then I want to 'set_gid()' to the three bubbles as 'C0', 'C1' ,'C2' respectively. How to do that ? As H0 is a single object <matplotlib.collections.PathCollection object at 0x0ADA6D30>, I don't know how to break down H0 and find the three 'bubble son' of H0 . Thanks for your tips.

推荐答案

所以，我知道这可能不是最有效的解决方案，但是循环呢?

So, I know that might not be the most efficient solution, but what about looping?

import matplotlib.pyplot as plt
import itertools as it

X = [2,4,6]
Y = [1,3,7]
radius = [650,890,320]
clr = ['r','b','g']
bubble_id = ['C0','C1','C2']

ax = plt.subplot(111)
for x, y, r, c, id in it.izip(X, Y, radius, clr, bubble_id):
    ax.scatter(x,y,s=r,c=c, gid=id)

在视觉上给出相同的情节.

visually gives the same plot.

在Ipython中，我介绍了 PathCollection 方法，看来没有简单的方法可以从中获取单个补丁.

In Ipython I've given a look at PathCollection methods, and it looks like there is no trivial way to get out the single patches from it.

在2020年进行编辑

由于现在您很可能使用python 3(如果不是，则应该使用)，因此可以使用内置的 zip 而不是 itertools.izip .谢谢@Guimoute回复了答案

Since nowadays your are very likely using python 3 (if not you should), it is possible to use the built-in zip instead of itertools.izip. Thank you @Guimoute for reviving the answer

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