如何在Matplotlib中以3D形式绘制热图颜色

Question

我使用Matplotlib 3D绘制我的数据集的3个维度,如下所示:

但是现在我还想将第四维(这是一个介于0到20之间的标量值)可视化为热图.所以基本上,我希望每个点都基于这个第四维的值来获取它的颜色.

Matplotlib中是否存在这样的东西？如何将[0-20]之间的一串数字转换为热图颜色？

我从这里获取代码:http://matplotlib.org/mpl_examples/mplot3d/scatter3d_demo.py

Answer 1

是的,像这样:

这里更新是带有颜色条的版本.

import numpy as np
from pylab import *
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt

def randrange(n, vmin, vmax):
    return (vmax-vmin)*np.random.rand(n) + vmin

fig = plt.figure(figsize=(8,6))

ax = fig.add_subplot(111,projection='3d')
n = 100

xs = randrange(n, 23, 32)
ys = randrange(n, 0, 100)
zs = randrange(n, 0, 100)

colmap = cm.ScalarMappable(cmap=cm.hsv)
colmap.set_array(zs)

yg = ax.scatter(xs, ys, zs, c=cm.hsv(zs/max(zs)), marker='o')
cb = fig.colorbar(colmap)

ax.set_xlabel('X Label')
ax.set_ylabel('Y Label')
ax.set_zlabel('Z Label')


plt.show()

好像:

更新以下是通过某些第四维属性对数据点着色的明确示例.

import numpy as np
from pylab import *
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt

def randrange(n, vmin, vmax):
    return (vmax-vmin)*np.random.rand(n) + vmin

fig = plt.figure(figsize=(8,6))

ax = fig.add_subplot(111,projection='3d')
n = 100

xs = randrange(n, 0, 100)
ys = randrange(n, 0, 100)
zs = randrange(n, 0, 100)
the_fourth_dimension = randrange(n,0,100)

colors = cm.hsv(the_fourth_dimension/max(the_fourth_dimension))

colmap = cm.ScalarMappable(cmap=cm.hsv)
colmap.set_array(the_fourth_dimension)

yg = ax.scatter(xs, ys, zs, c=colors, marker='o')
cb = fig.colorbar(colmap)

ax.set_xlabel('X Label')
ax.set_ylabel('Y Label')
ax.set_zlabel('Z Label')


plt.show()