当前位置：文江博客话题详情

如何制作 3D 散点图

发布于 2024-08-16 07:42:58 字数 115 浏览 8 评论 0原文

我目前有一个 nx3 矩阵数组。我想将三列绘制为三个轴。我怎样才能做到这一点？

我用谷歌搜索过，人们建议使用Matlab，但我真的很难理解它。我还需要它是一个散点图。

原文

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

梨涡少年 2024-08-23 07:42:59

您可以使用 matplotlib 来实现此目的。 matplotlib 有一个 mplot3d 模块，它可以完全满足您的需求。

import matplotlib.pyplot as plt
import random

fig = plt.figure(figsize=(12, 12))
ax = fig.add_subplot(projection='3d')

sequence_containing_x_vals = list(range(0, 100))
sequence_containing_y_vals = list(range(0, 100))
sequence_containing_z_vals = list(range(0, 100))

random.shuffle(sequence_containing_x_vals)
random.shuffle(sequence_containing_y_vals)
random.shuffle(sequence_containing_z_vals)

ax.scatter(sequence_containing_x_vals, sequence_containing_y_vals, sequence_containing_z_vals)
plt.show()

上面的代码生成一个如下图：

You can use matplotlib for this. matplotlib has a mplot3d module that will do exactly what you want.

import matplotlib.pyplot as plt
import random

fig = plt.figure(figsize=(12, 12))
ax = fig.add_subplot(projection='3d')

sequence_containing_x_vals = list(range(0, 100))
sequence_containing_y_vals = list(range(0, 100))
sequence_containing_z_vals = list(range(0, 100))

random.shuffle(sequence_containing_x_vals)
random.shuffle(sequence_containing_y_vals)
random.shuffle(sequence_containing_z_vals)

ax.scatter(sequence_containing_x_vals, sequence_containing_y_vals, sequence_containing_z_vals)
plt.show()

The code above generates a figure like:

回复收藏 0 原文

近箐 2024-08-23 07:42:59

使用对我有用的以下代码：

# Create the figure
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

# Generate the values
x_vals = X_iso[:, 0:1]
y_vals = X_iso[:, 1:2]
z_vals = X_iso[:, 2:3]

# Plot the values
ax.scatter(x_vals, y_vals, z_vals, c = 'b', marker='o')
ax.set_xlabel('X-axis')
ax.set_ylabel('Y-axis')
ax.set_zlabel('Z-axis')

plt.show()

虽然 X_iso 是我的 3-D 数组，但对于 X_vals、Y_vals、Z_vals，我从该数组复制/使用了 1 列/轴，并分别分配给这些变量/数组。

Use the following code it worked for me:

# Create the figure
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

# Generate the values
x_vals = X_iso[:, 0:1]
y_vals = X_iso[:, 1:2]
z_vals = X_iso[:, 2:3]

# Plot the values
ax.scatter(x_vals, y_vals, z_vals, c = 'b', marker='o')
ax.set_xlabel('X-axis')
ax.set_ylabel('Y-axis')
ax.set_zlabel('Z-axis')

plt.show()

while X_iso is my 3-D array and for X_vals, Y_vals, Z_vals I copied/used 1 column/axis from that array and assigned to those variables/arrays respectively.

回复收藏 0 原文

木森分化 2024-08-23 07:42:59

from mpl_toolkits import mplot3d
import numpy as np
import matplotlib.pyplot as plt
fig = plt.figure()
ax = plt.axes(projection='3d')

散点图

zdata = 15 * np.random.random(100)
xdata = np.sin(zdata) + 0.1 * np.random.randn(100)
ydata = np.cos(zdata) + 0.1 * np.random.randn(100)
ax.scatter3D(xdata, ydata, zdata);

Colab 笔记本

from mpl_toolkits import mplot3d
import numpy as np
import matplotlib.pyplot as plt
fig = plt.figure()
ax = plt.axes(projection='3d')

scatter plot

zdata = 15 * np.random.random(100)
xdata = np.sin(zdata) + 0.1 * np.random.randn(100)
ydata = np.cos(zdata) + 0.1 * np.random.randn(100)
ax.scatter3D(xdata, ydata, zdata);

Colab notebook

回复收藏 0 原文

弥繁 2024-08-23 07:42:59

使用plotly - 最简单、最实用且漂亮的绘图

import plotly.express as px
df = px.data.iris()
fig = px.scatter_3d(df, x='sepal_length', y='sepal_width', z='petal_width',
              color='species')
fig.show()

https://plotly.com/python/3d -散点图/

Using plotly - Easiest and most functional and nice plots

import plotly.express as px
df = px.data.iris()
fig = px.scatter_3d(df, x='sepal_length', y='sepal_width', z='petal_width',
              color='species')
fig.show()

https://plotly.com/python/3d-scatter-plots/

回复收藏 0 原文

~没有更多了~