30岁的程序员拥有20岁的浪漫，用Python画3D玫瑰送女朋友

Posted 2021-11-02 yunyun云芸

1.天生丽质

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

fig = plt.figure()
ax = fig.gca(projection='3d')
# 将相位向后移动了6*pi
[x, t] = np.meshgrid(np.array(range(25)) / 24.0, np.arange(0, 575.5, 0.5) / 575 * 20 * np.pi + 4*np.pi)
p = (np.pi / 2) * np.exp(-t / (8 * np.pi))
# 添加边缘扰动
change = np.sin(15*t)/150
# 将t的参数减少，使花瓣的角度变大
u = 1 - (1 - np.mod(3.3 * t, 2 * np.pi) / np.pi) ** 4 / 2 + change
y = 2 * (x ** 2 - x) ** 2 * np.sin(p)
r = u * (x * np.sin(p) + y * np.cos(p))
h = u * (x * np.cos(p) - y * np.sin(p))
c= cm.get_cmap('Reds')
surf = ax.plot_surface(r * np.cos(t), r * np.sin(t), h, rstride=1, cstride=1,
                       cmap= c, linewidth=0, antialiased=True)
plt.show()

2.姹紫嫣红

mport numpy as np
import matplotlib.pyplot as plt
from matplotlib import cm
from mpl_toolkits.mplot3d import Axes3D


fig = plt.figure()
ax = fig.gca(projection='3d')
[x, t] = np.meshgrid(np.array(range(25)) / 24.0, np.arange(0, 575.5, 0.5) / 575 * 17 * np.pi - 2 * np.pi)
p = (np.pi / 2) * np.exp(-t / (8 * np.pi))
u = 1 - (1 - np.mod(3.6 * t, 2 * np.pi) / np.pi) ** 4 / 2
y = 2 * (x ** 2 - x) ** 2 * np.sin(p)
r = u * (x * np.sin(p) + y * np.cos(p))
h = u * (x * np.cos(p) - y * np.sin(p))
surf = ax.plot_surface(r * np.cos(t), r * np.sin(t), h, rstride=1, cstride=1,
                       cmap=cm.gist_rainbow_r, linewidth=0, antialiased=True)
plt.show()

3.粉黛俏佳人

mport numpy as np
import matplotlib.pyplot as plt
from matplotlib import cm
from mpl_toolkits.mplot3d import Axes3D

fig = plt.figure()
ax = fig.gca(projection='3d')
[x, t] = np.meshgrid(np.array(range(25)) / 24.0, np.arange(0, 575.5, 0.5) / 575 * 30 * np.pi - 4*np.pi)
p = (np.pi / 2) * np.exp(-t / (8 * np.pi))
change = np.sin(20*t)/50
u = 1 - (1 - np.mod(3.3 * t, 2 * np.pi) / np.pi) ** 4 / 2 + change
y = 2 * (x ** 2 - x) ** 2 * np.sin(p)
r = u * (x * np.sin(p) + y * np.cos(p)) * 1.5
h = u * (x * np.cos(p) - y * np.sin(p))
c= cm.get_cmap('magma')
surf = ax.plot_surface(r * np.cos(t), r * np.sin(t), h, rstride=1, cstride=1,
                       cmap= c, linewidth=0, antialiased=True)
plt.show()

4.桃之夭夭

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
fig = plt.figure()
ax = fig.gca(projection='3d')
[x, t] = np.meshgrid(np.array(range(25)) / 24.0, np.arange(0, 575.5, 0.5) / 575 * 6 * np.pi - 4*np.pi)
p = (np.pi / 2) * np.exp(-t / (8 * np.pi))
change = np.sin(10*t)/20
u = 1 - (1 - np.mod(5.2 * t, 2 * np.pi) / np.pi) ** 4 / 2 + change
y = 2 * (x ** 2 - x) ** 2 * np.sin(p)
r = u * (x * np.sin(p) + y * np.cos(p)) * 1.5
h = u * (x * np.cos(p) - y * np.sin(p))
c= cm.get_cmap('spring_r')
surf = ax.plot_surface(r * np.cos(t), r * np.sin(t), h, rstride=1, cstride=1,
                       cmap= c, linewidth=0, antialiased=True)
plt.show()

5.思路步骤

以姹紫嫣红为例
1.
花的花瓣总是围绕中心轴生长，我们不妨将花看做一个中心旋转3D图。在此基础上建立柱状极坐标系。在极坐标系中，图像上的每个点有三个自由度(r,t,h)（其中t表示theta）
2.
观察花瓣生长规律，我们发现，其外边缘线在一条旋转内缩的曲线上。这条曲线里中心轴的距离r逐渐缩短，距离地平面h逐渐变高。且该曲线大概符合以下公式：

p = (np.pi / 2) * np.exp(-t / (8 * np.pi))
r = np.sin(p) 
h = np.cos(p)

3.
花瓣状乘类似于水滴状关于中心轴旋转。因此r和h都是关于模式u的函数：

u = 1 - (1 - np.mod(3.6 * t, 2 * np.pi) / np.pi) ** 4 / 2
r = u * np.sin(p) 
h = u * np.cos(p)

4.
为了图像更加真实，添加一个修正项y，使花瓣的形态向下凸，更接近真实生活：

y = 2 * (x ** 2 - x) ** 2 * np.sin(p)
r = u * (x * np.sin(p) + y * np.cos(p))
h = u * (x * np.cos(p) - y * np.sin(p))

5.
实际上我们画出来的不是花瓣的平面，而是一个个离散的点（总共25*1150 = 28750个），经过python的plot_surface逼近花瓣的面状：

[x, t] = np.meshgrid(np.array(range(25)) / 24.0, 
    np.arange(0, 575.5, 0.5) / 575 * 17 * np.pi - 2 * np.pi)

x,t是两个自变量，相当于参数方程中两个参数，最终表示为三维坐标系下的某点(rcos(θ),rsin(θ),h)，x是在[0,25)上取的25个点，t是-2pi到15pi上取的1150个点，在3D图像中共计28750个点，绘制出的散点图通过python自带的函数连成曲面状。

福利来啦

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