python 和 MATLAB 都能绘制的母亲节花束！！

本文主要是介绍python 和 MATLAB 都能绘制的母亲节花束！！，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

hey 母亲节快到了，教大家用python和MATLAB两种语言绘制花束~这段代码是我七夕节发的，我对代码进行了简化，同时自己整了个python版本

MATLAB 版本代码

在这里插入图片描述

function roseBouquet_M()
% @author : slandarer% 生成花朵数据
[xr,tr]=meshgrid((0:24)./24,(0:0.5:575)./575.*20.*pi+4*pi);
p=(pi/2)*exp(-tr./(8*pi));
ur=1-(1-mod(3.6*tr,2*pi)./pi).^4./2+sin(15*tr)/150;
yr=2*(xr.^2-xr).^2.*sin(p);
rr=ur.*(xr.*sin(p)+yr.*cos(p));
hr=ur.*(xr.*cos(p)-yr.*sin(p));tb=linspace(0,2,151);
rb=(0:.01:1)'*((abs((1-mod(tb*5,2))))/2+.3)./2.5;
xb=rb.*cos(tb*pi);
yb=rb.*sin(tb*pi);
hb=(-cos(rb*1.2*pi)+1).^.2;% 配色数据
cL=[.33 .33 .69;.68 .42 .63;.78 .42 .57;.96 .73 .44];
% cL=[.02 .04 .39;.02 .06 .69;.01 .26 .99;.17 .69 1];
cMr=sH(hr,cL);
cMb=sH(hb,cL.*.4+.6);
% 旋转矩阵生成
yz=72*pi/180;
Rz=@(n)[cos(yz/n),-sin(yz/n),0;sin(yz/n),cos(yz/n),0;0,0,1];
Rx=@(n)[1,0,0;0,cos(n),-sin(n);0,sin(n),cos(n)];
Rz1=Rz(1);Rz2=Rz(2);Rz3=Rz(3);
Rx1=Rx(pi/8);Rx2=Rx(pi/9);
% 图形绘制
hold on
cp={'EdgeAlpha',0.05,'EdgeColor','none','FaceColor','interp','CData',cMr};
surface(rr.*cos(tr),rr.*sin(tr),hr+0.35,cp{:})
[U,V,W]=rT(rr.*cos(tr),rr.*sin(tr),hr+0.35,Rx1);
V=V-.4;
for k=1:5[U,V,W]=rT(U,V,W,Rz1);surface(U,V,W-.1,cp{:})dS(U,V,W-.1)
end
[u1,v1,w1]=rT(xb,yb,hb./2.5+.32,Rx2);
v1=v1-1.35;
[u2,v2,w2]=rT(u1,v1,w1,Rz2);
[u3,v3,w3]=rT(u1,v1,w1,Rz3);
[u4,v4,w4]=rT(u3,v3,w3,Rz3);
U={u1,u2,u3,u4};
V={v1,v2,v3,v4};
W={w1,w2,w3,w4};
for k=1:5for b=1:4[ut,vt,wt]=rT(U{b},V{b},W{b},Rz1);U{b}=ut;V{b}=vt;W{b}=wt;surface(U{b},V{b},W{b},cp{3:7},cMb)dS(U{b},V{b},W{b})end
end
a=gca;axis off
a.Position=[0,0,1,1]+[-1,-1,2,2]./6;
axis equal
view(2,35);% 配色插值函数function c=sH(H,cL)X=rescale(H,0,1);x=rescale(1:size(cL,1),0,1);c=interp1(x,cL,X);end% 旋转矩阵应用至数据点function [U,V,W]=rT(X,Y,Z,R)U=X;V=Y;W=Z;for i=1:numel(X)v=[X(i);Y(i);Z(i)];n=R*v;U(i)=n(1);V(i)=n(2);W(i)=n(3);endend% 贝塞尔函数插值生成花杆并绘制function dS(X,Y,Z)[m,n]=find(Z==min(min(Z)));m=m(1);n=n(1);x1=X(m,n);y1=Y(m,n);z1=Z(m,n)+.03;x=[x1,0,(x1.*cos(pi/3)-y1.*sin(pi/3))./3].';y=[y1,0,(y1.*cos(pi/3)+x1.*sin(pi/3))./3].';z=[z1,-.7,-1.5].';P=[x,y,z];t=(1:50)/50;q=size(P,1)-1;c1=factorial(q)./factorial(0:q)./factorial(q:-1:0);c2=((t).^((0:q)')).*((1-t).^((q:-1:0)'));P=(P'*(c1'.*c2))';plot3(P(:,1),P(:,2),P(:,3),'Color',[88,130,126]./255,'LineWidth',1)end
end

python 版本代码

需要安装以下库：

matplotlib
numpy
scipy

import matplotlib.pyplot as plt
import numpy as np
from matplotlib.colors import LinearSegmentedColormap as lsc
from scipy.spatial.transform import Rotation as R
# @author : slandarer# 生成花朵数据
t1 = np.array(range(25))/24
t2 = np.arange(0, 575.5, 0.5)/575*20*np.pi + 4*np.pi
[xr, tr] = np.meshgrid(t1, t2)
pr = (np.pi/2)*np.exp(-tr/(8*np.pi))
ur = 1 - (1 - np.mod(3.6*tr, 2*np.pi)/np.pi)**4/2 + np.sin(15*tr)/150 + np.sin(15*tr)/150
yr = 2*(xr**2 - xr)**2*np.sin(pr)
rr = ur*(xr*np.sin(pr) + yr*np.cos(pr))
hr = ur*(xr*np.cos(pr) - yr*np.sin(pr))tb = np.resize(np.linspace(0, 2, 151), (1,151))
rb = np.resize(np.linspace(0, 1, 101), (101,1)) @ ((abs((1-np.mod(tb*5,2))))/2 + .3)/2.5
xb = rb*np.cos(tb*np.pi)
yb = rb*np.sin(tb*np.pi)
hb = np.power(-np.cos(rb*1.2*np.pi)+1, .2)cL = np.array([[.33,.33,.69], [.68,.42,.63], [.78,.42,.57], [.96,.73,.44]])
cL = np.array([[.02,.04,.39], [.02,.06,.69], [.01,.26,.99], [.17,.69,1]])
cMpr = lsc.from_list('slandarer', cL)
cMpb = lsc.from_list('slandarer', cL*.4 + .6)# 绕轴旋转数据点
def rT(X, Y, Z, T):SZ = X.shapeXYZ = np.hstack((X.reshape(-1, 1), Y.reshape(-1, 1), Z.reshape(-1, 1)))RMat = R.from_euler('xyz', T, degrees = True); XYZ = RMat.apply(XYZ)return XYZ[:,0].reshape(SZ), XYZ[:,1].reshape(SZ), XYZ[:,2].reshape(SZ)# 贝塞尔函数插值生成花杆并绘制
def dS(X, Y, Z):MN = np.where(Z == np.min(Z)); M = MN[0][0]; N = MN[1][0]x1 = X[M, N]; y1 = Y[M, N]; z1 = Z[M, N] + .03x = np.array([x1, 0, (x1*np.cos(np.pi/3) - y1*np.sin(np.pi/3))/3]).reshape((3,1))y = np.array([y1, 0, (y1*np.cos(np.pi/3) + x1*np.sin(np.pi/3))/3]).reshape((3,1))z = np.array([z1, -.7, -1.5]).reshape((3,1))P = np.hstack((x,y,z)).Tt = (np.array(range(50)) + 1)/50c1 = np.array([1, 2, 1]).reshape(3,1)c2 = np.power(t, np.array(range(3)).reshape(3,1))c3 = np.power(1 - t, np.array(range(2, -1, -1)).reshape(3,1))P = (P @ (c1*c2*c3))ax.plot(P[0], P[1], P[2], color = '#58827E')# 创建figure窗口及axis坐标区域
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
# 绘制花束
ax.plot_surface(rr*np.cos(tr), rr*np.sin(tr), hr + .35, rstride = 1, cstride = 1,facecolors = cMpr(hr), antialiased = True, shade = False)
U, V, W = rT(rr*np.cos(tr), rr*np.sin(tr), hr + .35, [180/8, 0, 0]); V = V - .4
for i in range(5):U, V, W = rT(U, V, W, [0, 0, 72])ax.plot_surface(U, V, W - .1, rstride = 1, cstride = 1,facecolors = cMpr(hr), antialiased = True, shade = False)dS(U, V, W - .1)u1, v1, w1=rT(xb, yb, hb/2.5 + .32, [180/9, 0, 0])
v1 = v1 - 1.35
u2, v2, w2 = rT(u1, v1, w1, [0, 0, 36])
u3, v3, w3 = rT(u1, v1, w1, [0, 0, 24])
u4, v4, w4 = rT(u3, v3, w3, [0, 0, 24])
for i in range(5):u1, v1, w1 = rT(u1, v1, w1, [0, 0, 72])u2, v2, w2 = rT(u2, v2, w2, [0, 0, 72])u3, v3, w3 = rT(u3, v3, w3, [0, 0, 72])u4, v4, w4 = rT(u4, v4, w4, [0, 0, 72])ax.plot_surface(u1, v1, w1, rstride = 1, cstride = 1,facecolors = cMpb(hb), antialiased = True, shade = False)ax.plot_surface(u2, v2, w2, rstride = 1, cstride = 1,facecolors = cMpb(hb), antialiased = True, shade = False)ax.plot_surface(u3, v3, w3, rstride = 1, cstride = 1,facecolors = cMpb(hb), antialiased = True, shade = False)ax.plot_surface(u4, v4, w4, rstride = 1, cstride = 1,facecolors = cMpb(hb), antialiased = True, shade = False)dS(u1, v1, w1)dS(u2, v2, w2)dS(u3, v3, w3)dS(u4, v4, w4)ax.set_position((-.215, -.3, 1.43, 1.43))
ax.set_box_aspect((1, 1, .8))
ax.view_init(elev = 50, azim = 2)
ax.axis('off')
plt.show()