各位，怎么用python画12个花瓣？

Z时代
2024-03-11
分类：IT

from turtle import *
circle(40,step = 12)
done

回答：

楼主的问题中已经定义好了turtle库，所以我在此基础上补充完善。思路是先定义一个花瓣

def draw_petal():
    for i in range(2):
        circle(40, 60)
        left(120)

然后循环生成12个


# 循环调用画花瓣的函数，共绘制12个花瓣
for i in range(12):
    draw_petal()
    left(30)done()

回答：

12 个随机颜色的花瓣，大小、位置、弧度等随机生成。

python">import turtle
import random
# 初始化
turtle.screensize(800, 600)
turtle.bgcolor("white")
turtle.speed(10)
turtle.hideturtle()
# 设置变量和函数
colors = ["red", "orange", "yellow", "green", "blue", "purple"]
petals = 12
radius = 100
amplitude = 20
period = 90
def draw_petal(petal_color):
    turtle.color(petal_color)
    turtle.begin_fill()
    for angle in range(0, 360, 5):
        x = radius * (1 + amplitude / 100.0 * 
            abs(math.sin(angle * math.pi / period)))
        y = x * math.sin(angle * math.pi / 180)
        turtle.goto(x, y)
    turtle.end_fill()
# 开始绘制花朵
for petal in range(petals):
    # 随机选择颜色并设置初始位置
    color = random.choice(colors)
    x = random.randint(-300, 300)
    y = random.randint(-200, 200)
    turtle.penup()
    turtle.goto(x, y)
    turtle.pendown()
    # 开始绘制花瓣
    size = random.randint(20, 60)
    turtle.pensize(size / 10)
    draw_petal(color)
# 显示完毕turtle.done()

回答：

import numpy as np
import matplotlib.pyplot as plt
theta = np.linspace(0, 2.*np.pi, 1000)  
r = np.cos(6*theta)  
fig = plt.figure()
ax = fig.add_subplot(111, polar=True)  
ax.plot(theta, r)  
ax.set_yticklabels([]) 
plt.show()

各位，怎么用python画12个花瓣？

import numpy as np
import matplotlib.pyplot as plt
def draw_petal(ax, radius, petal_num):
    theta = np.linspace(0, 2.*np.pi, 1000)
    r = radius * np.cos(petal_num*theta)
    ax.plot(theta, r)
fig = plt.figure()
ax = fig.add_subplot(111, polar=True)
ax.set_yticklabels([])
for i in range(5, 0, -1):
    draw_petal(ax, i/5.0, i*4)
plt.show()

各位，怎么用python画12个花瓣？