python七段数码管绘制
时间: 2023-10-14 19:20:03 浏览: 67
以下是 Python 代码,可以绘制七段数码管:
```python
import turtle
# 设置画笔
turtle.speed(0)
turtle.pensize(10)
turtle.penup()
# 定义每个数字的七段数码管
digits = {
0: [0, 1, 2, 3, 4, 5],
1: [1, 2],
2: [0, 1, 3, 4, 6],
3: [0, 1, 2, 3, 6],
4: [1, 2, 5, 6],
5: [0, 2, 3, 5, 6],
6: [0, 2, 3, 4, 5, 6],
7: [0, 1, 2],
8: [0, 1, 2, 3, 4, 5, 6],
9: [0, 1, 2, 3, 5, 6]
}
# 定义绘制七段数码管的函数
def draw_digit(number, x, y):
segments = digits[number]
turtle.setpos(x, y)
turtle.pendown()
if 0 in segments:
turtle.forward(50)
turtle.left(90)
if 1 in segments:
turtle.forward(50)
turtle.right(90)
turtle.forward(50)
turtle.right(90)
if 2 in segments:
turtle.forward(50)
turtle.right(90)
turtle.forward(50)
turtle.left(90)
if 3 in segments:
turtle.penup()
turtle.forward(50)
turtle.pendown()
turtle.left(90)
turtle.forward(50)
turtle.left(90)
turtle.forward(50)
if 4 in segments:
turtle.penup()
turtle.backward(50)
turtle.pendown()
turtle.right(90)
turtle.forward(50)
turtle.right(90)
turtle.forward(50)
if 5 in segments:
turtle.penup()
turtle.backward(50)
turtle.right(90)
turtle.forward(50)
turtle.left(90)
turtle.pendown()
turtle.forward(50)
if 6 in segments:
turtle.penup()
turtle.backward(50)
turtle.right(90)
turtle.forward(50)
turtle.left(90)
turtle.pendown()
turtle.forward(50)
# 绘制数字 1, 2, 3, 4, 5, 6, 7, 8, 9, 0
draw_digit(1, -150, 0)
draw_digit(2, -50, 0)
draw_digit(3, 50, 0)
draw_digit(4, 150, 0)
draw_digit(5, -150, -150)
draw_digit(6, -50, -150)
draw_digit(7, 50, -150)
draw_digit(8, 150, -150)
draw_digit(9, 0, -300)
draw_digit(0, 0, 150)
# 隐藏画笔
turtle.hideturtle()
# 显示绘制结果
turtle.done()
```
这段代码利用 turtle 库绘制七段数码管,定义了一个 `digits` 字典,其中包含每个数字的七段数码管的组成部分。然后定义了一个 `draw_digit` 函数,用于绘制单个数字的七段数码管。最后,调用 `draw_digit` 函数绘制数字 0 到 9。
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