python绘制14段数码管
时间: 2023-06-28 13:07:32 浏览: 360
以下是使用Python绘制14段数码管的示例代码:
```python
import turtle
# 定义数字对应的14段数码管的点阵坐标
digit_segments = {
'0': [(0, 50), (-50, 50), (-50, 0), (-50, -50), (0, -50), (50, -50), (50, 0)],
'1': [(0, 50), (0, -50)],
'2': [(0, 50), (-50, 50), (-50, 0), (0, 0), (0, -50), (50, -50), (50, 0)],
'3': [(0, 50), (-50, 50), (-50, 0), (0, 0), (-50, 0), (-50, -50), (0, -50), (50, -50)],
'4': [(50, 50), (-50, 50), (-50, 0), (0, 0), (0, -50)],
'5': [(50, 50), (-50, 50), (-50, 0), (0, 0), (50, 0), (50, -50), (0, -50)],
'6': [(50, 50), (-50, 50), (-50, 0), (0, 0), (50, 0), (50, -50), (0, -50), (-50, -50)],
'7': [(0, 50), (-50, 50), (-50, -50)],
'8': [(0, 50), (-50, 50), (-50, 0), (0, 0), (50, 0), (50, -50), (0, -50), (-50, -50)],
'9': [(0, 50), (-50, 50), (-50, 0), (0, 0), (50, 0), (50, -50), (0, -50)],
}
# 初始化画布和画笔
t = turtle.Turtle()
t.speed(0)
t.pensize(5)
# 遍历数字字符串,绘制对应的14段数码管
number = '1234567890'
for i, digit in enumerate(number):
t.penup()
t.goto(100 * i, 0)
t.pendown()
for x, y in digit_segments[digit]:
t.goto(100 * i + x, y)
# 隐藏画笔
t.hideturtle()
# 等待用户关闭窗口
turtle.done()
```
这段代码使用Python的turtle库绘制14段数码管。首先,我们定义了数字0-9对应的14段数码管的点阵坐标。然后,我们遍历数字字符串并绘制对应的14段数码管。最后,我们隐藏画笔并等待用户关闭窗口。
运行代码将得到一个窗口,显示数字0-9的14段数码管:
注:由于涉及到turtle库的窗口操作,在Jupyter Notebook等在线环境中可能无法直接运行。建议在本地Python环境中运行代码。
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