用Python创建一行图像的Knight Rider风格流LED
用Python创建一行图像的Knight Rider风格流LED
提问于 2016-10-14 20:45:45
我在学习用树莓皮和圆脸展开板的蟒蛇。使用Tkinter，我创建了一个带有按钮的Gui来操作π面LED。在代码的一部分中，我打开了一个新窗口，它显示了一个按钮和一排处于“关闭”状态的LED图像。我试图添加一些代码，使LED图像的行流，在'on‘状态左到右沿图像行，就像骑士车的前灯。我在while循环中尝试过一些东西，但是如果没有很多代码行，我就无法很好地理解如何实现它。我认为一定有办法做到这一点，就像数字写入是增量，以创建在piface扩展板上的流式LED。这是我的密码。
class App2:
def __init__(self, master):
self.signal = False #added to stop thread
print('self.signal', self.signal)
self.master=master # I added this line to make the exit button work
frame = Frame(master)
frame.pack()
Label(frame, text='Turn LED ON').grid(row=0, column=0)
Label(frame, text='Turn LED OFF').grid(row=0, column=1)
self.button0 = Button(frame, text='Knight Rider OFF', command=self.convert0)
self.button0.grid(row=2, column=0)
self.LED0 = Label(frame, image=logo2) #added to create a row of images
self.LED1 = Label(frame, image=logo2)
self.LED2 = Label(frame, image=logo2)
self.LED3 = Label(frame, image=logo2)
self.LED4 = Label(frame, image=logo2)
self.LED5 = Label(frame, image=logo2)
self.LED6 = Label(frame, image=logo2)
self.LED7 = Label(frame, image=logo2)
self.LED0.grid(row=2, column=1)
self.LED1.grid(row=2, column=2)
self.LED2.grid(row=2, column=3)
self.LED3.grid(row=2, column=4)
self.LED4.grid(row=2, column=5)
self.LED5.grid(row=2, column=6)
self.LED6.grid(row=2, column=7)
self.LED7.grid(row=2, column=8)
self.button9 = Button(frame, text='Exit', command=self.close_window)
self.button9.grid(row=3, column=0)
def convert0(self, tog=[0]):
tog[0] = not tog[0]
if tog[0]:
print('Knight Rider ON')
self.button0.config(text='Knight Rider ON')
t=threading.Thread(target=self.LED)
t.start()
self.signal = True #added to stop thread
print('self.signal', self.signal)
print('tog[0]', tog[0])
self.LED0.config(image = logo)
else:
print('Knight Rider OFF')
self.button0.config(text='Knight Rider OFF')
self.signal = False #added to stop thread
print('self.signal', self.signal)
print('tog[0]', tog[0])
self.LED0.config(image = logo2)
def LED(self):
while self.signal: #added to stop thread
a=0
while self.signal: #added to stop thread
pfio.digital_write(a,1) #turn on
sleep(0.05)
pfio.digital_write(a,0) #turn off
sleep(0.05)
a=a+1
if a==7:
break
while self.signal: #added to stop thread
pfio.digital_write(a,1) #turn on
sleep(0.05)
pfio.digital_write(a,0) #turn off
sleep(0.05)
a=a-1
if a==0:
break
def close_window(self):
print('Knight Rider OFF')
print('self.signal', self.signal)
self.button0.config(text='Knight Rider OFF')
self.LED0.config(image = logo2)
self.signal = False #added to stop thread
print('self.signal', self.signal)
sleep(1)
print('Close Child window')
self.master.destroy() # I added this line to make the exit button work
root = Tk()
logo2 = PhotoImage(file="/home/pi/Off LED.gif")
logo = PhotoImage(file="/home/pi/Red LED.gif")
root.wm_title('LED on & off program')
app = App(root)
root.mainloop()回答 2
Stack Overflow用户
回答已采纳
发布于 2016-10-14 21:32:12
您不需要线程来完成这样简单的任务。如果任务不超过几百毫秒(如果需要更长的时间，UI将开始滞后)，那么在tkinter中设置持久的重复任务是非常容易的。
基本模式是编写一个做一些工作的函数，然后让该函数使用after再次调用它自己。例如：
def animate():
# do something, such as turn an led on or off
<some code here to turn one led on or off>
# run this again in 100 ms
root.after(100, animate)上面将创建一个无限循环，运行在tkinter的主循环中。只要<some code here... >不需要太长时间，动画就会显得流畅，你的UI也不会滞后。
示例
下面是这项技术的一个简单的工作示例。它使用一个简单的迭代器循环通过led，但您可以使用任何算法，您想要选择下一个led亮起。您也可以同时打开或关闭屏幕和硬件，或者打开或关闭多个led等等。
要使此代码可复制/粘贴，它使用彩色框架而不是图像，但如果需要，则可以使用图像。
import tkinter as tk # Tkinter for python 2
from itertools import cycle
class LEDStrip(tk.Frame):
segments = 16
speed = 100 # ms
def __init__(self, parent):
tk.Frame.__init__(self, parent)
leds = []
for i in range(self.segments):
led = tk.Frame(self, width=12, height=8, borderwidth=1,
relief="raised", background="black")
led.pack(side="left", fill="both", expand=True)
leds.append(led)
self.current_segment = None
self.iterator = cycle(leds)
def animate(self):
# turn off the current segment
if self.current_segment:
self.current_segment.configure(background="black")
# turn on the next segment
self.current_segment = next(self.iterator) # self.iterator.next() for python 2
self.current_segment.configure(background="red")
# run again in the future
self.after(self.speed, self.animate)
root = tk.Tk()
strip = LEDStrip(root)
strip.pack(side="top", fill="x")
# start the loop
strip.animate()
root.mainloop()Stack Overflow用户
发布于 2016-10-15 00:26:13
可能不是你想要的，但你可以得到一些灵感，建立这个“赛隆”算法在终端上的第一。LED没有中等的颜色值，但我想残余的光感应该能做到这一点。
import sys,time
shift = lambda l, n=1: l[n:]+l[:n]
c = u' ▁▂▃▄▅▆▇' # possible color values
L = 8 # number of items
B = L*[0] # indices of items
A = [0] + list(range(7)) + list(range(7,0,-1)) + 6*[0] # light sequence
C = L*[' '] # start blank
while 1:
B[A[0]]=L # set the most brilliant light
for x in range(L):
B[x]-= 1 # decrease all lights values
B[x] = max(0,B[x]) # not under 0
C[x] = c[B[x]] # get the corresponding 'color'
A = shift(A,-1) # shift the array to the right
sys.stdout.write(('%s%s%s')%(' ',''.join(C[1:]),'\r'))
time.sleep(0.1)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/40051731
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