I have a simple GUI which run various scripts from another python file, everything works fine until the GUI is running a function which includes a while loop, at which point the GUI seems to crash and become in-active. Does anybody have any ideas as to how this can be overcome, as I believe this is something to do with the GUI being updated,Thanks. Below is a simplified version of my GUI.
GUI
#!/usr/bin/env python
# Python 3
from tkinter import *
from tkinter import ttk
from Entry import ConstrainedEntry
import tkinter.messagebox
import functions
AlarmCode = "2222"
root = Tk()
root.title("Simple Interface")
mainframe = ttk.Frame(root, padding="3 3 12 12")
mainframe.grid(column=0, row=0, sticky=(N, W, E, S))
mainframe.columnconfigure(0, weight=1)
mainframe.rowconfigure(0, weight=1)
ttk.Button(mainframe, width=12,text="ButtonTest",
command=lambda: functions.test()).grid(
column=5, row=5, sticky=SE)
for child in mainframe.winfo_children():
child.grid_configure(padx=5, pady=5)
root.mainloop()
functions
def test():
period = 0
while True:
if (period) <=100:
time.sleep(1)
period +=1
print(period)
else:
print("100 seconds has passed")
break
What will happen in the above is that when the loop is running the application will crash. If I insert a break in the else statement after the period has elapsed, everything will work fine. I want users to be able to click when in loops as this GUI will run a number of different functions.
Don't use time.sleep in the same thread than your Tkinter code: it freezes the GUI until the execution of test is finished. To avoid this, you should use after widget method:
# GUI
ttk.Button(mainframe, width=12,text="ButtonTest",
command=lambda: functions.test(root))
.grid(column=5, row=5, sticky=SE)
# functions
def test(root, period=0):
if period <= 100:
period += 1
print(period)
root.after(1000, lambda: test(root, period))
else:
print("100 seconds has passed")
Update:
In your comment you also add that your code won't use time.sleep, so your original example may not be the most appropiate. In that case, you can create a new thread to run your intensive code.
Note that I posted the alternative of after first because multithreading should be used only if it is completely necessary - it adds overhead to your applicacion, as well as more difficulties to debug your code.
from threading import Thread
ttk.Button(mainframe, width=12,text="ButtonTest",
command=lambda: Thread(target=functions.test).start())
.grid(column=5, row=5, sticky=SE)
# functions
def test():
for x in range(100):
time.sleep(1) # Simulate intense task (not real code!)
print(x)
print("100 seconds has passed")
Related
when I use threading.Thread to create new thread.it can not start. The code like this
import threading
import time
import sys
def worker():
count = 1
while True:
if count >= 6:
break
time.sleep(1)
count += 1
print("thread name = {}, thread id = {}".format(threading.current_thread().name,threading.current_thread().ident))
t1 = threading.Thread(target=worker,name="t1")
t2 = threading.Thread(target=worker,name='t2')
t1.start()
t2.start()
t1.join()
t2.join()
When I run this code. The windbg will not report error 、not print any thing and never return
enter image description here
I will to create new thread to run something
Don't use 'threading' within windbg. Windbg has own multithreading model and loop of debug events. It is near impossible to run all this threads together without bugs.
In fact I dont't recomend to use 'threading' also in standalone python program with pykd module. All my scripts always use 'multiprocessing' module.
I'm trying to build a regression function that call itself in a new process. The new process should not stop the parent process nor wait for it to finish, that is why I don't use join(). Do you have another way to create regression function with multi-process.
I use the following code:
import multiprocessing as mp
import concurrent.futures
import time
def do_something(c, seconds, r_list):
c += 1 # c is a counter that all processes should use
# such that no more than 20 processes are created.
print(f"Sleeping {seconds} second(s)...")
if c < 20:
P_V = mp.Value('d', 0.0, lock=False)
p = mp.Process(group=None, target=do_something, args=(c, 1, r_list,))
p.start()
if not p.is_alive():
r_list.append(P_V.value)
time.sleep(seconds)
print(f"Done Sleeping...{seconds}")
return f"Done Sleeping...{seconds}"
if __name__ == '__main__':
C = 0 # C is a counter that all processes should use
# such that no more than 20 processes are created.
Result_list = [] # results that come from all processes are saved here
Result_list.append(do_something(C, 1, Result_list))
Notice that results from all processes should be compared at the end.
In fact, this code is working well but the child processes, which are created in the recursive method, do not print anything, the list "Result_list" contains only one item from the first call, and C=0 at the end, any idea why?
Here's a simplified example of what I think you're trying to do (side note: launching processes recursively is a great way to accidentally create a "fork bomb". It is extremely more common to create multiple processes in some sort of loop instead)
from multiprocessing import Process, Queue
from time import sleep
from os import getpid
def foo(n_procs, return_Q, arg):
if __name__ == "__main__": #don't actually run the body of foo in the "main" process, just start the recursion
Process(target=foo, args=(n_procs, return_Q, arg)).start()
else:
n_procs -= 1
if n_procs > 0:
Process(target=foo, args=(n_procs, return_Q, arg)).start()
sleep(arg)
print(f"{getpid()} done sleeping {arg} seconds")
return_Q.put(f"{getpid()} done sleeping {arg} seconds") #put the result to a queue so we can get it in the main process
if __name__ == "__main__":
q = Queue()
foo(10, q, 2)
sleep(10) #do something else in the meantime
results = []
#while not q.empty(): #usually better to just know how many results you're expecting as q.empty can be unreliable
for _ in range(10):
results.append(q.get())
print("mp results:")
print("\n".join(results))
I am trying to learn Threading and Multiprocessing on a MacOS. I am unable to launch the processes though, with python giving the following error message.
Error
RuntimeError:
An attempt has been made to start a new process before the
current process has finished its bootstrapping phase.
This probably means that you are not using fork to start your
child processes and you have forgotten to use the proper idiom
in the main module:
if __name__ == '__main__':
freeze_support()
...
The "freeze_support()" line can be omitted if the program
is not going to be frozen to produce an executable.
my code:
parallel_processing.py
import multiprocessing
import time
start= time.perf_counter()
def do_something():
print('sleeping 1 sec....')
time.sleep(1)
return('done sleeping...')
# do_something()
p1 = multiprocessing.Process(target = do_something)
p2 = multiprocessing.Process(target = do_something)
p1.start()
p2.start()
finish= time.perf_counter()
print(f'finished in {finish-start} seconds')
It seems that you can resolve the issue by putting your code inside the block if __name__ == '__main__'. For example:
import multiprocessing as mp
import time
def do_something():
...
if __name__ == '__main__':
p1 = mp.Process(target=do_something)
p2 = mp.Process(target=do_something)
p1.start()
p2.start()
...
I'm not sure this is resulted from the recent macOS update or python, but it seems to have to do with how OS creates new threads, i.e. forking versus spwaning. Please see this blog post for details.
I am trying to allow a user to input multiple answers but only within an allocated amount of time. The problem is I have it running but the program will not interrupt the input. The program will only stop the user from inputing if the user inputs an answer after the time ends. Any ideas? Is what I am trying to do even possible in python?
I have tried using threading and the signal module however they both result in the same issue.
Using Signal:
import signal
def handler(signum, frame):
raise Exception
def answer_loop():
score = 0
while True:
answer = input("Please input your answer")
signal.signal(signal.SIGALRM, handler)
signal.alarm(5)
try:
answer_loop()
except Exception:
print("end")
signal.alarm(0)
Using Threading:
from threading import Timer
def end():
print("Time is up")
def answer_loop():
score = 0
while True:
answer = input("Please input your answer")
time_limit = 5
t = Timer(time_limit, end)
t.start()
answer_loop()
t.cancel()
Your problem is that builtin input does not have a timeout parameter and, AFAIK, threads cannot be terminated by other threads. I suggest instead that you use a GUI with events to finely control user interaction. Here is a bare bones tkinter example.
import tkinter as tk
root = tk.Tk()
label = tk.Label(root, text='answer')
entry = tk.Entry(root)
label.pack()
entry.pack()
def timesup():
ans = entry.get()
entry.destroy()
label['text'] = f"Time is up. You answered {ans}"
root.after(5000, timesup)
root.mainloop()
I would like a label to be .grid() then the program to wait 3 seconds and then .grid_forget(). I am very confused at the point .grid is executed. For example:
def remove_choice(self):
while True:
try:
get = int(self.entry_remove_choice.get())
except ValueError:
self.label_error_remove.grid(row=10,column=6) #A
time.sleep(3)
self.label_error_remove.grid_forget() #B
#Empty entry box
break
else:
#continue code
break
Once the button is pressed and remove_choice is executed, the button is displayed to be pressed in for three seconds then #A and #B are executed in one go and nothing is displayed.
If #B is removed then the error message is displayed after three seconds.
If #A and #B are swapped for print to terminal then program works how you would think, with one message, a wait of three seconds, then another message.
If you do a very sloppy solution (which Im not that bothered about for this program) and do this:
def remove_choice(self):
while True:
try:
get = int(self.entry_remove_choice.get())
except ValueError:
self.label_error_remove.grid(row=10,column=6) #A
for n in range (1,1000):
print("abc")
self.label_error_remove.grid_forget()
break
else:
#continue code
break
When executed "abc" is printed 1000 times taking around 1.5 seconds and then after this the program displays the grid.
Any suggestions to how to make TKinter wait please.
Also can someone explain why grid works like this, thanks.
Rather than trying 'forgetting' the label each time, why not just clear the error message text?
My example below will wait for the user to press the button and display the error message for 3 seconds. I'm using the .after method to schedule the hideError method 3 seconds (3000 ms) after the error message is displayed.
try:
import tkinter as tk
except:
import Tkinter as tk
import time
class App(tk.Frame):
def __init__(self,master=None,**kw):
tk.Frame.__init__(self,master=master,**kw)
self.errorMessage = tk.StringVar()
self.ErrorLabel = tk.Label(textvar=self.errorMessage)
self.ErrorLabel.grid()
self.button = tk.Button(text="Press Me",command=self.showError)
self.button.grid()
def showError(self):
# Disable the button and show the error message
self.button['state'] = tk.DISABLED
self.errorMessage.set("Error Message!!!!")
self.after(3000,self.hideError)
def hideError(self):
#Enable the button and clear the error message.
self.button['state'] = tk.NORMAL
self.errorMessage.set("")
if __name__ == '__main__':
root = tk.Tk()
App(root).grid()
root.mainloop()
It is considered bad practice to use while True loops or time.sleep inside GUI applications. They prevent the GUI from updating so in your code both actions appear to happen at the same time because the time.sleep operation is blocking the GUI and preventing the screen from being redrawn.
EDIT: Passing arguments from callbacks.
Current problem is that the after method expects to receive a reference to a function. self.hideError(3) returns NoneType not reference to a function call. We can solve this using anonymous functions and lambda.
I've started to use this snippet of code to help, its from guizero
def with_args( func_name, *args):
"""Helper function to make lambda functions easier
Thanks to guizero"""
return lambda: func_name(*args)
Then in your main section of code the line would look like this.
self.after(3000,with_args(self.hideError,3))
EDIT: There is an even simpler way. The .after method can take arguments itself.
self.after(3000,self.hideError,3)