I am running a function in a Jupyter notebook and I would like to know if it's possible to execute a bit of code if the user interrupts the kernel.
For example if you have this function:
import time
def time_sleep():
time.sleep(5)
print('hello')
Can I add a print('cell terminated') that runs if the cell is interrputed ?
Keyboard interrupts (ctrl+c) manifest as KeyboardInterrupt exceptions, so
try:
time.sleep(5)
print('That was a nice nap.')
except KeyboardInterrupt:
print('What a rude awakening!')
works if Jupyter sends a real interrupt signal to the kernel (and apparently it does!).
Related
Python 3.10 in macOS terminal
I kick off this simple script in macOS terminal, which runs infinitely (so please be warned when running it):
import itertools
for i in itertools.count(start=1):
try:
p = pow(2,i)
print(p)
except KeyboardInterrupt:
print("Loop terminated")
raise
Yet clicking on the keyboard does not terminate the loop.
What am I missing?
Thanks to everyone for the clarification. The problem is I was under the assumption that KeyboardInterrupt would respond to any key pressed. It actually only responds to Ctrl-C. Everything works as expected when using Ctrl-C.
from os.path import join
string=" Congratulations, you are about to embark upon one of life’s "
path=r"C:\Users\Nord.Kind\Desktop"
file="test.txt"
quit()
# This should not execute, but it does!!
with open(join(path,file),"w")as wfile:
wfile.write(string)
wfile.close()
In the above code example the code still executes the write in file command, even though its after a quit.
The same behavious occures when I use exit.
I am using Spyder 3.6
Also the kernel restarts each time I use exit or quit.
Any help?
(Spyder maintainer here) Your question contains this comment:
Also the kernel restarts each time I use exit or quit.
That's the behavior of the IPython kernel we use as a backend to execute users code. Those commands kill the kernel and that forces a kernel restart to maintain its associated console active. I'm afraid there's nothing you can do about it.
Note: The same happens in the Jupyter notebook.
One way is to use sys.exit() in lieu of quit()
import sys
... # code that executes
sys.exit()
... # this code won't execute
However, as noted by #AranFey in the comments, your code will throw an error if it attempts to execute the last part where the variable read is not defined.
You can use SystemExit:
# Code that will run
raise SystemExit
# Code that will not run
sys.exit() also raises this error but this doesn't require importing sys.
I have a python code that runs from the command line (Windows) and crashes on a MemoryError.
While I work on solving the error I want the code to re-run, with a flag, for example python c:\python_code.py --rerun, until I get a successful exit code.
Is this possible in Windows?
Also, it is key that the code restart, otherwise it is my understanding that the memory is not cleared.
Thank you
you can wrap your code around:
if flag==rerun:
while True:
try:
os.system('python python_code.py')
break
except MemoryError:
pass
I have multiple python files to run. How would I launch all those files within one .py script? This is what I came up with but it shows the screen action and really doesn't begin the other stuff unless I exit out of it. Here's the code, not much:
import os
print("Launching Bot, just for you.")
print("Loading up shard 0")
try:
os.system("screen python3.5 run_0.py > /dev/null")
except:
print("Shard 0 failed")
print("Loading up shard 1")
try:
os.system("screen python3.5 run_1.py > /dev/null")
except:
print("Shard 1 failed")
print("Done running shards...")
I was doing some research and they said to use subprocess but when I used it, it didn't run my command properly. (I don't have a copy of that code, I lost it).
The problem is that I want to run the python script and it works fine but I have to close the screen to start the other one and I just want it to run the command w/o showing the output. Can you help?
You should use import subprocess in a python file. You can then start other instance of other programs with :
subprocess.Popen([sys.executable, "newprogram.py"])
You can mix that with multiprocessing package to launch one thread by new program
p = multiprocessing.Process(target= mp_worker , args=( ))
p.start()
where mp_worker launches the other program.
My Python try/except loop does not seem to trigger a keyboard interrupt when Ctrl + C is pressed while debugging my code in PyCharm. (The same issue occurs when using Ctrl + C while running the program, but not in the PyCharm Python console.)
My code look like this:
try:
while loop:
print("busy")
except KeyboardInterrupt:
exit()
The full code can be viewed here. The code above produces the same error.
I know this is an old question, but I ran into the same problem and think there's an easier solution:
In PyCharm go to "Run"/"Edit Configurations" and check "Emulate terminal in output console".
PyCharm now accepts keyboard interrupts (make sure the console is focused).
Tested on:
PyCharm 2019.1 (Community Edition)
From your screen shot it appears that you are running this code in an IDE. The thing about IDEs is that they are not quite the same as running normally, especially when it comes to handling of keyboard characters. The way you press ctrl-c, your IDE thinks you want to copy text. The python program never sees the character. Pehaps it brings up a separate window when running? Then you would select that window before ctrl-c.
PyCharm's Python Console raises the exception console_thrift.KeyboardInterruptException on Ctrl-C instead of KeyboardInterrupt. The exception console_thrift.KeyboardInterruptException is not a subclass of KeyboardInterrupt, therefore not caught by the line except KeyboardInterrupt.
Adding the following lines would make your script compatible with PyCharm.
try:
from console_thrift import KeyboardInterruptException as KeyboardInterrupt
except ImportError:
pass
This would not break compatibility with running the script in a terminal, or other IDE, like IDLE or Spyder, since the module console_thrift is found only within PyCharm.
If that comment doesn't solve your problem, (from #tdelaney) you need to have your shell window focused (meaning you've clicked on it when the program is running.) and then you can use Control+C
You can also use PyCharm's Python console and use Ctrl + C, if you catch the exception that PyCharm raises when Ctrl + C is pressed. I wrote a short function below called is_keyboard_interrupt that tells you whether the exception is KeyboardInterrupt, including PyCharm's. If it is not, simply re-raise it. I paste a simplified version of the code below.
When it is run:
type 'help' and press Enter to repeat the loop.
type anything else and press Enter to check that ValueError is handled properly.
Press Ctrl + C to check that KeyboardInterrupt is caught, including in PyCharm's python console.
Note: This doesn't work with PyCharm's debugger console (the one invoked by "Debug" rather than "Run"), but there the need for Ctrl + C is less because you can simply press the pause button.
I also put this on my Gist where I may make updates: https://gist.github.com/yulkang/14da861b271576a9eb1fa0f905351b97
def is_keyboard_interrupt(exception):
# The second condition is necessary for it to work with the stop button
# in PyCharm Python console.
return (type(exception) is KeyboardInterrupt
or type(exception).__name__ == 'KeyboardInterruptException')
try:
def print_help():
print("To exit type exit or Ctrl + c can be used at any time")
print_help()
while True:
task = input("What do you want to do? Type \"help\" for help:- ")
if task == 'help':
print_help()
else:
print("Invalid input.")
# to check that ValueError is handled separately
raise ValueError()
except Exception as ex:
try:
# Catch all exceptions and test if it is KeyboardInterrupt, native or
# PyCharm's.
if not is_keyboard_interrupt(ex):
raise ex
print('KeyboardInterrupt caught as expected.')
print('Exception type: %s' % type(ex).__name__)
exit()
except ValueError:
print('ValueError!')
Here is working normally, since i put a variable "x" in your code and i use tabs instead spaces.
try:
def help():
print("Help.")
def doStuff():
print("Doing Stuff")
while True:
x = int(input())
if x == 1:
help()
elif x == 2:
doStuff()
else:
exit()
except KeyboardInterrupt:
exit()
Try shift + control + C. It worked for me.
Make sure the window is selected when you press ctrl+c. I just ran your program in IDLE and it worked perfectly for me.
One possible reason if <Strg+C> does not stop the program:
When a text is marked in the shell, <Strg+C> is interpreted as "copy the marked text to clipboard".
Just unmark the text and press <Strg+C> again.