print function variables if get exception - KeyboardInterrupt

Question

In the following simple example, I have a function foo() do something. There is a variable a inside the function. What I would like to do is that when the program catch the KeyboardInterrupt, the local functional variable a can be printed.

Clearly, the following way is not working. Is there anyway can do this?

def foo():
    a = 1
    for _ in xrange(10000):
        a += 1
    return a

try:
    a = foo()
except KeyboardInterrupt:
    print a

Updates (answer some comments below regarding why I want to Control-C):

Actually, my intention is to stop my Keras RNN training process earlier. Like the following code:

from keras.models import Sequential

def train():
    model = Sequential()
    ....
    model.build()
    ....
    for iteration in range(1, 200):
        ....
        model.fit(X_train, y_train, ...)
        ....
    ....


if __name__ == '__main__':
    try:
        model = train()
    except KeyboardInterrupt:
        model.save_weights("weights.hdf5")

If I stop the training process early, I would like to save the current model weights.

Well, from the docs: "Raised when the user hits the interrupt key (normally Control-C or Delete)". so do do you really want this or when the user press any key instead? — Eekan
– Eekan, Commented Apr 20, 2016 at 13:40
regarding why I want to Control-C, please see my question updates. — fluency03
– fluency03, Commented Apr 20, 2016 at 13:50

PM 2Ring · Accepted Answer · 2016-04-20 13:55:00Z

2

You just need to create your model outside the train function, and then pass model as an argument to train. That way you have a valid reference to model in case you interrupt train with Ctrl-C before it returns.

Eg,

from keras.models import Sequential

def train(model):
    model.build()
    ....
    for iteration in range(1, 200):
        ....
        model.fit(X_train, y_train, ...)
        ....
    ....


if __name__ == '__main__':
    try:
        model = Sequential()
        train(model)
    except KeyboardInterrupt:
        model.save_weights("weights.hdf5")

answered Apr 20, 2016 at 13:55

PM 2Ring

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Comments

user590028 · Accepted Answer · 2016-04-20 13:41:02Z

2

In your example, a is local to foo(). Either you need to make it global OR you want to add your exception catch inside foo(). Here's an example of the latter

def foo():
    try:
        a = 1
        for _ in xrange(10000):
            a += 1
        return a
    except KeyboardInterrupt:
        print a

answered Apr 20, 2016 at 13:41

user590028

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Comments

nephlm · Accepted Answer · 2016-04-20 13:42:30Z

0

Since foo() never completes a is never assigned. You'll need to break up your function so the caller knows it's last value at any given time.

def foo(a):
   return a + 1

try:
    for _ in xrange(10000):
        a = foo(a)
except KeyboardInterrupt:
    print a

Not sure there's much of a point of doing it that way, but for illustration. You could also use a global variable but I don't like recommending that for most situations.

answered Apr 20, 2016 at 13:42

nephlm

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