Return tuple with value in list

Question

I have this list of errors with their associated codes and descriptions:

Exception = namedtuple("Exception", "code name description")
exceptions = [
    Exception(1, "ILLEGAL FUNCTION", "Definition 1"),
    Exception(2, "ILLEGAL DATA ADDRESS", "Definition 2"),
    Exception(3, "ILLEGAL DATA VALUE", "Definition 2")
]

I would like a function that retrieves the exception with code==exception_code. I've searched around and the closest I could come up with is this:

# Returns the Exception tuple corresponding to the exception_code
def getException(self):
    return [item for item in exceptions if item[0] == self.exception_code]

This works but it's actually returning a list. My experience with Python is quite poor and I can't figure out how to simply return the tuple instead

NB: There will always be exactly one tuple with code == exception_code

Sample output of print x.getException with my current getException:

[Exception(code=2, name='ILLEGAL DATA ADDRESS', description='Definition 2')]

What? What use would that be? Also, if you don't want a list, why use a list comprehension? — jonrsharpe, Oct 03 '14 at 17:57

score 3 · Accepted Answer · answered Oct 03 '14 at 17:55

3

In that case, you're better off making exceptions a dict:

exceptions = {
    1: Exception(1, "ILLEGAL FUNCTION", "Definition 1"),
    2: Exception(2, "ILLEGAL DATA ADDRESS", "Definition 2"),
    3: Exception(3, "ILLEGAL DATA VALUE", "Definition 2")
}

From here, your getException function becomes trivial:

def getException(code):
    return exceptions[code]

... and you have to wonder whether the function call looks better than just inlining :-)

answered Oct 03 '14 at 17:55

mgilson

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Thanks this looks much better, will try it out and give you the point – Juicy Oct 03 '14 at 17:56

Return tuple with value in list

1 Answers1