How can this behavior be acomplished? Python "short circuting" test

Question

Hello all,

I have the following code:

def testGeodatabase(self):
    geodatabaseList = self.gp.ListWorkspaces("*","ALL")
    for x in geodatabaseList:
        if x == self.outputGeodatabase:
            return True
        else:
            pass
    return False

What i need to know the following: in case the if condition evaluates to true, will the function stop looking in the list and never return False? Or do i need a break statement?

def testGeodatabase(self):
    geodatabaseList = self.gp.ListWorkspaces("*","ALL")
    for x in geodatabaseList:
        if x == self.outputGeodatabase:
            return True
            break
        else:
            pass
    return False

If the following code does not solve my problem, what can i use to do simulate that behavior?

Thanks

Answer 1

The return statement will indeed cause the function to be exited at that point. No further code is executed in the function.

Here is a simple test which you could run to prove the point:

def someFunction(nums):
    for i in nums:
        if i == 1:
            return "Found 1!"
    return "Never found 1"

And running it:

>>> someFunction([2])  
'Never found 1'  
>>> someFunction([2,1,3])  
'Found 1!'

Answer 2

return is the end of the line, and nothing else will happen in that function afterwards. On the other hand, you could rewrite your function as

def testGeodatabase(self):
    return self.outputGeodatabase in self.gp.ListWorkspaces("*","ALL")

Answer 3

You don't need the break keyword in the code above. Actually, you don't need the

else:
   pass

either. The

return True

will exit the function.

Answer 4

I think that using any() is the best choice:

def testGeodatabase(self):
    geodatabaseList = self.gp.ListWorkspaces("*","ALL")
    return any(x == self.outputGeodatabase for x in geodatabaseList)