palewire wrote:
In my application, I'd like to have a function that compares two values,
either of which may be null, and then classify the result depending on
whether one is higher or lower than the other.
I find myself writing something clunky like this, and I'm curious whether
anyone might know a more elegant way for parsing this out.
def compare(a, b):
if not a and not b:
return "No data"
else:
if a == 0 and b == 0:
return "Never had"
else:
if a == b:
return "Stayed the same"
elif a < b:
return "Gained"
elif a > b and b > 0:
return "Lost Some"
elif a > b and b == 0:
return "Lost All"
If there's some obvious way to search for this solution online, feel free to
slap me with that. I tried digging around on Google and couldn't come up
with much.
Thanks much.
Before one can "optimize" a function, one needs to get the function
correct. Looks to me that if a and b are both zero, it'll say "No data"
and never get to the "Never had" test. And if a is positive and b is
None, it'll return None, rather than any string.
Anyway, any "optimized" construct is going to be nearly as big as this.
You'd have to define a list of strings, and then compute an index value
that gave you the appropriate value. While I could probably come up
with such, I hardly think it's worthwhile. The cases are confusing
enough to me that I'd not want to make the code any less readable by
some formula & lookup.
That brings us to the question of what is more elegant. To me the
elegant code is readable first, and efficient (space & time) second. I
suspect efficiency would be nearly optimal by careful ordering of the
nested ifs. Find the most likely cases, and test for them first.
If this is a programming puzzle, for entertainment purposes, here's the
way I'd tackle making it "efficient" obfuscated code. Start by checking
if either value is None. Something like:
table = ("Lost Some","Lost All","Gained","Gained","Stayed the
same","Never had, never will")
def compare2(a,b):
global table
if a is None or b is None:
return "Invalid data"
return table[4*(a==b)+2*(a<b)+(b==0)]
davea
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