Thanks for the explanation - very clear.

Cheers,
Bodsda 
Sent from my BlackBerry® wireless device

-----Original Message-----
From: Steven D'Aprano <[email protected]>
Sender: [email protected]
Date: Fri, 16 Sep 2011 20:43:45 
To: Tutor - python List<[email protected]>
Subject: Re: [Tutor] If statement optimization

[email protected] wrote:
> Hi,
> 
> In a normal if,elif,elif,...,else statement, are the conditions checked in a 
> linear fashion?

Yes.


> I am wondering if I should be making an effort to put the most likely true 
> condition at the beginning of the block

Probably not. The amount of time used in the average if...elif is 
unlikely to be significant itself. Don't waste your time trying to 
optimize something like this:


if n < 0:
    ...
elif n == 0:
    ...
else:
    ...


However, there are exceptions.

If the tests are very expensive, then it might be worthwhile putting the 
most likely case first. Or at least, put the cheapest cases first, leave 
the expensive ones for last:

if sum(mylist[1:]) > 1000 and mylist.count(42) == 3 and min(mylist) < 0:
     ...
elif len(mylist) < 5:
     ...

I probably should swap the order there, get the cheap len() test out of 
the way, and only perform the expensive test if that fails.


If you have LOTS of elif cases, like *dozens*, then firstly you should 
think very hard about re-writing your code, because that's pretty poor 
design... but if you can't change the design, then maybe it is 
worthwhile to rearrange the cases.

If you have something like this:


if s == "spam":
     func_spam(x)
elif s == "ham":
     func_ham(x)
elif s == "cheese":
     func_cheese(x)


you can often turn this into a dispatch table:


table = {"spam": func_spam, "ham": func_ham, "cheese": func_cheese}
func = table[s]  # lookup in the dispatch table
func(x)  # finally call the function


Note that inside table, you don't call the functions.

This pattern is especially useful, as lookup in a table in this manner 
takes close enough to constant time, whether there is one item or a million.



-- 
Steven

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