Thank you Kent, Michael, Tom and anyone else I'm forgetting who took time to 
reply.

I don't work quite so fast, very limited personal computer time means I only do 
it on weekends,

I read through your suggestions and eventually found a way to speed-up the 
proccess through sorting the Two lists, then manually iterating through each of 
them. This way I've completely canceled the need to compare Two lists: instead 
just ensuring I start from a point not taken in One list and having to only 
check whether Item not in BigList.

[If anyone's interested, I should have the script finished and thoroughly 
tested on, ah, next weekend, and I could post a link here.]

Again, Thx.
-Omer.

Message: 1
Date: Fri, 10 Aug 2007 08:11:47 -0400
From: Kent Johnson 
Subject: Re: [Tutor] Simple way to compare Two lists
To: Tom Fitzhenry , tutor@python.org
Message-ID: <[EMAIL PROTECTED]>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed

Tom Fitzhenry wrote:
> On Fri, Aug 10, 2007 at 02:54:44AM -0700, Jaggo wrote:
>> Can anyone think of any better way?
> 
> If SmallList and BigList are sorted (in order), there is a faster method:
> 
> def IsAPartOfList(SmallList,BigList):
>     for i in BigList:
>         for j in SmallList:
>             if i==j:
>                 return true
>             if i>j:
>                 break
>     return false
> 
> (I'm not sure how encouraged using break statements are, so wait for a tutor 
> to
> answer)

break is fine! If the list you are searching is sorted you can use the 
bisect module to do a binary search instead of the linear search above.

> If one list is already sorted but the other isn't, it may still be faster to
> sort the unsorted list then use the method above.

I don't think BigList has to be sorted in the above algorithm. If both 
lists are sorted I suppose you could write it like a merge sort, walking 
along both lists looking for a match.

Kent



       
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