Sorting both lists is unnecessary and not very scalable (order(NlogN)).
Assuming the lists do not contain duplicates,
just turn the longer one into a dict and check that each element of the
shorter list in that dict (e.g. "if j not in BigList: return false")
Since dict lookup is constant-time O(1), this approach is O(M)
i.e. speed is linear in the length of the shorter list;
and memory requirement is O(N+M) i.e. linear in the length
of the longer list. If M<<N this really saves you time.
Stephen
From: Jaggo <[EMAIL PROTECTED]>
Reply-To: [EMAIL PROTECTED]
To: tutor@python.org
Subject: Re: [Tutor] Simple way to compare Two lists
Date: Thu, 16 Aug 2007 10:11:14 -0700 (PDT)
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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