Learned list comprehension today, its cool:
>>> E1=[('a','g'),('r','s')]
>>> E2=[('g','a'),('r','q'),('f','h')]
>>> [x for x in E1 for y in E2 if x == y or (x[1],x[0])==y]
[('a', 'g')]
On Sunday 20 February 2005 09:24 am, [EMAIL PROTECTED] wrote:
> Hi,
> I have 2 lists of tuples that look like:
> E1=[('a','g'),('r','s')] and
> E2=[('g','a'),('r','q'),('f','h')].
> In this tuple, the ordering does not
> matter, i.e. (u,v) is the same as (v,u).
>
> What I want to do is the following:
> given 2 list of tuples, E1 and E2, I want to create another list with
> tuples that are common to both. So in the above example I would like
> to return ('a','g') as being common.
> So far I have the code below but it does not work out properly, I would
> be grateful if someone can help me out .
> thanks
>
> def list_of_tuples_intersect(E1,E2):
>
> S1=Set()
> S2=Set()
> for e in E1:
> S1.add((e[0],e[1]))
> S1.add((e[1],e[0]))
> for e in E2:
> S2.add((e[0],e[1]))
> S2.add((e[1],e[0]))
> S= S1 & S2
> SS=Set()
> done=Set()
>
> for e in S:
> if ((e[0],e[1]) in done) or ((e[1],e[0]) in done):
> continue
> else:
> SS.add((e[0],e[1]))
> done.add((e[0],e[1]))
> done.add((e[1],e[0]))
> return SS
--
James Stroud, Ph.D.
UCLA-DOE Institute for Genomics and Proteomics
Box 951570
Los Angeles, CA 90095
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