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# Problem involving sets...

Kind of a fun but confusing problem...

I have two lists. Each list contains elements of two-element lists.
l1 = [['c1',1],['c2',2],['c3',4]]
l2 = [['c1',1],['c2',2],['c4',4],['c3',3]]

Exactly in this format, where
superlist[0][0] is always a string
superlist[0][1] is always an integer

Now what I would like to do is find the intersect of those two
super-lists based on superlist[0][0] and then compare the integers to
find ultimately:
A list of strings of the intersect of l1/l2, where the l1[x][1] >
l2[x][1]
In the case of the above example, that would be simply:
['c3']
But since lists are unhashable, I can't make a set consisting of
lists... Maybe I am making this problem too difficult by thinking in
terms of sets. Does anyone have a simple solution????

-Thanks

Apr 14 '06 #1
4 3223
Em Sex, 2006-04-14 Ã*s 15:43 -0700, flamesrock escreveu:
Does anyone have a simple solution????

\$ python2.4
Python 2.4.3 (#2, Mar 30 2006, 21:52:26)
[GCC 4.0.3 (Debian 4.0.3-1)] on linux2
l1 = [['c1',1],['c2',2],['c3',4]]
l2 = [['c1',1],['c2',2],['c4',4],['c3',3]]
print [tuple(x) for x in l1] [('c1', 1), ('c2', 2), ('c3', 4)] print [tuple(x) for x in l2] [('c1', 1), ('c2', 2), ('c4', 4), ('c3', 3)] s1 = frozenset(tuple(x) for x in l1)
s2 = frozenset(tuple(x) for x in l2)
print s1 frozenset([('c2', 2), ('c1', 1), ('c3', 4)]) print s2 frozenset([('c2', 2), ('c4', 4), ('c3', 3), ('c1', 1)]) print s2-s1 frozenset([('c4', 4), ('c3', 3)]) print[list(x) for x in s2-s1]

[['c4', 4], ['c3', 3]]
--
Felipe.

Apr 14 '06 #2
flamesrock wrote:
Kind of a fun but confusing problem...

I have two lists. Each list contains elements of two-element lists.
l1 = [['c1',1],['c2',2],['c3',4]]
l2 = [['c1',1],['c2',2],['c4',4],['c3',3]]

Exactly in this format, where
superlist[0][0] is always a string
superlist[0][1] is always an integer

Now what I would like to do is find the intersect of those two
super-lists based on superlist[0][0] and then compare the integers to
find ultimately:
A list of strings of the intersect of l1/l2, where the l1[x][1] >
l2[x][1]
In the case of the above example, that would be simply:
['c3']

In [5]: l1 = [['c1',1],['c2',2],['c3',4]]

In [6]: l2 = [['c1',1],['c2',2],['c4',4],['c3',3]]

In [7]: d=dict(l1)

In [10]: [k for k,v in l2 if k in d and v < d[k]]
Out[10]: ['c3']

Kent
Apr 15 '06 #3
wowowow now THAT is what I call simple!

Apr 15 '06 #4
Kent Johnson wrote:
flamesrock wrote:
Kind of a fun but confusing problem...

I have two lists. Each list contains elements of two-element lists.
l1 = [['c1',1],['c2',2],['c3',4]]
l2 = [['c1',1],['c2',2],['c4',4],['c3',3]]

Exactly in this format, where
superlist[0][0] is always a string
superlist[0][1] is always an integer

Now what I would like to do is find the intersect of those two
super-lists based on superlist[0][0] and then compare the integers
to find ultimately:
A list of strings of the intersect of l1/l2, where the l1[x][1] >
l2[x][1]
In the case of the above example, that would be simply:
['c3']

In [5]: l1 = [['c1',1],['c2',2],['c3',4]]

In [6]: l2 = [['c1',1],['c2',2],['c4',4],['c3',3]]

In [7]: d=dict(l1)

In [10]: [k for k,v in l2 if k in d and v < d[k]]
Out[10]: ['c3']

Kent

The dict solution posted above is definitely better than the following,
but purely out of idle curiosity I was interested to see if this could
be done in one line using just list comprehensions ...

Python 2.4.2 (#1, Nov 15 2005, 15:54:06)
[GCC 3.3.6 (Gentoo 3.3.6, ssp-3.3.6-1.0, pie-8.7.8)] on linux2
l1 = [['c1',1],['c2',2],['c3',4]]
l2 = [['c1',1],['c2',2],['c4',4],['c3',3]]
[k2 for [k2,v2] in l2 if k2 in [k1 for k1,v1 in l1] and v2 < [v1

for k1,v1 in l1 if k1 == k2][0]]
['c3']

.... yup :-)
Dave.
--

Apr 15 '06 #5

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