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I'd like to compare the values in two different sets to test if any of
the positions in either set share the same value (e.g., if the third
element of each set is an 'a', then the test fails).
I have this:
def test_sets(original_set, trans_letters):
for pair in zip(original_set, trans_letters):
if pair[0] == pair[1]:
return False
return True
zip() was the first thing I thought of, but I was wondering if there's
some other way to do it, perhaps a builtin that actually does this kind
of testing.
Thanks.  
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John> I'd like to compare the values in two different sets to test if
John> any of the positions in either set share the same value (e.g., if
John> the third element of each set is an 'a', then the test fails).
Do you really mean "set" and not "list"? Note that they are unordered.
These two sets are equal:
set(['b', 'a', 'c'])
set(['a', 'b', 'c'])
Skip  
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Note that you are comparing ordered sequences, like lists, tuples,
strings, etc, and not sets. Something like this can be a little
improvement of your code, it avoids building the zipped list, and scans
the iterable unpacking it on the fly:
from itertools import izip
def test_sets(original_set, trans_letters):
for elem1, elem2 in izip(original_set, trans_letters):
if elem1 == elem2:
return False
return True
Bye,
bearophile  
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So you probably have to change the function test_sets name, because
it's not much useful on real sets.
Can't you use the == or != operators on those sequences?
Bye,
bearophile  
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John Salerno wrote: I'd like to compare the values in two different sets to test if any of the positions in either set share the same value (e.g., if the third element of each set is an 'a', then the test fails).
I have this:
def test_sets(original_set, trans_letters): for pair in zip(original_set, trans_letters): if pair[0] == pair[1]: return False return True
zip() was the first thing I thought of, but I was wondering if there's some other way to do it, perhaps a builtin that actually does this kind of testing.
There is no such concept as "position in [a] set". Sets in
math[s]/logic are *NOT* ordered. The order in which Python retrieves
elements when you do (for example) list(a_set) is a meaningless
artefact of the implementation du jour, and is not to be relied on. s = set(['xyzzy', 'plugh', 'sesame']) t = set(['xyzzy', 'plugh', 'mellon']) s
set(['sesame', 'plugh', 'xyzzy']) t
set(['plugh', 'mellon', 'xyzzy']) zip(s, t)
[('sesame', 'plugh'), ('plugh', 'mellon'), ('xyzzy', 'xyzzy')]
You may need one or more of these: s & t
set(['plugh', 'xyzzy']) s ^ t
set(['sesame', 'mellon']) s  t
set(['sesame', 'plugh', 'mellon', 'xyzzy']) (s  t)  t
set(['sesame']) (s  t)  s
set(['mellon'])
If that doesn't meet your needs:
back up a level and tell us what you are trying to achieve
If True:
read about sets in the Python docs
HTH,
John  
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> I'd like to compare the values in two different sets to test if any of the positions in either set share the same value (e.g., if the third element of each set is an 'a', then the test fails).
There's an inherant problem with this...sets by definition
are unordered, much like dictionaries. To compare them my
such means, you'd have to convert them to lists, sort the
lists by some ordering, and then compare the results.
Something like
s1 = set([1,3,5,7,9])
s2 = set([1,2,3])
list1 = list(s1)
list2 = list(s2)
list1.sort()
list2.sort()
if [(x,y) for x,y in zip(list1,list2) if x == y]:
print "There's an overlap"
else:
print "No matching elements"
Just to evidence matters, on my version of python (2.3.5 on
Debian), the following came back: set([1,3,5,7,9])
set([1,3,9,5,7])
That's not the original order, but the definition of a set
isn't hurt/changed by any ordering.
Thus, asking for the "position in a set" is an undefined
operation.
tkc
PS: for the above was done in 2.3.5 using this line:
from sets import Set as set  
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John Salerno <jo******@NOSPAMgmail.com> writes: I'd like to compare the values in two different sets to test if any of the positions in either set share the same value (e.g., if the third element of each set is an 'a', then the test fails).
I think by "sets" you mean "lists". Sets are unordered, as a few
people have mentioned.
I have this:
def test_sets(original_set, trans_letters): for pair in zip(original_set, trans_letters): if pair[0] == pair[1]: return False return True
That's fairly reasonable. You could use itertools.izip instead of
zip, which makes a generator instead of building up a whole new list
in memory. A more traditional imperativestyle version would be
something like:
def test_sets(original_set, trans_letters):
for i in xrange(len(original_set)):
if original_set[i] == trans_letters[i]:
return True
return False
You could even get cutesy and say something like (untested):
from itertools import izip
def test_sets(original_set, trans_letters):
return not sum(a==b for a,b in izip(original_set, trans_letters))
but that can be slower since it always scans both lists in entirety,
even if a matching pair of elements is found right away.
I don't offhand see a builtin function or nottooobscure oneliner
that shortcircuits, but maybe there is one.
Note that all the above examples assume the two lists are the same
length. Otherwise, some adjustment is needed.  
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John Salerno wrote: I'd like to compare the values in two different sets
Oops, I guess I was a little too loose in my use of the word 'set'. I'm
using sets in my program, but by this point they actually become
strings, so I'm really comparing strings.
Thanks for pointing that out to me, and I'll look into izip as well. I
was wondering if I could use an iterator for this somehow. :)  
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Paul Rubin wrote: You could even get cutesy and say something like (untested):
from*itertools*import*izip def*test_sets(original_set,*trans_letters): return*not*sum(a==b*for*a,b*in*izip(original_set,* trans_letters))
but that can be slower since it always scans both lists in entirety, even if a matching pair of elements is found right away.
Here's a variant that does performs only the necessary tests: from itertools import izip True not in (a == b for a, b in izip(range(3), range(3)))
False
A "noisy" equality test to demonstrate shortcircuiting behaviour:
def print_eq(a, b):
.... print "%r == %r > %r" % (a, b, a == b)
.... return a == b
.... True not in (print_eq(a, b) for a, b in izip(range(3), range(3)))
0 == 0 > True
False True not in (print_eq(a, b) for a, b in izip(["x", 1, 2], range(3)))
'x' == 0 > False
1 == 1 > True
False True not in (print_eq(a, b) for a, b in izip(["x", "x", "x"], range(3)))
'x' == 0 > False
'x' == 1 > False
'x' == 2 > False
True
Peter  
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Peter Otten <__*******@web.de> writes: Here's a variant that does performs only the necessary tests:
from itertools import izip True not in (a == b for a, b in izip(range(3), range(3)))
Cute!  
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John Salerno wrote: I'd like to compare the values in two different sets to test if any of the positions in either set share the same value (e.g., if the third element of each set is an 'a', then the test fails).
I have this:
def test_sets(original_set, trans_letters): for pair in zip(original_set, trans_letters): if pair[0] == pair[1]: return False return True
zip() was the first thing I thought of, but I was wondering if there's some other way to do it, perhaps a builtin that actually does this kind of testing.
Thanks.
'enumerate' is another possibility:
s1 = 'abcd'
s2 = 'zzzz'
s3 = 'zbzz'
s4 = 'zzbz'
def are_itemwise_different( L1, L2 ):
#if len(L1) != len(L2): return True
for idx, value in enumerate(L1):
if value == L2[idx]:
return False
return True
#after Peter Otten
def are_itemwise_different( L1, L2 ):
#if len(L1) != len(L2): return True
return True not in ( value == L2[idx] for idx, value in
enumerate(L1) )
assert are_itemwise_different(s1,s2)
assert not are_itemwise_different(s1,s3)
assert are_itemwise_different(s1,s4)
def itemwise_intersect( L1, L2 ):
#if len(L1) != len(L2): raise
for idx, value in enumerate(L1):
if value == L2[idx]:
yield value
assert list(itemwise_intersect(s1,s2)) == []
assert list(itemwise_intersect(s1,s3)) == ['b']
assert list(itemwise_intersect(s1,s4)) == []
Gerard  
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Gerard Flanagan wrote: John Salerno wrote: I'd like to compare the values in two different sets to test if any of the positions in either set share the same value (e.g., if the third element of each set is an 'a', then the test fails).
I have this:
def test_sets(original_set, trans_letters): for pair in zip(original_set, trans_letters): if pair[0] == pair[1]: return False return True
zip() was the first thing I thought of, but I was wondering if there's some other way to do it, perhaps a builtin that actually does this kind of testing.
Thanks.
'enumerate' is another possibility:
s1 = 'abcd' s2 = 'zzzz' s3 = 'zbzz' s4 = 'zzbz'
def are_itemwise_different( L1, L2 ): #if len(L1) != len(L2): return True for idx, value in enumerate(L1): if value == L2[idx]: return False return True
#after Peter Otten def are_itemwise_different( L1, L2 ): return True not in ( val == L2[idx] for idx, val in enumerate(L1) )
assert are_itemwise_different(s1,s2) assert not are_itemwise_different(s1,s3) assert are_itemwise_different(s1,s4)
s1 = 'abcd'
s2 = 'zzzz'
s3 = 'zbzz'
s4 = 'zzbz'
s5 = 'xbxx'
def itemwise_intersect( L1, L2 ):
return [value for idx, value in set(enumerate(L1)) &
set(enumerate(L2))]
assert itemwise_intersect(s1,s2) == []
assert itemwise_intersect(s1,s3) == ['b']
assert itemwise_intersect(s1,s4) == []
def itemwise_intersect( *args ):
s = set(enumerate(args[0]))
for t in ( set(enumerate(X)) for X in args[1:]):
s.intersection_update(t)
return [val for i,val in s]
assert itemwise_intersect(s1,s3,s5) == ['b']
Gerard   This discussion thread is closed Replies have been disabled for this discussion.   Question stats  viewed: 10606
 replies: 11
 date asked: May 14 '06
