comparing values in two sets

Discussion in 'Python' started by John Salerno, May 14, 2006.

1. John SalernoGuest

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.

John Salerno, May 14, 2006

2. Guest

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

, May 15, 2006

3. Guest

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

, May 15, 2006
4. Guest

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

, May 15, 2006
5. John MachinGuest

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/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:

HTH,
John

John Machin, May 15, 2006
6. Tim ChaseGuest

> 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

Tim Chase, May 15, 2006
7. Paul RubinGuest

John Salerno <> 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 imperative-style version would be
something like:

def test_sets(original_set, trans_letters):
for i in xrange(len(original_set)):
if original_set == trans_letters:
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 not-too-obscure one-liner
that short-circuits, but maybe there is one.

Note that all the above examples assume the two lists are the same
length. Otherwise, some adjustment is needed.

Paul Rubin, May 15, 2006
8. John SalernoGuest

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.

John Salerno, May 15, 2006
9. Peter OttenGuest

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 short-circuiting 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

Peter Otten, May 15, 2006
10. Paul RubinGuest

Peter Otten <> 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!

Paul Rubin, May 15, 2006
11. Gerard FlanaganGuest

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

Gerard Flanagan, May 15, 2006
12. Gerard FlanaganGuest

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

Gerard Flanagan, May 15, 2006