How come when I change the order of the two sets in the unions below, I get different results?
set1 = {1, 2, 3}
set2 = {True, False}
print(set1 | set2)
# {False, 1, 2, 3}
print(set2 | set1)
#{False, True, 2, 3}
The 1
and True
are equivalent and considered to be duplicates. Likewise the 0
and False
are equivalent as well:
>>> 1 == True
True
>>> 0 == False
True
When multiple equivalent values are encountered, sets keep the first one seen:
>>> {0, False}
{0}
>>> {False, 0}
{False}
To get them to be treated as distinct, just store them in a (value, type)
pair:
>>> set1 = {(1, int), (2, int), (3, int)}
>>> set2 = {(True, bool), (False, bool)}
>>> set1 | set2
{(3, <class 'int'>), (1, <class 'int'>), (2, <class 'int'>),
(True, <class 'bool'>), (False, <class 'bool'>)}
>>> set1 & set2
set()
Another way to make the values distinct is to store them as strings:
>>> set1 = {'1', '2', '3'}
>>> set2 = {'True', 'False'}
>>> set1 | set2
{'2', '3', 'False', 'True', '1'}
>>> set1 & set2
set()
Hope this clears up the mystery and shows the way forward :-)
Rescued from the comments:
This is the standard technique for breaking cross-type equivalence (i.e. 0.0 == 0
, True == 1
, and Decimal(8.5) == 8.5)
. The technique is used in Python 2.7's regular expression module to force unicode regexes to be cached distinctly from otherwise equivalent str regexes. The technique is also used in Python 3 for functools.lru_cache() when the typed parameter is true.
If the OP needs something other than the default equivalence relation, then some new relation needs to be defined. Depending the use case, that could be case-insensitivity for strings, normalization for unicode, visual appearance (things that look different are considered different), identity (no two distinct objects are considered equal), a value/type pair, or some other function that defines an equivalence relation. Given the OPs specific example, it would seem that he/she expected either distinction by type or visual distinction.