集合差与集合减法

Question

What distinguishes - and .difference() on sets? Obviously the syntax is not the same, one is a binary operator, the other is an instance method. What else?

s1 = set([1,2,3])
s2 = set([3,4,5])

>>> s1 - s2
set([1, 2])
>>> s1.difference(s2)
set([1, 2])

Answer 1

set.difference, set.union... can take any iterable as the second arg while both need to be sets to use - , there is no difference in the output.

Operation         Equivalent   Result
s.difference(t)   s - t        new set with elements in s but not in t

With .difference you can do things like:

s1 = set([1,2,3])

print(s1.difference(*[[3],[4],[5]]))

{1, 2}

It is also more efficient when creating sets using the *(iterable,iterable) syntax as you don't create intermediary sets, you can see some comparisons here

Answer 2

On a quick glance it may not be quite evident from the documentation but buried deep inside a paragraph is dedicated to differentiate the method call with the operator version

Note, the non-operator versions of union(), intersection(), difference(), and symmetric_difference(), issubset(), and issuperset() methods will accept any iterable as an argument. In contrast, their operator based counterparts require their arguments to be sets. This precludes error-prone constructions like set('abc') & 'cbs' in favor of the more readable set('abc').intersection('cbs') .

Answer 3

The documentation appears to suggest that difference can take multiple sets, so it is possible that it might be more efficient and clearer for things like:

s1 = set([1, 2, 3, 4])
s2 = set([2, 5])
s3 = set([3, 6])
s1.difference(s2, s3) # instead of s1 - s2 - s3

but I would suggest some testing to verify.