intersection
set.intersection_update
Updates s in place, keeping only elements found in s AND every given iterable. The method form accepts iterables; &= requires sets.
s.intersection_update(*others) | s &= otherParameters
| Parameter | Purpose |
|---|---|
| *others | Iterables (method) or sets (operator) to intersect with |
| returns | None — s is mutated |
Examples
>>> s = {1, 2, 3, 4}
>>> s.intersection_update([2, 3, 5])
>>> s
{2, 3}Retains only elements present in both.
>>> s = {1, 2, 3}
>>> s &= {2, 3, 4}
>>> s
{2, 3}&= augmented assignment form.
>>> s = {1, 2}
>>> s.intersection_update({3, 4})
>>> s
set()No overlap — s becomes empty.
>>> s = {1, 2, 3}
>>> s.intersection_update([1, 2], [2, 3])
>>> s
{2}Multiple iterables — must be in ALL of them.
Gotcha
Method form takes iterables; &= requires sets on both sides. To keep the original set intact, use intersection() which returns a new set.
Related methods
set.intersection
Returns a NEW set containing only elements found in s AND in every argument. Method form accepts any iterables; & requires sets.
set.update
Adds every element from each iterable argument to s in place. Equivalent to a union but mutates s instead of returning a new set.
set.difference_update
Removes from s in place every element found in any of the given iterables. Method form accepts iterables; -= requires sets.
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