add is faster for a single element because it is exactly for that purpose, adding a single element:
In [5]: timeit a.update([1])
10000000 loops, best of 3: 191 ns per loop
In [6]: timeit a.add(1)
10000000 loops, best of 3: 69.9 ns per loop
update expects an iterable or iterables so if you have a single hashable element to add then use add if you have an iterable or iterables of hashable elements to add use update.
s.add(x) add element x to set s
s.update(t) s |= t return set s with elements added from t
set.add adds an individual element to the set. So,
>>> a = set()
>>> a.add(1)
>>> a
set([1])
works, but it cannot work with an iterable, unless it is hashable. That is the reason why a.add([1, 2]) fails.
>>> a.add([1, 2])
Traceback (most recent call last):
File "<input>", line 1, in <module>
TypeError: unhashable type: 'list'
Here, [1, 2] is treated as the element being added to the set and as the error message says, a list cannot be hashed but all the elements of a set are expected to be hashables. Quoting the documentation,
Return a new set or frozenset object whose elements are taken from iterable. The elements of a set must be hashable.
In case of set.update, you can pass multiple iterables to it and it will iterate all iterables and will include the individual elements in the set. Remember: It can accept only iterables. That is why you are getting an error when you try to update it with 1
>>> a.update(1)
Traceback (most recent call last):
File "<input>", line 1, in <module>
TypeError: 'int' object is not iterable
But, the following would work because the list [1] is iterated and the elements of the list are added to the set.
>>> a.update([1])
>>> a
set([1])
set.update is basically an equivalent of in-place set union operation. Consider the following cases
>>> set([1, 2]) | set([3, 4]) | set([1, 3])
set([1, 2, 3, 4])
>>> set([1, 2]) | set(range(3, 5)) | set(i for i in range(1, 5) if i % 2 == 1)
set([1, 2, 3, 4])
Here, we explicitly convert all the iterables to sets and then we find the union. There are multiple intermediate sets and unions. In this case, set.update serves as a good helper function. Since it accepts any iterable, you can simply do
>>> a.update([1, 2], range(3, 5), (i for i in range(1, 5) if i % 2 == 1))
>>> a
set([1, 2, 3, 4])
a.update(1) in your code won't work. add accepts an element and put it in the set if it is not already there but update takes an iterable and makes a unions of the set with that iterable. It's kind of like append and extend for the lists.
I guess no one mentioned about the good resource from Hackerrank. I'd like to paste how Hackerrank mentions the difference between update and add for set in python.
Sets are unordered bag of unique values. A single set contains values of any immutable data type.
CREATING SET
myset = {1, 2} # Directly assigning values to a set
myset = set() # Initializing a set
myset = set(['a', 'b']) # Creating a set from a list
print(myset) ===> {'a', 'b'}
add method directly adds elements to the set while the update method converts first argument into set then it adds
the list is hashable therefore we cannot add a hashable list to unhashable set.