What am I doing wrong here?
a = set().add(1)
print a # Prints `None`
I'm trying to add the number 1
to the empty set.
The set add() method adds a given element to a set if the element is not present in the set. Syntax: set. add(elem) The add() method doesn't add an element to the set if it's already present in it otherwise it will get added to the set.
Use the Concatenation Operator to Append to a Set We can combine two sets together using this operator, which results in a set that contains the items of the first and the second set.
It is a convention in Python that methods that mutate sequences return None
.
Consider:
>>> a_list = [3, 2, 1]
>>> print a_list.sort()
None
>>> a_list
[1, 2, 3]
>>> a_dict = {}
>>> print a_dict.__setitem__('a', 1)
None
>>> a_dict
{'a': 1}
>>> a_set = set()
>>> print a_set.add(1)
None
>>> a_set
set([1])
Some may consider this convention "a horrible misdesign in Python", but the Design and History FAQ gives the reasoning behind this design decision (with respect to lists):
Why doesn’t
list.sort(
) return the sorted list?In situations where performance matters, making a copy of the list just to sort it would be wasteful. Therefore,
list.sort()
sorts the list in place. In order to remind you of that fact, it does not return the sorted list. This way, you won’t be fooled into accidentally overwriting a list when you need a sorted copy but also need to keep the unsorted version around.In Python 2.4 a new built-in function –
sorted()
– has been added. This function creates a new list from a provided iterable, sorts it and returns it.
Your particular problems with this feature come from a misunderstanding of good ways to create a set rather than a language misdesign. As Lattyware points out, in Python versions 2.7 and later you can use a set literal a = {1}
or do a = set([1])
as per Sven Marnach's answer.
Parenthetically, I like Ruby's convention of placing an exclamation point after methods that mutate objects, but I find Python's approach acceptable.
The add()
method adds an element to the set, but it does not return the set again -- it returns None
.
a = set()
a.add(1)
or better
a = set([1])
would work.
Because add()
is modifing your set in place returning None
:
>>> empty = set()
>>> print(empty.add(1))
None
>>> empty
set([1])
Another way to do it that is relatively simple would be:
a = set()
a = set() | {1}
this creates a union between your set a and a set with 1 as the element
print(a) yields {1} then because a would now have all elements of both a and {1}
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