Add number to set
Question:
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.
Answers:
Because add() is modifing your set in place returning None:
>>> empty = set()
>>> print(empty.add(1))
None
>>> empty
set([1])
You should do this:
a = set()
a.add(1)
print a
Notice that you’re assigning to a the result of adding 1, and the add operation, as defined in Python, returns None – and that’s what is getting assigned to a in your code.
Alternatively, you can do this for initializing a set:
a = set([1, 2, 3])
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.
The add method updates the set, but returns None.
a = set()
a.add(1)
print a
You are assigning the value returned by set().add(1) to a. This value is None, as add() does not return any value, it instead acts in-place on the list.
What you wanted to do was this:
a = set()
a.add(1)
print(a)
Of course, this example is trivial, but Python does support set literals, so if you really wanted to do this, it’s better to do:
a = {1}
print(a)
The curly brackets denote a set (although be warned, {} denotes an empty dict, not an empty set due to the fact that curly brackets are used for both dicts and sets (dicts are separated by the use of the colon to separate keys and values.)
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.
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}
Alternatively to a = set() | {1} consider “in-place” operator:
a = set()
a |= {1}
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.
Because add() is modifing your set in place returning None:
>>> empty = set()
>>> print(empty.add(1))
None
>>> empty
set([1])
You should do this:
a = set()
a.add(1)
print a
Notice that you’re assigning to a the result of adding 1, and the add operation, as defined in Python, returns None – and that’s what is getting assigned to a in your code.
Alternatively, you can do this for initializing a set:
a = set([1, 2, 3])
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.
The add method updates the set, but returns None.
a = set()
a.add(1)
print a
You are assigning the value returned by set().add(1) to a. This value is None, as add() does not return any value, it instead acts in-place on the list.
What you wanted to do was this:
a = set()
a.add(1)
print(a)
Of course, this example is trivial, but Python does support set literals, so if you really wanted to do this, it’s better to do:
a = {1}
print(a)
The curly brackets denote a set (although be warned, {} denotes an empty dict, not an empty set due to the fact that curly brackets are used for both dicts and sets (dicts are separated by the use of the colon to separate keys and values.)
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.
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}
Alternatively to a = set() | {1} consider “in-place” operator:
a = set()
a |= {1}