Understanding the behavior of Python's set

The documentation for the built-in type set says:

class set([iterable])

Return a new set or frozenset object whose elements are taken from iterable. The elements of a set must be hashable.

That is all right but why does this work:

>>> l = range(10)
>>> s = set(l)
>>> s
set([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

And this doesn't:

 >>> s.add([10])

    Traceback (most recent call last):
      File "<pyshell#7>", line 1, in <module>
        s.add([10])
    TypeError: unhashable type: 'list'

Both are lists开发者_如何学Go. Is some magic happening during the initialization?

When you initialize a set, you provide a list of values that must each be hashable.

s = set()
s.add([10])

is the same as

s = set([[10]])

which throws the same error that you're seeing right now.

In [13]: (2).__hash__
Out[13]: <method-wrapper '__hash__' of int object at 0x9f61d84>

In [14]: ([2]).__hash__ # nothing.

The thing is that set needs its items to be hashable, i.e. implement the __hash__ magic method (this is used for ordering in the tree as far as I know). list does not implement that magic method, hence it cannot be added in a set.

In this line:

s.add([10])

You are trying to add a list to the set, rather than the elements of the list. If you want ot add the elements of the list, use the update method.

Think of the constructor being something like:

class Set:
    def __init__(self,l):
        for elem in l:
            self.add(elem)

Nothing too interesting to be concerned about why it takes lists but on the other hand add(element) does not.

It behaves according to the documentation: set.add() adds a single element (and since you give it a list, it complains it is unhashable - since lists are no good as hash keys). If you want to add a list of elements, use set.update(). Example:

>>> s = set([1,2,3])
>>> s.add(5)
>>> s
set([1, 2, 3, 5])
>>> s.update([8])
>>> s
set([8, 1, 2, 3, 5])

s.add([10]) works as documented. An exception is raised because [10] is not hashable.

There is no magic happening during initialisation.

set([0,1,2,3,4,5,6,7,8,9]) has the same effect as set(range(10)) and set(xrange(10)) and set(foo()) where

def foo():
    for i in (9,8,7,6,5,4,3,2,1,0):
        yield i

In other words, the arg to set is an iterable, and each of the values obtained from the iterable must be hashable.