How to Construct a Set Out of List Items in Python

How to construct a set out of list items in python?

If you have a list of hashable objects (filenames would probably be strings, so they should count):

lst = ['foo.py', 'bar.py', 'baz.py', 'qux.py', Ellipsis]

you can construct the set directly:

s = set(lst)

In fact, set will work this way with any iterable object! (Isn't duck typing great?)

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If you want to do it iteratively:

s = set()
for item in iterable:
    s.add(item)

But there's rarely a need to do it this way. I only mention it because the set.add method is quite useful.

How to make a set of lists

You can create a set of tuples, a set of lists will not be possible because of non hashable elements as you mentioned.

>>> l = [[1, 2, 3], [2, 4, 5], [1, 2, 3], [2, 4, 5]]
>>> set(tuple(i) for i in l)
{(1, 2, 3), (2, 4, 5)}

set of list of lists in python

Since the lists are mutable, they cannot be hashed. The best bet is to convert them to a tuple and form a set, like this

>>> mat = [[1,2,3],[4,5,6],[1,2,3],[7,8,9],[4,5,6]]
>>> set(tuple(row) for row in mat)
set([(4, 5, 6), (7, 8, 9), (1, 2, 3)])

We iterate through the mat, one list at a time, convert that to a tuple (which is immutable, so sets are cool with them) and the generator is sent to the set function.

If you want the result as list of lists, you can extend the same, by converting the result of set function call, to lists, like this

>>> [list(item) for item in set(tuple(row) for row in mat)]
[[4, 5, 6], [7, 8, 9], [1, 2, 3]]

Add list to set

You can't add a list to a set because lists are mutable, meaning that you can change the contents of the list after adding it to the set.

You can however add tuples to the set, because you cannot change the contents of a tuple:

>>> a.add(('f', 'g'))
>>> print a
set(['a', 'c', 'b', 'e', 'd', ('f', 'g')])

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Edit: some explanation: The documentation defines a set as an unordered collection of distinct hashable objects. The objects have to be hashable so that finding, adding and removing elements can be done faster than looking at each individual element every time you perform these operations. The specific algorithms used are explained in the Wikipedia article. Pythons hashing algorithms are explained on effbot.org and pythons __hash__ function in the python reference.

Some facts:

• Set elements as well as dictionary keys have to be hashable
• Some unhashable datatypes:
  
  
  • list: use tuple instead
  • set: use frozenset instead
  • dict: has no official counterpart, but there are some
    recipes
• Object instances are hashable by default with each instance having a unique hash. You can override this behavior as explained in the python reference.

Convert a list of lists into a set of lists in Python 3

lists are not hashable in python. Instead, map to tuples, which are hashable:

>>> mylist = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> set(map(tuple, mylist))
{(4, 5, 6), (7, 8, 9), (1, 2, 3)}
>>> set(tuple(x) for x in mylist)
{(4, 5, 6), (7, 8, 9), (1, 2, 3)}

You can also use a clean set comprehension here as well:

>>> {tuple(x) for x in mylist}
{(4, 5, 6), (7, 8, 9), (1, 2, 3)}

Converting a list to a set changes element order

1. A set is an unordered data structure, so it does not preserve the insertion order.

2. This depends on your requirements. If you have an normal list, and want to remove some set of elements while preserving the order of the list, you can do this with a list comprehension:

   >>> a = [1, 2, 20, 6, 210]
   >>> b = set([6, 20, 1])
   >>> [x for x in a if x not in b]
   [2, 210]
   

   If you need a data structure that supports both fast membership tests and preservation of insertion order, you can use the keys of a Python dictionary, which starting from Python 3.7 is guaranteed to preserve the insertion order:

   >>> a = dict.fromkeys([1, 2, 20, 6, 210])
   >>> b = dict.fromkeys([6, 20, 1])
   >>> dict.fromkeys(x for x in a if x not in b)
   {2: None, 210: None}
   

   b doesn't really need to be ordered here – you could use a set as well. Note that a.keys() - b.keys() returns the set difference as a set, so it won't preserve the insertion order.

   In older versions of Python, you can use collections.OrderedDict instead:

   >>> a = collections.OrderedDict.fromkeys([1, 2, 20, 6, 210])
   >>> b = collections.OrderedDict.fromkeys([6, 20, 1])
   >>> collections.OrderedDict.fromkeys(x for x in a if x not in b)
   OrderedDict([(2, None), (210, None)])
   

Pythonic way to create union of all values contained in multiple lists

set.union does what you want:

>>> results_list = [[1,2,3], [1,2,4]]
>>> results_union = set().union(*results_list)
>>> print(results_union)
set([1, 2, 3, 4])

You can also do this with more than two lists.

How to convert a set to a list in python?

It is already a list:

>>> type(my_set)
<class 'list'>

Do you want something like:

>>> my_set = set([1, 2, 3, 4])
>>> my_list = list(my_set)
>>> print(my_list)
[1, 2, 3, 4]

EDIT:
Output of your last comment:

>>> my_list = [1,2,3,4]
>>> my_set = set(my_list)
>>> my_new_list = list(my_set)
>>> print(my_new_list)
[1, 2, 3, 4]

I'm wondering if you did something like this:

>>> set = set()
>>> set([1, 2])

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'set' object is not callable

How to get a list from a set?

>>> s = set([1, 2, 3])
>>> list(s)
[1, 2, 3]

Note that the list you get doesn't have a defined order.

Create an empty list with certain size in Python

You cannot assign to a list like xs[i] = value, unless the list already is initialized with at least i+1 elements. Instead, use xs.append(value) to add elements to the end of the list. (Though you could use the assignment notation if you were using a dictionary instead of a list.)

Creating an empty list:

>>> xs = [None] * 10
>>> xs
[None, None, None, None, None, None, None, None, None, None]

Assigning a value to an existing element of the above list:

>>> xs[1] = 5
>>> xs
[None, 5, None, None, None, None, None, None, None, None]

Keep in mind that something like xs[15] = 5 would still fail, as our list has only 10 elements.

range(x) creates a list from [0, 1, 2, ... x-1]

# 2.X only. Use list(range(10)) in 3.X.
>>> xs = range(10)
>>> xs
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Using a function to create a list:

>>> def display():
...     xs = []
...     for i in range(9): # This is just to tell you how to create a list.
...         xs.append(i)
...     return xs
... 
>>> print display()
[0, 1, 2, 3, 4, 5, 6, 7, 8]

List comprehension (Using the squares because for range you don't need to do all this, you can just return range(0,9) ):

>>> def display():
...     return [x**2 for x in range(9)]
... 
>>> print display()
[0, 1, 4, 9, 16, 25, 36, 49, 64]

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