Python Language Set: Difference between revisions

From NovaOrdis Knowledge Base
Jump to navigation Jump to search
 
(4 intermediate revisions by the same user not shown)
Line 2: Line 2:
* [[Python_Language#Set|Python Language]]
* [[Python_Language#Set|Python Language]]
=Overview=
=Overview=
A set is an unordered collection of unique elements.
A set is an unordered collection of unique elements. Like dictionary keys, set elements must be immutable and [[Eq_()_and_hash_()_in_Python#Hashability|hashable]].


=Sets and the <tt>__hash__()</tt> Function=
=Sets and the <tt>__hash__()</tt> Function=
Line 29: Line 29:
s = set('ABC')
s = set('ABC')
print(s)  # {'C', 'B', 'A'}
print(s)  # {'C', 'B', 'A'}
</syntaxhighlight>
=Set Equality=
Sets are equal only if their contents are equal:
<syntaxhighlight lang='py'>
assert {1, 2, 3} == {3, 2, 1}
</syntaxhighlight>
</syntaxhighlight>


Line 70: Line 75:


If the element does not exist, throws <code>KeyError: 'a'</code>.
If the element does not exist, throws <code>KeyError: 'a'</code>.
<syntaxhighlight lang='py'>
s.pop()
</syntaxhighlight>
Remove an arbitrary element from the set, raising <code>KeyError</code> if the set is empty.


=Convert to List=
=Convert to List=
Line 77: Line 87:
</syntaxhighlight>
</syntaxhighlight>
=Set Operations=
=Set Operations=
==Union==
See [[#Add_Multiple_Elements_with_union()|above]].
==Inclusion==
==Inclusion==
Check whether a set is included in another set with <code>issubset()</code>:
Check whether a set is included in another set with <code>issubset()</code>:
Line 97: Line 109:


<syntaxhighlight lang='py'>
<syntaxhighlight lang='py'>
set_a = {'a', 'b', 'c', 1, 2, 3}
a = {'a', 'b', 'c', 1, 2, 3}
set_b = {'x', 'y', 'c', 1, 10, 20}
b = {'x', 'y', 'c', 1, 10, 20}
assert set_a.intersection(set_b) == {1, 'c'}
assert a.intersection(b) == {1, 'c'}
assert a & b == {1, 'c'}
</syntaxhighlight>
=Clear=
Reset the set to an empty state, discarding all elements:
<syntaxhighlight lang='py'>
a.clear()
</syntaxhighlight>
</syntaxhighlight>



Latest revision as of 17:12, 17 May 2024

Internal

Overview

A set is an unordered collection of unique elements. Like dictionary keys, set elements must be immutable and hashable.

Sets and the __hash__() Function

Sets and the __hash__() Function

Creation and Initialization

A set can be created in two ways, via the set() function or with a set literal.

A set can be created and initialization with a curly braces set literal:

s = {1, 2, 3}

Sets can also be created with the set() function: Set from a tuple:

s = set((1, 2, 3))

Set from a list:

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

Set from a string:

s = set('ABC')
print(s)  # {'C', 'B', 'A'}

Set Equality

Sets are equal only if their contents are equal:

assert {1, 2, 3} == {3, 2, 1}

Shallow Copy

s = set()
s.add('a')
s2 = s.copy()

Add Elements

Add One Element

s = set()
s.add('a')
assert 'a' in s

Add Multiple Elements

Update the set with the union of itself and others, where others can be any iterable. update() eliminates duplicates.

s = set()
s.update(['a', 'b', 'a'])
assert len(s) == 2
assert 'a' in s
assert 'b' in s

Add Multiple Elements with union()

union() merges the elements of the iterable given as argument into the current set, and returns a new set, which is the union of the current set and the argument.

a = {1, 2}
b = {5, 6}
assert a | b == {1, 2, 5, 6}
assert a.union(b) == {1, 2, 5, 6}

Remove an Element from a Set

s.remove('a')

Does not return anything, simply removes the element.

If the element does not exist, throws KeyError: 'a'.

s.pop()

Remove an arbitrary element from the set, raising KeyError if the set is empty.

Convert to List

s=set()
l=list(s)

Set Operations

Union

See above.

Inclusion

Check whether a set is included in another set with issubset():

set_a = {'a', 'b', 'c', 'd'}
set_b = {'a', 'c'}
assert set_b.issubset(set_a)

The function also works for non-homogenous sets, where the elements have different types:

set_a = {'a', 'b', 'c', 1, 5, 9}
set_b = {1, 'c'}
assert set_b.issubset(set_a)

Intersection

a = {'a', 'b', 'c', 1, 2, 3}
b = {'x', 'y', 'c', 1, 10, 20}
assert a.intersection(b) == {1, 'c'}
assert a & b == {1, 'c'}

Clear

Reset the set to an empty state, discarding all elements:

a.clear()

frozenset()

The fronzenset() returns an immutable frozenset object initialized with the elements from a given iterable. frozenset objects can be used as dictionary keys.

If the iterable contains duplicate elements, they are ignored (the iterable is handled like a set):

f = frozenset(['a', 'b', 'c'])
f2 = frozenset(['a', 'b', 'c', 'a'])
assert f == f2