Set

In Python programs we often want to collect unique strings or ints. A set is ideal—it is a dictionary but with no linked values. A frozenset is immutable and can be a dictionary key.

A built-in

The set() function is built-in, as Python is "batteries included." With it we can remove duplicate elements. We can initialize it with curly brackets.

First example

This program initializes a set. When we initialize a set, we do not include the values in the syntax as with a dictionary. We specify only the keys.

# Create a set.
items = {"arrow", "spear", "arrow", "arrow", "rock"}

# Print set.
print(items)
print(len(items))

# Use in-keyword.
if "rock" in items:
    print("Rock exists")

# Use not-in keywords.
if "clock" not in items:
    print("Cloak not found")
{'spear', 'arrow', 'rock'}
3
Rock exists
Cloak not found

Add

Elements can be added to a set with the add method. Here we create an empty set with the set built-in method. We then add 3 elements to the set.

# An empty set.
items = set()

# Add three strings.
items.add("cat")
items.add("dog")
items.add("gerbil")
print(items)
{'gerbil', 'dog', 'cat'}

Create, list

We can create a set from a list (or other iterable collection). Consider this example. We pass a list with six different elements in it to set(). The duplicates are ignored.

# Create a set from this list.
# ... Duplicates are ignored.
numbers = set([10, 20, 20, 30, 40, 50])
print(numbers)
{40, 10, 20, 50, 30}

Subset, superset

In set theory, we determine relations between sets of elements. And with the Python set type, we can compute these with built-in methods.

numbers1 = {1, 3, 5, 7}
numbers2 = {1, 3}

# Is subset.
if numbers2.issubset(numbers1):
    print("Is a subset")

# Is superset.
if numbers1.issuperset(numbers2):
    print("Is a superset")

# Intersection of the two sets.
print(numbers1.intersection(numbers2))
Is a subset
Is a superset
{1, 3}

Union

Another set operation that is available is union(). This combines two sets. Any element in either set is retained in the return value of union. But duplicates are eliminated.

# Two sets.
set1 = {1, 2, 3}
set2 = {6, 5, 4, 3}

# Union the sets.
set3 = set1.union(set2)
print(set3)
{1, 2, 3, 4, 5, 6}

Subtract, difference

A set can be subtracted from another set. The difference() method is used in this case. The syntax that is clearer is the best choice.

a = {"new york", "connecticut", "new jersey"}
b = {"connecticut", "pennsylvania", "maine"}

# Subtract.
c = a - b
print(c)

# Difference.
c = a.difference(b)
print(c)

# Subtract in opposite order.
c = b - a
print(c)

# Difference in opposite order.
c = b.difference(a)
print(c)
{'new jersey', 'new york'}
{'new jersey', 'new york'}
{'pennsylvania', 'maine'}
{'pennsylvania', 'maine'}

Discard

We pass discard() the value of an element we want to remove. If the element does not exist, discard will cause no error—it does nothing. Remove, however, will cause a KeyError.

animals = {"cat", "dog", "parrot", "walrus"}
print(animals)

# Discard nonexistent element, nothing happens.
animals.discard("zebra")
print(animals)

# Discard element that exists.
animals.discard("cat")
print(animals)

# Remove element that exists.
animals.remove("parrot")
print(animals)

# Remove causes an error if the element is not found.
animals.remove("buffalo")
{'walrus', 'dog', 'parrot', 'cat'}
{'walrus', 'dog', 'parrot', 'cat'}
{'walrus', 'dog', 'parrot'}
{'walrus', 'dog'}
Traceback (most recent call last):
  File "...", line 16, in <module>
    animals.remove("buffalo")
KeyError: 'buffalo'

Get keys, dictionary

A dictionary contains only unique keys. With the set() built-in, we can get these keys and convert them into a set.

# This dictionary contains key-value pairs.
dictionary = {"cat": 1, "dog": 2, "bird": 3}
print(dictionary)

# This set contains just the dictionary's keys.
keys = set(dictionary)
print(keys)
{'bird': 3, 'dog': 2, 'cat': 1}
{'cat', 'bird', 'dog'}

Map

Methods such as map can be used to transform collections. The result of map() is not a set. It is a "map object" which we can enumerate in a for-loop.

values = {10, 20, 30}

# Multiply all values in the set by 100.
result = map(lambda x: x * 100, values)

# Display our results.
for value in result:
    print(value)
1000
2000
3000

Benchmark, set

How does a set compare in performance to a dictionary? Logically the performance should be similar. I test how an in-keyword test runs.

import time

set1 = {"a", "b", "c", "z"}
dict1 = {"a": 1, "b": 2, "c": 3, "z": 4}

print(time.time())

# Version 1: use set.
i = 0
while i < 10000000:
    a = "z" in set1
    i += 1

print(time.time())

# Version 2: use dictionary.
i = 0
while i < 10000000:
    a = "z" in dict1
    i += 1

print(time.time())
1346615677.741
1346615679.7   (Set        = 1.959 s)
1346615681.732 (Dictionary = 2.032 s)

1346615958.731
1346615960.692 (Set        = 1.961 s)
1346615962.736 (Dictionary = 2.044 s)

Sometimes, the existence of keys is our main consideration. The keys have no specific value. Here a set avoids confusion with having unused values in a dictionary.