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Creating Nested Dictionary Values using `setdefault()` 🔥
When grouping data, it's often necessary to check if a key exists, create a container for it, and then add a value.
For example, distributing users by role. Without special methods, this usually involves a separate key check.
The
The result is the same structure without a separate key existence check:
It's important to note that the expression of the second argument is evaluated every time
In this code,
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When grouping data, it's often necessary to check if a key exists, create a container for it, and then add a value.
For example, distributing users by role. Without special methods, this usually involves a separate key check.
users = [
("admin", "alex"),
("user", "max"),
("admin", "kate"),
]
groups = {}
for role, name in users:
if role not in groups:
groups[role] = []
groups[role].append(name)
The
setdefault() method allows you to perform this operation directly when accessing the dictionary. If the key exists, it returns its current value. If the key is missing, the provided value is written to the dictionary and then returned:groups = {}
for role, name in users:
groups.setdefault(
role,
[],
).append(name)The result is the same structure without a separate key existence check:
print(groups)
# {
# 'admin': ['alex', 'kate'],
# 'user': ['max']
# }
It's important to note that the expression of the second argument is evaluated every time
setdefault() is called, even if the key already exists. Therefore, you should avoid creating expensive objects or performing functions with side effects there:value = cache.setdefault(
key,
build_value(),
)
In this code,
build_value() will be called before the method itself is executed. If the value creation should only happen when the key is missing, it's better to use an explicit check or a suitable data structure, such as defaultdict.setdefault() is well-suited for compactly initializing simple mutable containers when grouping and aggregating data. However, it's important to remember that the provided value is evaluated regardless of whether the key exists.#Python #Coding #Dicts #Programming #CodeTips #DevLife
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