Python Language Dictionary
Internal
TODO
- TO PROCESS PyOOP "Dictionaries", "Dictionary use cases", "Using defaultdict", "Counter"
- Handle Missing Keys with setdefault() and defaultdict()
- Ordered Dictionary
Overview
A dictionary is a mutable collection of key-value pairs. The pairs can be accessed and modified. Each key is unique within the key set, and can be an instance of any immutable type: boolean, integer, float, tuple, string, etc., including None. There is a difference between a key that does not exist in the dictionary and a key that has been explicitly set to None - see Access with [] Syntax for more details. In other programming languages, the same data structure is referred to as "associative array" or "hash tables" or "hash maps".
The same underlying dictionary data structure can be referred by multiple variable names, which can be assigned with =, and if the structure is changed, the change will be reflected if any of the variables is used for access. To avoid this, make a copy of the dictionary with copy().
Dictionary type()
The function type() applied to a dictionary returns:
<class 'dict'>
To check whether an instance is a dictionary:
i = ...
if type(i) is dict:
...
For dict subclasses:
i = ...
if isinstance(i, dict):
...
Key Discussion
While the values of a dictionary can be any Python object, the keys have to be immutable object like scalar types (int, float, string) or tuples that contains immutable objects. They need to be hashable. See Dictionaries and the _hash_() Function below.
The keys 1 and True are equivalent. Why?
None as Key
None is a valid key:
d = {}
d[None] = 10
d['A'] = 20
assert len(d.keys()) == 2
assert d[None] == 10
assert d['A'] == 20
Dictionaries and the __hash__() Function
Dictionaries and the == Operator
It would be nice if == recursively deep-compares maps, and it seems to be working, Research this and come up with a definitive conclusion.
Create a Dictionary
A new dictionary instance is declared using the {...} syntax. The curly braces are placed around comma-separated key: value pairs. The dictionary can be empty
d = {}
or it can be populated with values:
d = {'a': 'b', 1: 2}
It is good form to insert a space after :. A comma is tolerated after the last pair.
Converting Other Data Structures to a Dictionary
Any two-value sequences can be converted into a dictionary using the dict() function.
Examples:
# list of lists
l = [['a', 'b'], ['c', 'd'], ['e', 'f']]
assert dict(l) == {'a': 'b', 'c': 'd', 'e': 'f'}
# list of tuples
l = [('a', 'b'), ('c', 'd'), ('e', 'f')]
assert dict(l) == {'a': 'b', 'c': 'd', 'e': 'f'}
# tuple of two-item lists
l = (['a', 'b'], ['c', 'd'], ['e', 'f'])
assert dict(l) == {'a': 'b', 'c': 'd', 'e': 'f'}
# list of two-character strings
l = ['ab', 'cd', 'ef']
assert dict(l) == {'a': 'b', 'c': 'd', 'e': 'f'}
# tuple of two-character strings
l = ('ab', 'cd', 'ef')
assert dict(l) == {'a': 'b', 'c': 'd', 'e': 'f'}
From a Key Sequence and a Value Sequence with zip()
The sequences must not be of the same type, and neither have the same length:
l = ['a', 'b', 'c', 'd']
t = (10, 22, 77, 99, 1010)
assert dict(zip(l, t)) == {'a':10, 'b':22, 'c':77, 'd':99}
Also see:
With Comprehensions
Access a Dictionary
Test for Empty Dictionary
Empty dictionaries evaluate to False:
d1 = {}
if not d1:
print("dictionary is empty")
Size of a Dictionary
The number of key is given by the len() function:
d = {...}
print(len(d))
Access Individual Elements
Individual dictionary elements can be accessed with the [] syntax or with the get() function.
Access with [] Syntax
The [] syntax can only be used with keys that exist in the dictionary:
d = {'a': 'b'}
print(d['a'])
An attempt to access an inexistent key will throw a KeyError exception. To avoid the exception, use get() instead, or test the existence of the key first.
However, a key can be explicitly set to None, and in this case, square bracket access to the key will return None:
d = {'a': None}
assert d['a'] is None
get() Function
The get() will return the associated value, or None if the key does not exist.
d = {'a': 'b'}
assert d.get('a') == 'b'
assert d.get('no-such-key') is None
The get() function allows for a second argument which will be returned instead of None in case the key does not exist:
assert d.get('no-such-key', 'alternative') == 'alternative'
Suggestions on how to safely recursively navigate a complex data structure:
Test the Existence of a Key
The existence of the key can be tested with in:
if 'some-key' in d:
print("key exists")
The non-existence is tested with this idiom:
if 'some-key' not in d:
print("key does not exist")
get() can also be used:
if d.get('some-key') is not None:
print("key exists")
Get All Keys with keys() Function
d = {'a': 'b', 'c': 'd'}
print(d.keys())
In Python 3, the keys() will return a dict_keys(), which is an iterable view of keys. This is useful with large dictionaries because the runtime does not use the time and the memory to create and store a list that might not be used. In case you need a list, use list() to wrap the result of keys().
list(d.keys())
Get All Values with values() Function
d = {'a': 'b', 'c': 'd'}
print(d.values())
In Python 3, the values() will return a dict_values(), which is an iterable view of values. This is useful with large dictionaries because the runtime does not use the time and the memory to create and store a list that might not be used. In case you need a list, use list() to wrap the result of values().
list(d.values())
Get All Key-Value Pairs with items() Function
d = {'a': 'b', 'c': 'd'}
print(d.items())
In Python 3, the items() will return a dict_items(), which is an iterable view of items. This is useful with large dictionaries because the runtime does not use the time and the memory to create and store a list that might not be used. Use use list() to wrap the result of items(). The list contains tuples of key and values.
for i in list(d.items()):
print('key:', i[0])
print('value:', i[1])
Equivalent:
for (k, v) in list(d.items()):
print('key:', k)
print('value:', v)
Iterate over a Dictionary
Use a for loop:
Iterate over Keys
Use the dictionary itself as argument of in. This is equivalent with iterating over the results of the keys() function.
d = {'a':'A', 'b':'B', 'c':'C'}
for k in d:
print(k)
This will display:
a
b
c
Iterate over Values
Apply the in operator to the result of the values() function:
d = {'a':'A', 'b':'B', 'c':'C'}
for v in d.values():
print(v)
This will display:
A
B
C
Iterate over Both Keys and Values
Apply the in operator to the result of the items() function. Each element is a tuple containing the key and the corresponding value:
d = {'a':'A', 'b':'B', 'c':'C'}
for kvt in d.items():
print(kvt)
This will display:
('a', 'A')
('b', 'B')
('c', 'C')
Variables corresponding to the elements of the tuple can be assigned in one step, operation known as tuple unpacking, so the following syntax where we assign the key and the value to their corresponding variable is valid:
d = {'a':'A', 'b':'B', 'c':'C'}
for k, v in d.items():
print('key:', k, 'value:', v)
This will display:
key: a value: A
key: b value: B
key: c value: C
Modify a Dictionary
Add an Element
Individual elements can be added with the [] syntax:
d = {}
d['a'] = 'b'
assert d['a'] == 'b'
Handle Missing Keys with setdefault() and defaultdict()
If the values are containers that need initialization when a key is first added to the dictionary, setdefault() is a dictionary method that set a value to the given default or returns the value if the key already exists:
d = {}
d.setdefault('a', []).append('b')
assert d['a'] == ['b']
TODO IPy Page 116.
Modify Individual Elements
Modification with [] Syntax
Individual elements can be modified with the [] syntax. If the key does not exist, the key-value pair will be added to the dictionary. If the key exists, the associated value will be updated:
d = {}
d['a'] = 'b'
assert d['a'] == 'b'
d['a'] = 'c'
assert d['a'] == 'c'
The syntax can be used with both literals and variables:
d = {}
d['a'] = 'b'
k = 'c'
d[k] = 'd'
assert d['c'] == 'd'
Delete Individual Element
With del
d = {'a': 'b', 'c': 'd'}
del d['a']
print(d) # will display {'c': 'd'}
Note that if the key being deleted does not exist, the [] syntax will throw an KeyError exception. The solution is to guard with an if:
if 'a' in d:
del d['a']
With pop()
pop() returns the value for the key provided as argument, and removes the key, if the key is in the dictionary. Throws KeyError if the key is not in the dictionary.
d = {'a': 'A'}
assert d.pop('a') == 'A'
assert len(d) == 0
The missing key can be recovered from the KeyError instance. The name of the missing key, as string, is propagated as the exception message as the first element of the args tuple, and can be accessed as such:
try:
...
except KeyError as e:
print(e.args[0])
Delete All Items
Use the clear() function:
d = {...}
d.clear()
Combine Dictionaries
Two dictionaries can be combined with update():
d1 = {'a': 'b'}
d2 = {'c': 'd'}
d1.update(d2) # d1 will become {'a': 'b', 'c': 'd'}, d2 will remain unchanged
If the same key exists in both dictionaries, the value associated with the key in the second dictionary will take precedence.
Copy a Dictionary
d1 = {'a': 'b'}
d2 =d1.copy()
d1['a'] = 'c'
print(d1['a']) # will display 'c'
print(d2['a']) # will display 'b'
Recursively Merge Dictionaries
Use update(). This only works on shallow trees, it fails to do deep recursive merge. Any existing keys in the data passed to update will have their old values discarded.
dict1 = {
'a': 'A',
}
dict2 = {
'b': 'B',
'c': ['C', 'D']
}
dict1.update(dict2)
assert dict1 == {
'a': 'A',
'b': 'B',
'c': ['C', 'D']
}
Ordered Dictionary
TODO IPy Page 120.
Subclass a Dictionary
class StorageClass(dict):
def __init__(self, storage_class_config):
super().__init__()
name = None
if storage_class_config:
name = storage_class_config.get('name')
if not name:
raise ValueError("invalid 'config.csi_driver.storage_classes' element: missing 'name'")
self['name'] = name
self['volumeBindingMode'] = 'WaitForFirstConsumer'
self['reclaimPolicy'] = 'Delete'
self['allowVolumeExpansion'] = True
self['parameters'] = {
'type': 'gp3',
'encrypted': 'true',
'csi.storage.k8s.io/fstype': 'ext4',
'throughput': '125',
'iops': '3000'
}