Python dict()

Different forms of dict() constructors are:

class dict(**kwarg)
class dict(mapping, **kwarg)
class dict(iterable, **kwarg)

Note: **kwarg let you take an arbitrary number of keyword arguments.

A keyword argument is an argument preceded by an identifier (eg. name=). Hence, the keyword argument of the form kwarg=value is passed to dict() constructor to create dictionaries.

dict() doesn't return any value (returns None).


Example 1: Create Dictionary Using keyword arguments only

numbers = dict(x=5, y=0)
print('numbers =', numbers)
print(type(numbers))

empty = dict()
print('empty =', empty)
print(type(empty))

Output

numbers = {'y': 0, 'x': 5}
<class 'dict'>
empty = {}
<class 'dict'>

Example 2: Create Dictionary Using Iterable

# keyword argument is not passed
numbers1 = dict([('x', 5), ('y', -5)])
print('numbers1 =',numbers1)

# keyword argument is also passed
numbers2 = dict([('x', 5), ('y', -5)], z=8)
print('numbers2 =',numbers2)

# zip() creates an iterable in Python 3
numbers3 = dict(zip(['x', 'y', 'z'], [1, 2, 3]))
print('numbers3 =',numbers3)

Output

numbers1 = {'y': -5, 'x': 5}
numbers2 = {'z': 8, 'y': -5, 'x': 5}
numbers3 = {'z': 3, 'y': 2, 'x': 1}

Example 3: Create Dictionary Using Mapping

numbers1 = dict({'x': 4, 'y': 5})
print('numbers1 =',numbers1)

# you don't need to use dict() in above code
numbers2 = {'x': 4, 'y': 5}
print('numbers2 =',numbers2)

# keyword argument is also passed
numbers3 = dict({'x': 4, 'y': 5}, z=8)
print('numbers3 =',numbers3)

Output

numbers1 = {'x': 4, 'y': 5}
numbers2 = {'x': 4, 'y': 5}
numbers3 = {'x': 4, 'z': 8, 'y': 5}

Also Read:

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