Use the fromkeys function to initialize a dictionary with any default value. In your case, you will initialize with None since you don't have a default value in mind.

Copyempty_dict = dict.fromkeys(['apple','ball'])

this will initialize empty_dict as:

Copyempty_dict = {'apple': None, 'ball': None}

As an alternative, if you wanted to initialize the dictionary with some default value other than None, you can do:

Copydefault_value = 'xyz'
nonempty_dict = dict.fromkeys(['apple','ball'],default_value)

Curly braces. Passing keyword arguments into dict(), though it works beautifully in a lot of scenarios, can only initialize a map if the keys are valid Python identifiers.

This works:

Copya = {'import': 'trade', 1: 7.8}

Copya = dict({'import': 'trade', 1: 7.8})

This won't work:

Copya = dict(import='trade', 1=7.8)

It will result in the following error:

Copy    a = dict(import='trade', 1=7.8)
             ^
SyntaxError: invalid syntax

dict.fromkeys directly solves the problem:

Copy>>> dict.fromkeys([1, 2, 3, 4])
{1: None, 2: None, 3: None, 4: None}

This is actually a classmethod, so it works for dict-subclasses (like collections.defaultdict) as well.

The optional second argument, which defaults to None, specifies the value to use for the keys. Note that the same object will be used for each key, which can cause problems with mutable values:

Copy>>> x = dict.fromkeys([1, 2, 3, 4], [])
>>> x[1].append('test')
>>> x
{1: ['test'], 2: ['test'], 3: ['test'], 4: ['test']}

If this is unacceptable, see How can I initialize a dictionary whose values are distinct empty lists? for a workaround.