Is there a better way to convert a list to a dictionary in Python with keys but no values?

Question

I was sure that there would be a one liner to convert a list to a dictionary where the items in the list were keys and the dictionary had no values.

The only way I could find to do it was argued against.

"Using list comprehensions when the result is ignored is misleading and inefficient. A for loop is better"

myList = ['a','b','c','d']
myDict = {}
x=[myDict.update({item:None}) for item in myList]

>>> myDict
{'a': None, 'c': None, 'b': None, 'd': None}

It works, but is there a better way to do this?

Answer 1

Use dict.fromkeys:

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

Values default to None, but you can specify them as an optional argument:

>>> my_list = [1, 2, 3]
>>> dict.fromkeys(my_list, 0)
{1: 0, 2: 0, 3: 0}

From the docs:

a.fromkeys(seq[, value]) Creates a new dictionary with keys from seq and values set to value.

dict.fromkeys is a class method that returns a new dictionary. value defaults to None. New in version 2.3.

Answer 2

You could use a set instead of a dict:

>>> myList=['a','b','c','d']
>>> set(myList)
set(['a', 'c', 'b', 'd'])

This is better if you never need to store values, and are just storing an unordered collection of unique items.