CODEWITHSUNDEEP
pythonlistdictionary
algorhythm
algorhythm

Reputation: 3426

Creating a dictionary from a list and fixed variable in Python

I am trying to create a dictionary from a list and a fixed variable. I know that zip() can create a dictionary from two lists but how can I adapt it to intake a fixed variable?

What works:

countries = ["Italy", "Germany", "Spain", "USA", "Switzerland"]
dishes = ["pizza", "sauerkraut", "paella", "Hamburger"]
country_specialities = zip(countries, dishes)

What I need:

dishes = ["pizza", "sauerkraut", "paella", "Hamburger"]
country = "uk"

To output:

menu = {"pizza": "uk", "sauerkraut": "uk", "paella": "uk", "Hamburger": "uk"}

Upvotes: 4

Views: 2659

Answers (4)

Martijn Pieters
Martijn Pieters

Reputation: 1121764

You can use the dict.fromkeys() class method here:

dict.fromkeys(dishes, country)

Note that you should not use this where the default is a mutable value (unless you meant to share that mutable value among all the keys).

From the documentation:

Create a new dictionary with keys from seq and values set to value.

fromkeys() is a class method that returns a new dictionary. value defaults to None.

This is going to be faster than using a dictionary comprehension; the looping over dishes is done entirely in C code, while a dict comprehension would execute the loop in Python bytecode instead.

If your list is going to contain 200.000 values, then that makes a huge difference:

>>> from timeit import timeit
>>> import random
>>> from string import ascii_lowercase
>>> values = [''.join([random.choice(ascii_lowercase) for _ in range(random.randint(20, 40))]) for _ in range(200000)]
>>> def dictcomp(values):
...     return {v: 'default' for v in values}
... 
>>> def dictfromkeys(values):
...     return dict.fromkeys(values, 'default')
... 
>>> timeit('f(values)', 'from __main__ import values, dictcomp as f', number=100)
4.984026908874512
>>> timeit('f(values)', 'from __main__ import values, dictfromkeys as f', number=100)
4.159089803695679

That's almost a second of difference between the two options when repeating the operation 100 times.

Upvotes: 7

fixxxer
fixxxer

Reputation: 16134

Expanding on with what you were doing with zip

In [54]: dict(zip(dishes, [country] * len(dishes)) )
Out[54]: {'Hamburger': 'uk', 'paella': 'uk', 'pizza': 'uk', 'sauerkraut': 'uk'}

Upvotes: 1

geckon
geckon

Reputation: 8754

You can do this:

dishes = ["pizza", "sauerkraut", "paella", "Hamburger"]
country = "uk"
menu = dict((dish, country) for dish in dishes)

Upvotes: 2

John Kugelman
John Kugelman

Reputation: 361595

{dish: country for dish in dishes}

Upvotes: 7

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