Destructuring-Bind Dictionary Contents

Destructuring-bind dictionary contents

If you are afraid of the issues involved in the use of the locals dictionary and you prefer to follow your original strategy, Ordered Dictionaries from python 2.7 and 3.1 collections.OrderedDicts allows you to recover you dictionary items in the order in which they were first inserted

Destructuring dicts and objects in Python

You can use operator module from standard library as follows:

from operator import attrgetter
id, email, gender, username = attrgetter('id', 'email', 'gender', 'username')(current_user)
print(id, email, gender, username)

In case you have a dict like from your example

currentUser = {
  "id": 24,
  "name": "John Doe",
  "website": "http://mywebsite.com",
  "description": "I am an actor",
  "email": "example@example.com",
  "gender": "M",
  "phone_number": "+12345678",
  "username": "johndoe",
  "birth_date": "1991-02-23",
  "followers": 46263,
  "following": 345,
  "like": 204,
  "comments": 9
}

just use itemgetter instead of attrgetter:

from operator import itemgetter
id, email, gender, username = itemgetter('id', 'email', 'gender', 'username')(currentUser)
print(id, email, gender, username)

Divide a dictionary into variables

Problem is that dicts are unordered, so you can't use simple unpacking of d.values(). You could of course first sort the dict by key, then unpack the values:

# Note: in python 3, items() functions as iteritems() did
#       in older versions of Python; use it instead
ds = sorted(d.iteritems())
name0, name1, name2..., namen = [v[1] for v in ds]

You could also, at least within an object, do something like:

for k, v in dict.iteritems():
    setattr(self, k, v)

Additionally, as I mentioned in the comment above, if you can get all your logic that needs your unpacked dictionary as variables in to a function, you could do:

def func(**kwargs):
    # Do stuff with labeled args

func(**d)

ES6-like Python dict spread

No this isn't really possible. You can't have

a, b, c = spread(d)

and

a, c, b = spread(d)

give the same value to b. This is because the right side of an assignment statement is evaluated first. So spread executes and returns its values before your code knows which order you put them in on the left.

Some googling leads be to believe that by "spread-like syntax for dicts", you're looking for the **dict syntax. See What does ** (double star/asterisk) and * (star/asterisk) do for parameters?

Return copy of dictionary excluding specified keys

You were close, try the snippet below:

>>> my_dict = {
...     "keyA": 1,
...     "keyB": 2,
...     "keyC": 3
... }
>>> invalid = {"keyA", "keyB"}
>>> def without_keys(d, keys):
...     return {x: d[x] for x in d if x not in keys}
>>> without_keys(my_dict, invalid)
{'keyC': 3}

Basically, the if k not in keys will go at the end of the dict comprehension in the above case.

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