Divide a dictionary into variables

The existing answers will work, but they're all essentially re-implementing a function that already exists in the Python standard library: operator.itemgetter()

From the docs:

Return a callable object that fetches item from its operand using the operand’s __getitem__() method. If multiple items are specified, returns a tuple of lookup values. For example:

After f = itemgetter(2), the call f(r) returns r[2].

After g = itemgetter(2, 5, 3), the call g(r) returns (r[2], r[5], r[3]).

---

In other words, your destructured dict assignment becomes something like:

from operator import itemgetter

d = {'key_1': 'value_a', 'key_2': 'value_b'}
key_1, key_2 = itemgetter('key_1', 'key_2')(d)

# prints "Key 1: value_a, Key 2: value_b"
print("Key 1: {}, Key 2: {}".format(key_1, key_2))

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)