python - Destructuring-bind dictionary contents

Question

I am trying to 'destructure' a dictionary and associate values with variables names after its keys. Something like

params = {'a':1,'b':2}
a,b = params.values()

But since dictionaries are not ordered, there is no guarantee that params.values() will return values in the order of (a, b). Is there a nice way to do this?

Answer 1

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

Answer 2

from operator import itemgetter

params = {'a': 1, 'b': 2}

a, b = itemgetter('a', 'b')(params)

Instead of elaborate lambda functions or dictionary comprehension, may as well use a built in library.

Answer 3

One way to do this with less repetition than Jochen's suggestion is with a helper function. This gives the flexibility to list your variable names in any order and only destructure a subset of what is in the dict:

pluck = lambda dict, *args: (dict[arg] for arg in args)

things = {'blah': 'bleh', 'foo': 'bar'}
foo, blah = pluck(things, 'foo', 'blah')

Also, instead of joaquin's OrderedDict you could sort the keys and get the values. The only catches are you need to specify your variable names in alphabetical order and destructure everything in the dict:

sorted_vals = lambda dict: (t[1] for t in sorted(dict.items()))

things = {'foo': 'bar', 'blah': 'bleh'}
blah, foo = sorted_vals(things)

Answer 4

Python is only able to "destructure" sequences, not dictionaries. So, to write what you want, you will have to map the needed entries to a proper sequence. As of myself, the closest match I could find is the (not very sexy):

a,b = [d[k] for k in ('a','b')]

This works with generators too:

a,b = (d[k] for k in ('a','b'))

Here is a full example:

>>> d = dict(a=1,b=2,c=3)
>>> d
{'a': 1, 'c': 3, 'b': 2}
>>> a, b = [d[k] for k in ('a','b')]
>>> a
1
>>> b
2
>>> a, b = (d[k] for k in ('a','b'))
>>> a
1
>>> b
2

Answer 5

Here's another way to do it similarly to how a destructuring assignment works in JS:

params = {'b': 2, 'a': 1}
a, b, rest = (lambda a, b, **rest: (a, b, rest))(**params)

What we did was to unpack the params dictionary into key values (using **) (like in Jochen's answer), then we've taken those values in the lambda signature and assigned them according to the key name - and here's a bonus - we also get a dictionary of whatever is not in the lambda's signature so if you had:

params = {'b': 2, 'a': 1, 'c': 3}
a, b, rest = (lambda a, b, **rest: (a, b, rest))(**params)

After the lambda has been applied, the rest variable will now contain: {'c': 3}

Useful for omitting unneeded keys from a dictionary.

Hope this helps.

Answer 6

Maybe you really want to do something like this?

def some_func(a, b):
  print a,b

params = {'a':1,'b':2}

some_func(**params) # equiv to some_func(a=1, b=2)

Answer 7

How come nobody posted the simplest approach?

params = {'a':1,'b':2}

a, b = params['a'], params['b']

Answer 8

Warning 1: as stated in the docs, this is not guaranteed to work on all Python implementations:

CPython implementation detail: This function relies on Python stack frame support in the interpreter, which isn’t guaranteed to exist in all implementations of Python. If running in an implementation without Python stack frame support this function returns None.

Warning 2: this function does make the code shorter, but it probably contradicts the Python philosophy of being as explicit as you can. Moreover, it doesn't address the issues pointed out by John Christopher Jones in the comments, although you could make a similar function that works with attributes instead of keys. This is just a demonstration that you can do that if you really want to!

def destructure(dict_):
    if not isinstance(dict_, dict):
        raise TypeError(f"{dict_} is not a dict")
    # the parent frame will contain the information about
    # the current line
    parent_frame = inspect.currentframe().f_back

    # so we extract that line (by default the code context
    # only contains the current line)
    (line,) = inspect.getframeinfo(parent_frame).code_context

    # "hello, key = destructure(my_dict)"
    # -> ("hello, key ", "=", " destructure(my_dict)")
    lvalues, _equals, _rvalue = line.strip().partition("=")

    # -> ["hello", "key"]
    keys = [s.strip() for s in lvalues.split(",") if s.strip()]

    if missing := [key for key in keys if key not in dict_]:
        raise KeyError(*missing)

    for key in keys:
        yield dict_[key]

In [5]: my_dict = {"hello": "world", "123": "456", "key": "value"}                                                                                                           

In [6]: hello, key = destructure(my_dict)                                                                                                                                    

In [7]: hello                                                                                                                                                                
Out[7]: 'world'

In [8]: key                                                                                                                                                                  
Out[8]: 'value'

This solution allows you to pick some of the keys, not all, like in JavaScript. It's also safe for user-provided dictionaries

Answer 9

Look for other answers as this won't cater to the unexpected order in the dictionary. will update this with a correct version sometime soon.

try this

data = {'a':'Apple', 'b':'Banana','c':'Carrot'}
keys = data.keys()
a,b,c = [data[k] for k in keys]

result:

a == 'Apple'
b == 'Banana'
c == 'Carrot'

Answer 10

Well, if you want these in a class you can always do this:

class AttributeDict(dict):
    def __init__(self, *args, **kwargs):
        super(AttributeDict, self).__init__(*args, **kwargs)
        self.__dict__.update(self)

d = AttributeDict(a=1, b=2)

Answer 11

Based on @ShawnFumo answer I came up with this:

def destruct(dict): return (t[1] for t in sorted(dict.items()))

d = {'b': 'Banana', 'c': 'Carrot', 'a': 'Apple' }
a, b, c = destruct(d)

(Notice the order of items in dict)

Answer 12

With Python 3.10, you can do:

d = {"a": 1, "b": 2}

match d:
    case {"a": a, "b": b}:
        print(f"A is {a} and b is {b}")

but it adds two extra levels of indentation, and you still have to repeat the key names.

Answer 13

Since dictionaries are guaranteed to keep their insertion order in Python >= 3.7, that means that it's complete safe and idiomatic to just do this nowadays:

params = {'a': 1, 'b': 2}
a, b = params.values()
print(a)
print(b)

Output:

1
2

Answer 14

I don't know whether it's good style, but

locals().update(params)

will do the trick. You then have a, b and whatever was in your params dict available as corresponding local variables.