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.         


        

Answer 1

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 2

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 3

try this

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

result:

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

Answer 4

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 5

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 6

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.