This is the following problem:
main_module.py
from collections import OrderedDict
from my_other_module import foo
a = OrderedDict([
('a', 1),
('b', 2),
('c', 3),
('d', 4),
])
foo(**a)
my_other_module.py
def foo(**kwargs):
for k, v in kwargs.items():
print k, v
When i run main_module.py
I'm expecting to get printout with the order I specified:
a 1
b 2
c 3
d 4
But instead I'm getting:
a 1
c 3
b 2
d 4
I do understand that this has something to do with the way **
operator is implemented and somehow it looses order how dictionary pairs are passed in. Also I do understand that dictionaries in python are not ordered as lists are, because they're implemented as hash tables. Is there any kind of 'hack' that I could apply so I get the behaviour that is needed in this context?
P.S. - In my situation I can't sort the dictionary inside foo function since there are no rules which could be followed except strict order that values are passed in.
By using **a
you're unpacking the ordered dictionary into an argument dictionary.
So when you enter in foo
, kwargs
is just a plain dictionary, with order not guaranteed (unless you're using Python 3.6+, but that's still an implementation detail in 3.6 - the ordering becomes official in 3.7: Are dictionaries ordered in Python 3.6+?)
You could just lose the packing/unpacking in that case so it's portable for older versions of python.
from collections import OrderedDict
def foo(kwargs):
for k, v in kwargs.items():
print(k, v)
a = OrderedDict([
('a', 1),
('b', 2),
('c', 3),
('d', 4),
])
foo(a)
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