Replace object reference in function

Question

def modify(a_dict):
    alt_dict = a_dict.copy()
    alt_dict['s'] = 2
    a_dict = alt_dict

a_dictionary = {}
a_dictionary['f'] = 5

modify(a_dictionary)
print(a_dictionary)

The expected result is that a_dictionary will now also containt the key 's'. I know lists and dictionaries are passed by reference as mutables, but is there a way to update the reference within a function as I'm trying to do above?.. Or is there a better way?..

To clarify, I do indeed need to make a copy, because the changes I make in inside my actual function in code I might discard if an exception happens.

moooeeeep · Accepted Answer · 2020-04-29 13:35:02Z

2

Assigning a new object to the local variable doesn't help you here. But as dictionaries are mutable, you can clear and update the original dictionary given the new data:

def modify(a_dict):
    alt_dict = a_dict.copy()
    alt_dict['s'] = 2
    a_dict.clear()
    a_dict.update(alt_dict)

edited Apr 29, 2020 at 13:35

answered Apr 29, 2020 at 12:21

moooeeeep

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user2390182 · Accepted Answer · 2020-04-29 12:19:50Z

0

You can do:

def modify(a_dict):
    alt_dict = a_dict.copy()
    alt_dict['s'] = 2
    # do lots of stuff

    # make sure all changes are "backported"
    a_dict.update(alt_dict)

answered Apr 29, 2020 at 12:19

user2390182

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Comments

Dave Costa · Accepted Answer · 2020-04-29 12:22:59Z

0

How about returning the modified copy?

def modify(a_dict):
    alt_dict = a_dict.copy()
    alt_dict['s'] = 2
    return alt_dict

a_dictionary = {}
a_dictionary['f'] = 5

a_dictionary = modify(a_dictionary)
print(a_dictionary)

answered Apr 29, 2020 at 12:22

Dave Costa

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2 Comments

Jack Avante Over a year ago

Unfortunately I'm returning other things in my actual code, that's why I needed to do it by reference, but this is a good suggestion

Dave Costa Over a year ago

@JackAvante A function in Python can return an arbitrary number of objects.

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