Adding multiple Values to a dictionary key in python?

Question

I would like to add multiple values to a dictonary key in python. The value is again a dict with some paraments changing..

I have written the following code but that seem to overwrite. any ways to do it effectively?

from collections import defaultdict
node = 'node 1'
d = defaultdict(dict)
d[node] = {'interface':{'Eth1/48':'10.10.10.1'}} --> Here Eth1/48 is dynamically populated.

d[node] = {'interface':{'Eth1/47':'10.10.11.1'}} --> here Eth1/47 is again dynamically populated.
>>> d
defaultdict(<class 'dict'>, {'node 1': {'interface': {'Eth1/47': '10.10.11.1'}}})

I want the output something like this:

{'node 1':{'interface': {'Eth1/48':'10.10.10.1'}
                        {'Eth1/47':'10.10.11.1'}

Scrubster · Accepted Answer · 2018-12-20 19:04:24Z

1

Make the dict not a dict of dicts, but a dict where a key corresponds to a list of other dicts.

d = defaultdict(dict)
d[node] = {'interface': [{'Eth1/48':'10.10.10.1'}]
d[node]['interface'].append({'Eth1/47':'10.10.11.1'})

Dan's comment below was an excellent one. He pointed out that it makes more sense to have IPs that can correspond to the same key rather than having a list of dicts of Eth*:ip pairs

d = defaultdict(dict)
d[node] = {'interface': {'Eth1/47':['10.10.10.1']}
d[node]['interface']['Eth1/48'] = ['10.10.10.1']

Now you can do

d[node]['interface']['Eth1/47'].append('new.ip.address.ballin')
d[node]['interface']['Eth1/48'].append('new.ip.address.ballin')

edited Dec 20, 2018 at 19:04

answered Dec 20, 2018 at 18:22

Scrubster

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3 Comments

erik258 Over a year ago

Since the interface ID is the key and the address is the value, I think it makes less sense to put each interfce key/value pair in a different list element. Better to combine to one dict with key of interface ID and value of address in my opinion. Interfaces can hold more than one address so the place for the list, I'd say, is in the dictionary value {'Eth1/48':['10.0.0.1', '10.10.10.1']}

Scrubster Over a year ago

I believe ingvar's answer is the closest to what OP wanted and would go with that. I gave a way in how I would do it for multiple values for the same key.

Scrubster Over a year ago

Dan, that is an excellent Idea. I'll edit my answers to note what you have said. I agree with your comment. It is a better idea.

timgeb · Accepted Answer · 2018-12-20 18:23:50Z

This does not exactly match your output, but what I would recommend (as dicts of only a single (key, value) pair are not so useful).

Go one level deeper with a defaultdict(lambda: defaultdict(dict)).

>>> from collections import defaultdict                                                                                
>>>                                                                                                                    
>>> node = 'node 1'                                                                                                    
>>> d = defaultdict(lambda: defaultdict(dict))                                                                         
>>> d[node]['interface']['Eth1/48'] = '10.10.10.1'                                                                     
>>> d[node]['interface']['Eth1/47'] = '10.10.11.1'
>>>                                                                                                                    
>>> d                                                                                                                  
defaultdict(<function __main__.<lambda>()>,
{'node 1': defaultdict(dict,
    {'interface': {'Eth1/47': '10.10.11.1',
                   'Eth1/48': '10.10.10.1'}})})

ingvar · Accepted Answer · 2018-12-20 18:25:39Z

0

Change inner dict instead of adding new key to outer one:

d = defaultdict(dict)
d[node] = {'interface':{'Eth1/48':'10.10.10.1'}}
d[node]['interface']['Eth1/47'] = '10.10.11.1'

If you have new dict of values you can add in this way too:

d[node]['interface'] = {**d[node]['interface'], **{'Eth1/47':'10.10.11.1'}}  # joins current d[node]['interface'] and {'Eth1/47':'10.10.11.1'} into one dict

edited Dec 20, 2018 at 18:25

answered Dec 20, 2018 at 18:19

ingvar

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1 Comment

erik258 Over a year ago

or you could define the inner dict all at once: {'interface':{'Eth1/48':'10.10.10.1', 'Eth1/47': '10.10.11.1'}'

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