Specific list format Python

Question

I have researched a lot of similar questions but I can't seem to find a solution on them or on my own. What I want to achieve is to get two dictionaries of this form each:

{'apple': ['5', '65'], 'blue': ['9', '10', '15', '43'],
'is': ['5', '6', '13', '45', '96'], 'yes': ['1', '2', '3', '11'], 
'zone': ['5', '6', '9', '10', '12', '14', '18', '19', '29', '45']}

{'apple': ['43'], 'appricote': ['1', '2', '3', '4', '5', '6'],
'candle': ['1', '2', '4', '5', '6', '9'], 'delta': ['14', '43', '47'], 
'dragon': ['23', '24', '25', '26'], 'eclipse': ['11', '13', '15', '19'], 
'island': ['1', '34', '35']}

And I want to have to following format:

apple ['5', '43', '65']
apricot ['1', '2', '3', '4', '5', '6']
blue ['9', '10', '15', '43']
candle ['1', '2', '4', '5', '6', '9']
delta ['14', '43', '47']
dragon ['23', '24', '25', '26']
eclipse ['11', '13', '15', '19']
is ['5', '6', '13', '45', '96']
island ['1', '34', '35']
yes ['1', '2', '3', '11']
zone ['5', '6', '9', '10', '12', '14', '18', '19', '29', '45']

But this is the format I get:

apple [['5', '65'], ['43']]
blue [['9', '10', '15', '43']]
is [['5', '6', '13', '45', '96']]
yes [['1', '2', '3', '11']]
zone [['5', '6', '9', '10', '12', '14', '18', '19', '29', '45']]
appricote [['1', '2', '3', '4', '5', '6']]
candle [['1', '2', '4', '5', '6', '9']]
delta [['14', '43', '47']]
dragon [['23', '24', '25', '26']]
eclipse [['11', '13', '15', '19']]
island [['1', '34', '35']]

This is my code so far

def merge_dictionaries(dict1, dict2): 
    from itertools import chain
    from collections import defaultdict

    dict3 = defaultdict(list)
    for k, v in chain(dict1.items(), dict2.items()):
        dict3[k].append(v)

    for k, v in dict3.items():
        print(k, v)

So my question is should I further process the format I have now or is it possible to achieve the format I want straight away. For example if you tell me to process the format I already have achieved I can merge the sub-lists into one list for each key and then sort, and the sort the keys also. But I want to know if there is a more direct way.

Answer 1

You need to use list.extend and not list.append:

>>> d = defaultdict(list)
>>> for k,v in chain(dict1.items(), dict2.items()):
...     d[k].extend(v)
... 
>>> d
defaultdict(<class 'list'>, {'apple': ['5', '65', '43'], 'blue': ['9', '10', '15', '43'], 'is': ['5', '6', '13', '45', '96'], 'yes': ['1', '2', '3', '11'], 'zone': ['5', '6', '9', '10', '12', '14', '18', '19', '29', '45'], 'appricote': ['1', '2', '3', '4', '5', '6'], 'candle': ['1', '2', '4', '5', '6', '9'], 'delta': ['14', '43', '47'], 'dragon': ['23', '24', '25', '26'], 'eclipse': ['11', '13', '15', '19'], 'island': ['1', '34', '35']})

You could sort the resulting lists if their order is important.

Answer 2

Your code is too complex.

a = {'apple': ['5', '65'], 'blue': ['9', '10', '15', '43'],
     'is': ['5', '6', '13', '45', '96'], 'yes': ['1', '2', '3', '11'],
     'zone': ['5', '6', '9', '10', '12', '14', '18', '19', '29', '45']}

b = {'apple': ['43'], 'appricote': ['1', '2', '3', '4', '5', '6'],
     'candle': ['1', '2', '4', '5', '6', '9'], 'delta': ['14', '43', '47'],
     'dragon': ['23', '24', '25', '26'], 'eclipse': ['11', '13', '15', '19'],
     'island': ['1', '34', '35']}

output = []
for key in a:
    temp = a[key]
    if key in b:
        temp.extend(b[key])
    output.append('{} {}'.format(key, sorted(temp)))

print('\n'.join(output))

# is ['13', '45', '5', '6', '96']
# blue ['10', '15', '43', '9']
# zone ['10', '12', '14', '18', '19', '29', '45', '5', '6', '9']
# apple ['43', '5', '65']
# yes ['1', '11', '2', '3']

Caveats:

• Since a is a dictionary the order of the elements in the outputted string is not persistent (in Python <= 3.6)

• Since your values are strings they are sorted lexicographically.

Both of these points can easily be fixed if they are relevant.