CODEWITHSUNDEEP
python-2.7dictionarycopy
Subhakar K S
Subhakar K S

Reputation: 87

copying multi-level Python dictionaries

I have below Python dictionary as source

d1 = {
  'a': 1,
  'b': 2,
  'c': [{'d': 3, 'e': 4, 'un': 'wanted1', 'dont': 'needthis1'},
        {'d': 5, 'e': 6, 'un': 'wanted2', 'dont': 'needthis2'}]
  'xyz': 'abc',
  'zxy': 'cab',
  'wva': 'xyw'
}

And I want to copy the values of some specific keys to a different dict to form below target dictionary

d2 = {
  'some_attr_1': 1,  
  'some_attr_x': 2,
  'attr_some_z': [{'attr_x': 3, 'attrib': 4},
                  {'attr_x': 5, 'attrib': 6}]
}

Note:

My current approach is as below to have mapping between source and target dictionary keys.

attr_map1 = {
   'some_attr_1': 'a',
   'some_attr_x': 'b'
}

attr_map2 = {
  'attr_x': 'd',
  'attrib': 'e',
}

d2 = dict()
for k, v in attr_map1.items():
   d2[k] = d1[v]

l1 = list()
for d_elem in d1['c']:
   temp_dict = dict()
   for k, v in attr_map2.items():
       temp_dict[k] = d_elem[v]
   l1.append(temp_dict)
d2['attr_some_z'] = l1

Is there any alternate, better and speedy approach to achieve this?

I am looking for a solution in Python 2.7.

thanks,

Upvotes: 0

Views: 105

Answers (1)

Ajax1234
Ajax1234

Reputation: 71461

You can use recursion:

d1 = {'a': 1, 'b': 2, 'c': [{'d': 3, 'e': 4}, {'d': 5, 'e': 6}]}
def build(d):
  return {f't_{a}':b if not isinstance(b, (dict, list)) else 
           list(map(build, b)) if isinstance(b, list) else build(b) for a, b in d.items()}

print(build(d1))

Output:

{
't_a': 1, 
 't_b': 2, 
 't_c': [
    {'t_d': 3, 't_e': 4}, 
    {'t_d': 5, 't_e': 6}
   ]
}

Edit: to run this solution in Python2, replace the f-string with simple concatenation:

d1 = {'a': 1, 'b': 2, 'c': [{'d': 3, 'e': 4}, {'d': 5, 'e': 6}]}
def build(d):
   return {'t_'+a:b if not isinstance(b, (dict, list)) else 
       list(map(build, b)) if isinstance(b, list) else build(b) for a, b in d.items()}

Upvotes: 2

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