Shorten length of key in dictionary-CodePudding

I'm stuck at trying to shorten the keys of my dictionary.

This is d1:

{
    'A0Z40': [ 'A14160,A14161,A14162,A14163,A14164,A14165,A14166' ],
    'A0Z41': [ 'A1403,A1407,A1408,A1409,A1410,A1411' ],
    'A0Z42': [ 'A1201,A1205', 'A12041,A12042,A12043,A12044,A12045,A12046' ],
    'A0Z43': [ 'A1101' ]
}

And I'm trying to get this:

{
    'A0Z' : [ 'A14160,A14161,A14162,A14163,A14164,A14165,A14166' ],  
            [ 'A1403,A1407,A1408,A1409,A1410,A1411' ], 
            [ 'A1201,A1205', 'A12041,A12042,A12043,A12044,A12045,A12046' ],   
            ['A1101']
}

I've tried with a dictionary comprehension

d = {k[:3] : v for k, v in d.items() }

but it erased the value since a dictionary can only have unique keys...

Do you have any other solution?

Thank you very much

CodePudding user response：

You can do it with regular for loop using get and default value

d = {}
for k, v in d1.items():
    d[k[:3]] = d.get(k[:3], [])   v

this will combine the lists in the values to one list under the key 'A0Z'

{'A0Z': ['A14160,A14161,A14162,A14163,A14164,A14165,A14166', 'A1403,A1407,A1408,A1409,A1410,A1411', 'A1201,A1205', 'A12041,A12042,A12043,A12044,A12045,A12046', 'A1101']}

If you would like to have the value as a list of lists just put the value v in a list in each iteration

d[k[:3]] = d.get(k[:3], [])   [v]

outout:

{'A0Z': [['A14160,A14161,A14162,A14163,A14164,A14165,A14166'], ['A1403,A1407,A1408,A1409,A1410,A1411'], ['A1201,A1205', 'A12041,A12042,A12043,A12044,A12045,A12046'], ['A1101']]}

CodePudding user response：

Your expected output is not valid dict. Maybe you want the value to be list of lists? Or combine all lists in single list?

from collections import defaultdict

data = {'A0Z40': ['A14160,A14161,A14162,A14163,A14164,A14165,A14166'],
  'A0Z41': ['A1403,A1407,A1408,A1409,A1410,A1411'],
  'A0Z42': ['A1201,A1205', 'A12041,A12042,A12043,A12044,A12045,A12046'],
  'A0Z43': ['A1101']}

result =  defaultdict(list)
for key, value in data.items():
    result[key[:3]].extend(value) # or .append(value)

print(result)

output:

defaultdict(<class 'list'>, {'A0Z': ['A14160,A14161,A14162,A14163,A14164,A14165,A14166', 'A1403,A1407,A1408,A1409,A1410,A1411', 'A1201,A1205', 'A12041,A12042,A12043,A12044,A12045,A12046', 'A1101']})

or when using .append():

defaultdict(<class 'list'>, {'A0Z': [['A14160,A14161,A14162,A14163,A14164,A14165,A14166'], ['A1403,A1407,A1408,A1409,A1410,A1411'], ['A1201,A1205', 'A12041,A12042,A12043,A12044,A12045,A12046'], ['A1101']]})

CodePudding user response：

D = {'A0Z40': ['A14160,A14161,A14162,A14163,A14164,A14165,A14166'],
     'A0Z41': ['A1403,A1407,A1408,A1409,A1410,A1411'],
     'A0Z42': ['A1201,A1205', 'A12041,A12042,A12043,A12044,A12045,A12046'],
     'A0Z43': ['A1101']}

# to merge this lists you could do this
ML = []
for v in D.values():
    ML  = v
ND = {'A0Z': ML}
print(ND)
# if you want a list of lists you could do this
ML = []
for v in D.values():
    ML.append(v)
ND = {'A0Z': ML}
print(ND)