CODEWITHSUNDEEP
pythonlistcsvdictionary
vizakshat
vizakshat

Reputation: 221

Convert ordered dictionary to different format

I need to convert the given list of ordered dictionaries in a different format with minimal time complexity. Code for reproducing example:

import csv
from collections import OrderedDict

list_of_dicts = [OrderedDict([('key_a','value_a'),('key_b','value_b')]), 
                 OrderedDict([('key_a','value_c'),('key_b','value_d')]),
                 OrderedDict([('key_a','value_e'),('key_b','value_f')])]

Need to convert the above to the following (WITHOUT USING EXPLICIT FOR LOOP):

convertedDictionary = [OrderedDict([('value_a','value_b')]),
                       OrderedDict([('value_c','value_d')]),
                       OrderedDict([('value_e','value_f')])]

Upvotes: 3

Views: 46

Answers (1)

Jean-François Fabre
Jean-François Fabre

Reputation: 140266

first step: get the lists: simple enough mapping dict.values to the dicts:

>>> list(map(dict.values,list_of_dicts))
[['value_a', 'value_b'], ['value_c', 'value_d'], ['value_e', 'value_f']]

Now for the final result, without loops, the only alternative is map with lambda but it's ugly

>>> list(map(lambda x : OrderedDict((x,)),map(dict.values,list_of_dicts)))
[OrderedDict([('value_a', 'value_b')]),
 OrderedDict([('value_c', 'value_d')]),
 OrderedDict([('value_e', 'value_f')])]

which is much better (and probably faster) with an outer list comprehension:

[OrderedDict((x,)) for x in map(dict.values,list_of_dicts)]

note: why using OrderedDict if you only have 1 value per dictionary? why storing only one value per dictionary? Unless this is an initialization step, the data model looks dubious.

Upvotes: 3

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