comparison between nested two OrderedDict

Question

I want to compare between two nested orderedDict dict, where i have to compare each and every key of dict1 Vs dict2.

>>> dict1 = OrderedDict([('a', OrderedDict([('b', '20'),
        ('c', '30'),('d', OrderedDict([('e', '40')]))]))])
>>> dict2 = OrderedDict([('a', OrderedDict([('b', '20'),
        ('c', '30'),('d', OrderedDict([('e', '50')]))]))])

for k,v in dict1.iteritems():
    for k1,v1 in v.iteritems():
        print "key %s, value  %s" %(k1,v1)
        for k2,v2 in dict2.iteritems():
            for k3,v3 in v2.iteritems():
               print "key %s, value  %s" %(k3,v3)

Output:


key b, value  20
key b, value  20
key c, value  30
key d, value  OrderedDict([('e', '50')])
key c, value  30
key b, value  20
key c, value  30
key d, value  OrderedDict([('e', '50')])
key d, value  OrderedDict([('e', '40')])
key b, value  20
key c, value  30
key d, value  OrderedDict([('e', '50')])

As the code is quite long, please help in writting in sort with comparision between (k1,v1) vs (k3,v3)

I'm not able to follow. Are you comparing whole dictionaries or just certain keys? — kokeen, Aug 17 '19 at 05:53
@kokeen I want to compare each "key" & "value" of dict 1 with "key and "value" of dict 2.Since its nested ordereddict , so finding difficult to compare — STripathy, Aug 17 '19 at 05:57
That's essentially comparing two whole dictionaries. [This](https://stackoverflow.com/questions/27265939/comparing-python-dictionaries-and-nested-dictionaries) might be a good start. — kokeen, Aug 17 '19 at 06:00
In my case, i have to iterate through and compare between(k1,v1) and (k3,v3) , as my output shown. — STripathy, Aug 17 '19 at 06:07
Check the link I've added in my previous comment. It helps in comparing nested dictionaries. — kokeen, Aug 17 '19 at 06:09

Trenton McKinney · Answer 1 · 2019-08-17T07:24:53.760

Data:

dict1 = OrderedDict([('a',
              OrderedDict([('b', '20'),
                           ('c', '30'),
                           ('d', OrderedDict([('e', '40')]))]))])

dict2 = OrderedDict([('a',
              OrderedDict([('b', '20'),
                           ('c', '30'),
                           ('d', OrderedDict([('e', '50')]))]))])

Perhaps unpacking the `OrderedDicts` into a `DataFrame` is a better option:

import pandas as pd
from pandas.io.json import json_normalize

dict_list = [dict1, dict2]  # more OrderedDicts can be added

df = pd.concat([json_normalize(x) for x in dict_list], sort=True).reset_index(drop=True)

df =

Also works if the `keys` are different:

print the `product` of `df` rows:

from itertools import product

dict_product = [list(product(df.iloc[y-1], df.iloc[y])) for y in df.index if y != 0]

print(dict_product)

[[('20', '20'),
  ('20', '30'),
  ('20', '50'),
  ('30', '20'),
  ('30', '30'),
  ('30', '50'),
  ('40', '20'),
  ('40', '30'),
  ('40', '50')]]

comparison between nested two OrderedDict

1 Answers1

Data:

Perhaps unpacking the OrderedDicts into a DataFrame is a better option:

Also works if the keys are different:

print the product of df rows:

Perhaps unpacking the `OrderedDicts` into a `DataFrame` is a better option:

Also works if the `keys` are different:

print the `product` of `df` rows: