CODEWITHSUNDEEP
pythondictionary
e.iluf
e.iluf

Reputation: 1659

how to compare two dictionaries in python

I want to compare two dictionaries and find a matching value

RFQDict dictionary is in this format

'9905-01-485-2102': {'Add_method': 'Dibbs Extract', 'NSN': '9905-01-485-2102', 'Nomenclature': 'REFLECTOR, INDICATIN', 'RFQ_Date': '10-2-2018', 'RFQ_QTY': ' 4200', 'Time': '2018-10-13 22:19:01'}

AwardsDict dictionary is in this format 

'W91WAW07C0023': {'Awarddate': '08-13-2007', 'Awardid': 'W91WAW07C0023', 'Cage': '7A293', 'Deliverynumber': '»', 'NSN': '0001', 'Nomenclature': 'INSTITUTE FOR  ANALYSES', 'PR': '0014447158', 'Price': 'See Award Doc', 'Time': '2018-05-17 17:41:54'}

I want to compare the NSN values to see if there is a match

I used this script to get the two dictionaries 

RFQDict = {}
AwardsDict = {}

# Fetch RFQ 

RFQref = db.reference('TestRFQ')

snapshot = RFQref.get()
for key, val in snapshot.items():
    RFQDict[key] = val
    print('{0} => {1}'.format(key, val))


Awardsref = db.reference('DibbsAwards')

dsnapshot = Awardsref.get()
for key, val in dsnapshot.items():
    AwardsDict[key] = val
    print('{0} => {1}'.format(key, val))

print(RFQDict)
print(AwardsDict)

Each dictionary contains thousands of key values. How could I compare the NSN values ?

Upvotes: 0

Views: 755

Answers (2)

slider
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Reputation: 12990

Here's a simplified version of your problem. You can group your AwardsDict "value" dictionaries by NSN and then compare and match.

RFQDict = {
    'a': {'Add_method': 'D1', 'NSN': '9905'},
    'b': {'Add_method': 'D2', 'NSN': '9906'},
    'c': {'Add_method': 'D3', 'NSN': '9907'},
    'd': {'Add_method': 'D4', 'NSN': '9908'}
}
AwardsDict = {
    'W21': {'Awarddate': '08-13-2007', 'Awardid': '1', 'NSN': '9906'},
    'W22': {'Awarddate': '08-14-2007', 'Awardid': '2', 'NSN': '9905'},
    'W23': {'Awarddate': '08-15-2007', 'Awardid': '3', 'NSN': '9908'},
    'W24': {'Awarddate': '08-16-2007', 'Awardid': '4', 'NSN': '9907'},
}

# First create a new dictionary with "NSN" as keys and awards as matches 
nsn_awards = {v['NSN']: v for v in AwardsDict.values()}

# go through all values of RFQDict and find a match by
# looking up the its NSN in the nsn_awards_dict
matches = [(rfq, nsn_awards[rfq['NSN']]) for rfq in RFQDict.values()]

print(matches)

prints:

[({'NSN': '9908', 'Add_method': 'D4'}, {'NSN': '9908', 'Awardid': '3', 'Awarddate': '08-15-2007'}), ({'NSN': '9907', 'Add_method': 'D3'}, {'NSN': '9907', 'Awardid': '4
', 'Awarddate': '08-16-2007'}), ({'NSN': '9906', 'Add_method': 'D2'}, {'NSN': '9906', 'Awardid': '1', 'Awarddate': '08-13-2007'}), ({'NSN': '9905', 'Add_method': 'D1'}
, {'NSN': '9905', 'Awardid': '2', 'Awarddate': '08-14-2007'})]

Note that this gives us an O(n) solution but the tradeoff is space for the dict we're creating.

If there's a possibility that you can have NSNs that don't have a match in AwardsDict, you can perform a simple if nsn in nsn_awards check before creating a match. To explain it better, here's a version without list comprehensions:

matches = []
for rfq in RFQDict.values():
    if rfq['NSN'] in nsn_awards: # only append if there's a match
        nsn = rfq['NSN']
        matches.append((rfq, nsn_awards[nsn]))

Upvotes: 1

Recessive
Recessive

Reputation: 1949

if dict1["NSN"]==dict2["NSN"]:
    print("Equal")

Upvotes: 1

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