CODEWITHSUNDEEP
pythondictionary
F. Elliot
F. Elliot

Reputation: 71

Create a new list from two dictionaries

This is a question about Python. I have the following list of dictionaries:

listA = [
          {"t": 1, "tid": 2, "gtm": 3, "c1": 4, "id": "111"},
          {"t": 3, "tid": 4, "gtm": 3, "c1": 4, "c2": 5, "id": "222"},
          {"t": 1, "tid": 2, "gtm": 3, "c1": 4, "c2": 5, "id": "333"},
          {"t": 5, "tid": 6, "gtm": 3, "c1": 4, "c2": 5, "id": "444"}
        ]

and a dictionary I wanted to compare with:

dictA = {"t": 1, "tid": 2, "gtm": 3}

I wanted to create a list of dicts that match all the items in dictA from listA and to include the "id" field as well:

listB = [
          {"t": 1, "tid": 2, "gtm": 3, "id": "111"},
          {"t": 1, "tid": 2, "gtm": 3, "id": "333"}
        ]

I tried doing this:

for k in listA:
    for key, value in k.viewitems() & dictA.viewitems():
        print key, value

But it's matching any item in dictA.

Upvotes: 3

Views: 673

Answers (4)

Padraic Cunningham
Padraic Cunningham

Reputation: 180391

You would need to check the length of the intersection, just checking if dct.viewitems() & dictA.viewitems() would evaluate to True for any intersection :

[dct for dct in listA if len(dct.viewitems() & dictA.viewitems()) == len(dictA)]

Or just check for a subset, if the items from dictA are a subset of each dict:

[dct for dct in listA if dictA.viewitems() <= dct.viewitems()]

Or reverse the logic looking for a superset:

 [dct for dct in listA if dct.viewitems() >= dictA.viewitems()]

Upvotes: 3

Rahul K P
Rahul K P

Reputation: 16081

Try this,all will check the existence of dictA in listA.

[i for i in listA if all(j in i.items() for j in dictA.items())]

Result 

[{'c1': 4, 'gtm': 3, 'id': '111', 't': 1, 'tid': 2},
 {'c1': 4, 'c2': 5, 'gtm': 3, 'id': '333', 't': 1, 'tid': 2}]

Upvotes: 1

Kalpesh Dusane
Kalpesh Dusane

Reputation: 1485

For python 2.7 : 

listA = [
              {"t": 1, "tid": 2, "gtm": 3, "c1": 4, "id": "111"},
              {"t": 3, "tid": 4, "gtm": 3, "c1": 4, "c2": 5, "id": "222"},
              {"t": 1, "tid": 2, "gtm": 3, "c1": 4, "c2": 5, "id": "333"},
              {"t": 5, "tid": 6, "gtm": 3, "c1": 4, "c2": 5, "id": "444"}
            ]
dictA = {"t": 1, "tid": 2, "gtm": 3}
for k in listA:
    if all(x in k.viewitems() for x in dictA.viewitems()):
        print k

It gives output as :

{'tid': 2, 'c1': 4, 'id': '111', 't': 1, 'gtm': 3}
{'gtm': 3, 't': 1, 'tid': 2, 'c2': 5, 'c1': 4, 'id': '333'}

And if you want to create list then instead of print, add dictionary to list As follows:

listA = [
          {"t": 1, "tid": 2, "gtm": 3, "c1": 4, "id": "111"},
          {"t": 3, "tid": 4, "gtm": 3, "c1": 4, "c2": 5, "id": "222"},
          {"t": 1, "tid": 2, "gtm": 3, "c1": 4, "c2": 5, "id": "333"},
          {"t": 5, "tid": 6, "gtm": 3, "c1": 4, "c2": 5, "id": "444"}
        ]
dictA = {"t": 1, "tid": 2, "gtm": 3}
ans =[]
for k in listA:
    if all(x in k.viewitems() for x in dictA.viewitems()):
        ans.append(k)
        #print k
print ans

It gives output:

[{'tid': 2, 'c1': 4, 'id': '111', 't': 1, 'gtm': 3}, {'gtm': 3, 't': 1, 'tid': 2, 'c2': 5, 'c1': 4, 'id': '333'}]

Upvotes: 1

Phidelux
Phidelux

Reputation: 2271

You can use a dictionary view:

listA = [
          {"t": 1, "tid": 2, "gtm": 3, "c1": 4, "id": "111"},
          {"t": 3, "tid": 4, "gtm": 3, "c1": 4, "c2": 5, "id": "222"},
          {"t": 1, "tid": 2, "gtm": 3, "c1": 4, "c2": 5, "id": "333"},
          {"t": 5, "tid": 6, "gtm": 3, "c1": 4, "c2": 5, "id": "444"}
        ]

dictA = {"t": 1, "tid": 2, "gtm": 3}

for k in listA:
    if dictA.viewitems() <= k.viewitems():
        print k

And for python 3 use:

if dictA.items() <= k.items():
    print(k)

Upvotes: 2

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