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, 'gt

Solution 1:

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 iflen(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()]

Solution 2:

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)

Solution 3:

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'}]

Solution 4:

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

[i foriin listA ifall(j in i.items() forjin 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}]