Intersection of two list of dictionaries based on a key

Question

I have two different list of dictionaries,

list1 = [{'count': 351, 'att_value': 'one'},
         {'count': 332,  'att_value': 'two'},
         {'count': 336,  'att_value': 'six'},
         {'count': 359,  'att_value': 'nine'},
         {'count': 304,  'att_value': 'four'}]

list2 = [{'count': 359,'person_id' : 4},
         {'count': 351, 'person_id' : 12},
         {'count': 381, 'person_id' : 8}]

How to find an intersection of list_A with list_B based on "count" key by including rest of the key as like list_C?

list3 = [{'count':359, 'att_value' : 'nine', 'person_id':4},
         {'count':351, 'att_value' : 'one', 'person_id':12},
         {'count':381, 'att_value' : '-', 'person_id':8}] 

I would like to retain the keys from list2, but with missing values from list1 represented by "-".

Answer 1

You can also use pandas library for this:

In [102]: df1 = pd.DataFrame(list1)
In [104]: df2 = pd.DataFrame(list2)

In [106]: pd.merge(df2,df1, on='count', how='left').fillna('-')
Out[106]: 
     count att_value
0    359      nine
1    351       one
2    381         -

Answer 2

You can do this with a list comprehension. First, build a set of all counts from list2, and then filter out dictionaries based on constant time set membership check.

counts = {d2['count'] for d2 in list2}
list3 = [d for d in list1 if d['count'] in counts]

print(list3)
# [{'count': 351, 'att_value': 'one', 'person_id': 12}, 
#  {'count': 359, 'att_value': 'nine', 'person_id': 4}]

---

(Re:Edit) To handle other keys (besides just "att_value") appropriately, giving a default value of '-' in the same way, you can use:

keys = list1[0].keys() - {'count'}
idx = {d['count'] : d for d in list1}
list3 = []
for d in list2:
    d2 = idx.get(d['count'], dict.fromkeys(keys, '-'))
    d2.update(d)
    list3.append(d2)

print(list3)
# [{'count': 359, 'att_value': 'nine', 'person_id': 4},
#  {'count': 351, 'att_value': 'one', 'person_id': 12},
#  {'person_id': 8, 'att_value': '-', 'count': 381}]