CODEWITHSUNDEEP
pythondictionary
Rina
Rina

Reputation: 159

Find common elements and their frequency in list of dictionaries

I have multiple (~40) lists that contain dictionaries, that I would like to find

  1. which are those list items (in this case dictionaries) that are common in all lists
  2. how many times each unique item appears across all lists.

Some examples of the lists are:

a = [{'A': 0, 'B': 0},
     {'A': 0, 'C': 1},
     {'D': 1, 'C': 0},
     {'D': 1, 'E': 0}]

b = [{'A': 0},
     {'B': 0, 'C': 1},
     {'D': 1, 'C': 0},
     {'D': 1, 'E': 0}]

c = [{'C': 0},
     {'B': 1},
     {'D': 1, 'C': 0, 'E': 0},
     {'D': 1, 'E': 0}]

What I tried so far, it is the following code, but it returned values that were not common in all lists...

def flatten(map_groups):
    items = []
    for group in map_groups:
        items.extend(group)
    return items


def intersection(map_groups):
    unique = []
    items = flatten(map_groups)
    for item in items:
        if item not in unique and items.count(item) > 1:
            unique.append(item)
    return unique

all_lists = [a,b,c]
intersection(all_lists)

What I would expect to get as a result would be:


1. {'D': 1, 'E': 0} as a common item in all lists

2.   {'D': 1, 'E': 0}, 3   
     {'D': 1, 'C': 0}, 2
     {'A': 0, 'B': 0}, 1
     {'A': 0, 'C': 1}, 1
     {'A': 0}, 1
     {'B': 0, 'C': 1}, 1
     {'C': 0},
     {'B': 1},
     {'D': 1, 'C': 0, 'E': 0}

Upvotes: 0

Views: 888

Answers (2)

mensi
mensi

Reputation: 9826

To count things, python comes with a nice class: collections.Counter. Now the question is: What do you want to count?

For example, if you want to count the dictionaries that have the same keys and values, you can do something like this:

>>> count = Counter(tuple(sorted(x.items())) for x in a+b+c)
>>> count.most_common(3)
[((('C', 0), ('D', 1)), 2), ((('D', 1), ('E', 0)), 2), ((('A', 0), ('B', 0)), 1)]

The dictionaries here are converted to tuples with sorted items to make them comparable and hashable. Getting for example the 3 most common back as a list of dictionaries is also not too hard:

>>> [dict(x[0]) for x in count.most_common(3)]
[{'C': 0, 'D': 1}, {'D': 1, 'E': 0}, {'A': 0, 'B': 0}]

Upvotes: 2

Red
Red

Reputation: 27547

You can use a nested for loop:

a = [{'A': 0, 'B': 0},
     {'A': 0, 'C': 1},
     {'D': 1, 'C': 0},
     {'D': 1, 'E': 1}]

b = [{'A': 0},
     {'B': 0, 'C': 1},
     {'D': 1, 'C': 0},
     {'D': 1, 'E': 0}]

c = [{'C': 0},
     {'B': 1},
     {'D': 1, 'C': 0, 'E': 0},
     {'D': 1, 'E': 0}]

abc_list = [*a, *b, *c]
abc = list()

for d in abc_list:
    for i in abc:
        if d == i[0]:
            abc[abc.index(i)] = (d, i[1] + 1)
            continue
    abc.append((d, 1))

print(abc)

Output:

[({'A': 0, 'B': 0}, 1),
 ({'A': 0, 'C': 1}, 1),
 ({'D': 1, 'C': 0}, 2),
 ({'D': 1, 'E': 1}, 1),
 ({'A': 0}, 1),
 ({'B': 0, 'C': 1}, 1),
 ({'D': 1, 'E': 0}, 2),
 ({'C': 0}, 1),
 ({'B': 1}, 1),
 ({'D': 1, 'C': 0, 'E': 0}, 1)]

Explanation:

The line

[*a, *b, *c]

unpacks all the values in lists a, b and c into a single list, which \i named abc_list.

The continue statement where I put it means to directly continue to the next iteration of the inner for loop, without reaching abc.append((d, 1)).

The above output answers question 2. For question 1, we can use the built-in max() method on the abc list, with a custom key:

print(max(ABC, key=lambda x:x[1])[0])

Of course, it will only return one dictionary, {'D': 1, 'C': 0}. If you want to print out multiple dictionaries that appear the most frequently:

m = max(abc, key=lambda x:x[1])[1]
for d in abc:
    if d[1] == m:
      print(d[0])

Output:

{'D': 1, 'C': 0}
{'D': 1, 'E': 0}

Upvotes: 1

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