Need to find top 10 used surnames in a files. Made a dictonary but need to sort it the rest

As mentioned in a comment, your dict is actually a list.

Try using the Counter object from the collections library. In the below example, I have edited your list so that it contains a few duplicates.

from collections import Counter

surnames = ['KRISTIANSEN', 'OLDERVIK', 'GJERSTAD', 'VESTLY SKIVIK', 'NYMANN', 'ØSTBY', 'LINNERUD', 'REMLO', 'SKARSHAUG', 'ELI', 'ADOLFSEN', 'OLDERVIK', 'ØSTBY', 'ØSTBY']

counter = Counter(surnames)

for name in counter.most_common(3):
    print(name)

The result becomes:

('ØSTBY', 3)
('OLDERVIK', 2)
('KRISTIANSEN', 1)

Change the integer argument to most_common to 10 for your use case.

You could place all the values in a dictionary where the value is the number of times it appears in the dataset, and filter through your newly created dictionary and push any result that has a value count > 10 to your final array.

edit: your surname_dict was initialized as an array, not a dictionary.

surname_dict = {}
top_ten = []
for index in data_list:
    if index['lastname'] not in surname_dict.keys():
        surname_dict[index['lastname']] = 1
    else:
        surname_dict[index['lastname']] += 1

for k, v in sorted(surname_dict.items()):
    if v >= 10:
        top_ten.append(k)
return top_ten

The best approach to answer your question is to consider the top ten categories : for example : category of names that are used 9 times and category of names that are used 200 times and so . Because , we could have a case where 100 of users use different usernames but all of them have to be on the top 10 used username. So to implement my approach here is the script :

def counter(file : list):
    L = set(file)
    i = 0
    M = {}
    for j in L :
        for k in file :
            if j == k:
                i+=1
        M.update({i : j})
        i = 0
    D = list(M.keys())
    D.sort()
    F = {}
    if len(D)>= 10:
        K = D[0:10]
        for i in K:
            F.update({i:D[i]})
        return F
    else :
        return M

Note: my script calculate the top ten categories .

Just use a standard dictionary. I've added some duplicates to your data, and am using a threshold value to grab any names with more than 2 occurences. Use threshold = 10 for your actual code.

names = ['KRISTIANSEN', 'OLDERVIK', 'GJERSTAD', 'VESTLY SKIVIK', 'NYMANN', 'ØSTBY','ØSTBY','ØSTBY','REMLO', 'LINNERUD', 'REMLO', 'SKARSHAUG', 'ELI', 'ADOLFSEN']

# you need 10 in your code, but I've only added a few dups to your sample data
threshold = 2

di = {}
for name in names:
    #grab name count, initialize to zero first time
    count = di.get(name, 0)
    di[name] = count + 1


#basic filtering, no sorting
unsorted = {name:count for name, count in di.items() if count >= threshold}
print(f"{unsorted=}")


#sorting by frequency: filter out the ones you don't want
bigenough = [(count, name) for name, count in di.items() if count >= threshold]

tops = sorted(bigenough, reverse=True)

print(f"{tops=}")

#or as another dict

tops_dict = {name:count for count, name in tops}
print(f"{tops_dict=}")

Output:

unsorted={'ØSTBY': 3, 'REMLO': 2}
tops=[(3, 'ØSTBY'), (2, 'REMLO')]
tops_dict={'ØSTBY': 3, 'REMLO': 2}

Update. Wanted to share what code I made in the end. Thank you guys so much. The feedback really helped.

Code:

etternavn_dict = {}

for index in data_list:
    if index['etternavn'] not in etternavn_dict.keys():
        etternavn_dict[index['etternavn']] = 1
    else:
        etternavn_dict[index['etternavn']] += 1

print("\nTopp 10 etternavn:")
count = 0

for k, v in sorted(etternavn_dict.items(), key=lambda item: item[1]):
    if count < 10:
        print(k)
        count += 1
    else:
        break