如何获取只有唯一数字的列表？ （Python）

Question

I want to write a program that checks for duplicate values and removes them. So for example I want from this:

list = [2,6,8,2,9,8,8,5,2,2]

To get only this:

uniques = [6,9,5]

I can't seem to find a good way to compare every item with each other to see if they are equal and them remove them. Any help will be very appreciated!

Answer 1

Use collections.Counter :

>>> from collections import Counter                                         
>>> lst = [2,6,8,2,9,8,8,5,2,2]                                             
>>> c = Counter(lst)                                                        
>>> [x for x in c if c[x] == 1]                                             
[6, 9, 5]

Other than using count or in for each element, this should be O(n) instead of O(n²).

Answer 2

If you want to preserve the order of the elements too, follow this best method :

uniques = list(dict.fromkeys(list))

print(uniques)

Answer 3

Given that you want a list of values that appear only once in your list, this should work. I used a hashmap to store the count of values in the the list and then added the keys which have a count of 1 into unqiue .

l = [2,6,8,2,9,8,8,5,2,2]
unique = []
count = {}

for i in l:
    if count.get(i) is None:
        count[i] = 1
    else:
        count[i]+=1

for i in count.keys():
    if count[i] == 1:
        unique.append(i)

print(unique)

Output

[6, 9, 5]

Answer 4

def unique(x: list):                                                                                                                                                                                                  
    a = set()                                                                                                                                                                                                            
    for i in range(len(x)):                                                                                                                                                                                                  
        if x.index(x[i]) == (len(x) - 1) - x[::-1].index(x[i]):                                                                                                                                                                  
            a.add(x[i])                                                                                                                                                                                                  
    return list(a)
                                                                                                                                                                                                                                                                                                                                                                                                                   
print(unique([1, 2, 1, 11, 4, 4, 6, 3, 2]))                                                                                                                                                                          
# [3, 11, 6]

Answer 5

list = [2,6,8,2,9,8,8,5,2,2] uniques = [] for i in range(len(list)): if list.count(list[i])==1: uniques.append(list[i])

Answer 6

 list = [2,6,8,2,9,8,8,5,2,2] uniques = [] for i in list: if i not in uniques: uniques.append(i)