Delete unique elements from a list

Question

I faced some problem with solving the next problem:

We have a list of elements (integers), and we should return a list consisting of only the non-unique elements in this list. Without changing order of the list I think the best way is to delete or remove all unique element.

Take note that I just start to learn python and would like only the simplest solutions.

Here is my code:

def checkio(data):
    for i in data:
        if data.count(i) == 1:    #if element seen in the list just ones, we delet this el
            ind = data.index(i)
            del data[ind]
    return data

Rather than deleting the items, what about creating a new list with the unique values? — JoErNanO, Nov 11 '14 at 17:26
@JoErNanO I need a list with NON-unique elements (all of them with original order) — Nikolay, Nov 11 '14 at 18:44
just convert a list into set by using a set() function, you will get the desired output. If you want a final answer in a list, just convert that set into a list using list() function. — Kshitij Labhasetwar, Jan 21 '20 at 10:37

score 5 · Answer 1 · edited Mar 31 '18 at 15:44

You can implement a OrderedCounter, eg:

from collections import OrderedDict, Counter

class OrderedCounter(Counter, OrderedDict): 
    pass

data = [1, 3, 1, 2, 3, 5, 8, 1, 5, 2]

duplicates = [k for k, v in OrderedCounter(data).items() if v > 1]
# [1, 3, 2, 5]

So you count the occurrence of each value, then filter on if it has a frequency of more than one. Inheriting from OrderedDict means the order of the original elements is preserved.

Going by comments, you want all duplicated elements reserved, so you can pre-build a set of the duplicate entries, then re-iterate your original list, eg:

from collections import Counter

data = [1, 3, 1, 2, 3, 5, 8, 1, 5, 2]
duplicates = {k for k, v in Counter(data).items() if v > 1}
result = [el for el in data if el in duplicates]
# [1, 3, 1, 2, 3, 5, 1, 5, 2]

The first one looks like wishful thinking. Inherit from two classes, hoping that the result does what you want it to. Maybe it does, but it's not obvious why. — Kelly Bundy, Apr 01 '23 at 21:35

score 5 · Accepted Answer · answered Nov 11 '14 at 18:55

5

Your function can be made to work by iterating over the list in reverse:

def checkio(data):
    for index in range(len(data) - 1, -1, -1):
        if data.count(data[index]) == 1:
            del data[index]
    return data

print(checkio([3, 3, 5, 8, 1, 4, 5, 2, 4, 4, 3, 0]))
[3, 3, 5, 4, 5, 4, 4, 3]
print(checkio([1, 2, 3, 4]))
[]

This works, because it only deletes numbers in the section of the list that has already been iterated over.

answered Nov 11 '14 at 18:55

ekhumoro

115,249
20
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336

This has a quadratic run time complexity. The better (`O(n)`) solution is to [use a `Counter`](https://stackoverflow.com/a/26870897/1222951). – Aran-Fey Mar 31 '18 at 15:47

score 5 · Answer 3 · answered Jun 02 '16 at 16:55

5

Just I used list Comprehensions.

def checkio(data):
    a=[i for i in data if data.count(i)>1]
    return a
print checkio([1,1,2,2,1,1,1,3,4,5,6,7,8])

answered Jun 02 '16 at 16:55

Satish Kumar Reddy

111
1
2
5

Irshad Bhat · Answer 4 · 2014-11-11T19:02:54.767

2

Try this:

>>> a=[1,2,3,3,4,5,6,6,7,8,9,2,0,0]
>>> a=[i for i in a if a.count(i)>1]
>>> a
[2, 3, 3, 6, 6, 2, 0, 0]
>>> a=[1, 2, 3, 1, 3]
>>> a=[i for i in a if a.count(i)>1]
>>> a
[1, 3, 1, 3]
>>> a=[1, 2, 3, 4, 5]
>>> a=[i for i in a if a.count(i)>1]
a
[]

edited Nov 11 '14 at 19:02

answered Nov 11 '14 at 17:50

Irshad Bhat

8,479
1
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i ned all of non-unique elements. Example 1: [1, 2, 3, 1, 3] -->> 1 and 3 non-unique elements and result will be [1, 3, 1, 3]. Example 2: [1, 2, 3, 4, 5] -->> non unique elements and result will be [] – Nikolay Nov 11 '14 at 18:47

score 2 · Answer 5 · answered Feb 17 '17 at 15:53

def checkio(data):
    lis = []
    for i in data:
        if data.count(i)>1:
            lis.append(i)
    print(lis)
checkio([1,2,3,3,2,1])

Yeah it's a bit late to contribute to this thread but just wanted to put it there on the net for anyone else use.

score 1 · Answer 6 · answered Nov 11 '14 at 19:02

Following what you have started, iterating on the list of integers, but not counting or deleting elements, try just testing if the element has already been seen, append it to a list of duplicated elements:

def checkio(data):
    elements = []
    duplicates = []
    for i in data:
        if i not in elements:
            elements.append(i)
        else:
            if i not in duplicates:
                duplicates.append(i)
    return duplicates

d = [1, 3, 1, 2, 3, 5, 8, 1, 5, 2]

print (checkio(d))
#[1, 3, 5, 2]

d = [1, 3, 1, 2, 3, 5, 8, 1, 5, 2] result should be --->>> [1, 3, 1, 2, 3, 5, 1, 5, 2] (deleted onlu unique '8') — Nikolay, Nov 11 '14 at 19:07

score 1 · Answer 7 · answered Oct 18 '19 at 12:37

1

 numbers = [1, 1, 1, 1, 3, 4, 9, 0, 1, 1, 1]
 x=set(numbers)
 print(x)

You can use the set key word too to get the desired solution.

answered Oct 18 '19 at 12:37

Pole_Star

107
7

score 0 · Answer 8 · answered Dec 08 '19 at 00:14

I used an integer and bool to check every time the list was modified within a while loop.

rechecks = 1
runscan = True
while runscan == True:
    for i in data:
         if data.count(i) <2:
             data.remove(i)
             rechecks+=1
            #need to double check now
    if rechecks >0:
        runscan = True
        rechecks-=1
    else:
        runscan = False       
return data

score 0 · Answer 9 · answered Apr 03 '20 at 12:06

Would it not be easier to generate a new list?

def unique_list(lst):
    new_list = []
    for value in lst:
        if value not in new_list:
            new_list.append(value)
    return new_list

lst = [1,2,3,1,4,5,1,6,2,3,7,8,9]
print(unique_list(lst))

Prints [1,2,3,4,5,6,7,8,9]

Delete unique elements from a list

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