CODEWITHSUNDEEP
pythonlistunique
Nikolay
Nikolay

Reputation: 90

Delete unique elements from a list

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

Upvotes: 3

Views: 17994

Answers (9)

Jace999
Jace999

Reputation: 831

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]

Upvotes: 0

Kris
Kris

Reputation: 1

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

Upvotes: 0

Pole_Star
Pole_Star

Reputation: 117

 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.

Upvotes: 2

Jon Clements
Jon Clements

Reputation: 142146

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]

Upvotes: 5

Max
Max

Reputation: 23

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.

Upvotes: 2

Satish Kumar Reddy
Satish Kumar Reddy

Reputation: 111

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])

Upvotes: 5

chapelo
chapelo

Reputation: 2562

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]

Upvotes: 1

Irshad Bhat
Irshad Bhat

Reputation: 8709

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
[]

Upvotes: 2

ekhumoro
ekhumoro

Reputation: 120598

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.

Upvotes: 5

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