Delete unique elements from a list

问题

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

回答1:

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.

回答2:

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]

回答3:

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
[]

回答4:

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])

回答5:

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.

回答6:

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]

回答7:

 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.

回答8:

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

来源：https://stackoverflow.com/questions/26870699/delete-unique-elements-from-a-list

标签

python

list

unique