如何在python的列表中找到唯一元素？ （不使用 set）

Question

Write a function that accepts an input list and returns a new list which contains only the unique elements (Elements should only appear one time in the list and the order of the elements must be preserved as the original list. ).

def unique_elements (list):
    new_list = []
    length = len(list)
    i = 0
    while (length != 0):
        if (list[i] != list [i + 1]):
            new_list.append(list[i])
        i = i + 1
        length = length - 1
    '''new_list = set(list)'''
    return (new_list)

#Main program
n = int(input("Enter length of the list: "))
list = []
for i in range (0, n):
    item = int(input("Enter only integer values: "))
    list.append(item)
print ("This is your list: ", list)
result = unique_elements (list)
print (result)

I am stuck with this error:

IndexError: list index out of range

Answer 1

This is the simplest way to do it:

a = [1, 2, 2, 3]
b = []
for i in a:
    if i not in b:
        b.append(i)
print (b)
[1, 2, 3]

Answer 2

The problem with your code is that you are looping length times but checking list[i] with list[i+1] , thus accessing an element past the end of the input list (eg in a list with 6 elements there are 6-1=5 pairs of consecutive elements).

A second issue with your code is that an input with only one element [1] should give as output [1] even if this element is not different from any other. The input text means you should remove elements that are equal to other elements already present, not that you should keep elements that are different from the next one.

Another issue is that you're only checking for consecutive duplicates ie given the input list [1, 2, 1, 2] your logic wouldn't detect any duplication... looks like the exercise instead requires in this case as output of [1, 2] .

A trace for a simple algorithm to do this is

for each element in input
   if the element has not been included in output
       add the element to the end of output

Note also that to check if an element is present in a list Python provides the in operator (eg if x in output: ... ) that can save you an explicit loop for that part.

As a side note naming an input parameter list is considered bad practice in Python because list is the name of a predefined function and your parameter is hiding it.

Answer 3

O(n) solution without using a set:

>>> from collections import Counter, OrderedDict
>>> class OrderedCounter(Counter, OrderedDict):
...     pass
... 
>>> lst = [1, 2, 2, 3, 4, 5, 4]
>>> [x for x,c in OrderedCounter(lst).items() if c==1]
[1, 3, 5]

Answer 4

One line implementation:

list = [100, 3232, 3232, 3232, 57, 57, 90]
new_list = []

[new_list.append(x) for x in list if x not in new_list]

print(new_list)

Prints:

[100, 3232, 57, 90]

Answer 5

The problematic line is > if (list[i] != list [i + 1]): < (6th line in your code).

Reason: Imagine your list has got 4 elements.

Eg: mylist = [ 1, 2,2,3].

mylist[i] != mylist [i + 1]

In the last iteration 'i' will be 4 , so i + 1 will be 5.

There is no such 5th index in that list, because list indexes are counted from zero.

mylist[0] = 1

mylist[1] = 2

mylist[2] = 2

mylist[3] = 3

mylist[4] = No Index

mylist[5] = No Index

    def unique_elements (list):
        new_list = []

   # Replace the while with a for loop** 

        for i in list:
          if i not in new_list:
            new_list.append(i)


        return (new_list)

    #Main program
    n = int(input("Enter length of the list: "))
    list = []
    for i in range (0, n):
        item = int(input("Enter only integer values: "))
        list.append(item)
    print ("This is your list: ", list)
    result = unique_elements (list)
    print (result)

Answer 6

l = [1, 2, 2, 3,4,5,6,5,7,8]
myList = []
[ myList.append(item) for item in l if item not in myList]
print(myList)

Answer 7

Use of collections is best IMO, Accepted answer is simplest, Adding another approach using dict where you can check frequency as well,

text = [100, 3232, 3232, 3232, 57, 57, 90]
# create empty dictionary
freq_dict = {}
 
# loop through text and count words
for word in text:
    # set the default value to 0
    freq_dict.setdefault(word, 0)
    # increment the value by 1
    freq_dict[word] += 1
 
unique_list = [key for key,value in freq_dict.items()]

print(unique_list )

print(freq_dict )

[100, 3232, 57, 90]
{100: 1, 3232: 3, 57: 2, 90: 1}

[Program finished] 

You can also print by values based on requirements.

Answer 8

You can work this out using sets and set comprehension syntax is in most cases overlooked. Just as lists sets can also be generated using comprehension.

elements = [1, 2, 3, 3, 5, 7, 8, 7, 9]
unique_elements = {element for element in elements}
print(unique_elements)

Answer 9

Found an easier approach for people on Python 3.7.

Using a dictionary :

list(dict.fromkeys(mylist))