Get common elements from two lists without repetitions

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Get common elements from two lists without repetitions

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#1 by (1 votes)

1

I have to implement the following statement:

Design a function that receives two lists and returns common elements to both, without repeating any (intersection of sets). Example: if you receive the lists [1, 2, 1] and [2, 3, 2, 4], you will return the list [2].

I tried the following:

def listas(a,b):
    lista_final=[]
    for i in a:
        for x in b:
            if i == b:
                lista_final.append(i)
    print (lista_final)

lista=[]
lista2=[]

i=0
while i == 0:
    num=int(input("Escribe una lista de numeros "))
    lista.append(num)
    a=str(input("Pulsa x si quieres acabar: "))
    if a == 'x':
        i=i+1
i=0
while i == 0:
    num2=int(input("Escribe una lista de numeros "))
    lista2.append(num2)
    a=str(input("Pulsa x si quieres acabar: "))
    if a == 'x':
        i=i+1

listas(lista,lista2)

But I always get an empty list and I do not know how to continue.

python python-3.x

asked by Pedro 08.01.2018 в 13:47

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1 answer

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score 1 · Answer 1

The main idea is yours, but some observations:

The comparison if i == b should actually be if i == x . With this you would find only the common elements. But if you have repetitions due to the for nested, for each element present in both lists you will get n * m times that element in the final list (where n is the number of times that element appears in the first list and m which appears in the second).
To avoid the above you should check for each iteration that this element is not already in the list of results.
To know if an item is in a list in Python, the simplest thing is to use in in a conditional: if elemento in lista: . As you only look for the common elements, it is enough that you iterate on one of the lists.

With this, your code could be:

def listas(a, b):
    lista_final = []
    for i in a:
        if (i not in lista_final) and (i in b):
            lista_final.append(i)
    return lista_final

Example:

>>> lista = [1,2,3,2,4,5]
>>> lista2 = [7,2,5,8,10]
>>> res = listas(lista, lista2)
>>> print(res) 
[2, 5]

Of course, in "real" life it is an extremely inefficient method compared to the intersection of sets already implemented in Python using hash tables:

res = list(set(lista) & set(lista2))