Htykuut

>>> list1 = [1,2,3,4,5,6]

>>> list2 = [3, 5, 7, 9]

>>> list(set(list1).intersection(list2))

[3, 5]

You can also use sets and get the commonalities in one line: subtract the set containing the differences from one of the sets.

A = [1,2,3,4]

B = [2,4,7,8]

commonalities = set(A) - (set(A) - set(B))

The solutions suggested by S.Mark and SilentGhost generally tell you how it should be done in a Pythonic way, but I thought you might also benefit from knowing why your solution doesn't work. The problem is that as soon as you find the first common element in the two lists, you return that single element only. Your solution could be fixed by creating a result list and collecting the common elements in that list:

def common_elements(list1, list2):

    result = 

    for element in list1:

        if element in list2:

            result.append(element)

    return result

An even shorter version using list comprehensions:

def common_elements(list1, list2):

    return [element for element in list1 if element in list2]

However, as I said, this is a very inefficient way of doing this -- Python's built-in set types are way more efficient as they are implemented in C internally.

use set intersections, set(list1) & set(list2)

>>> def common_elements(list1, list2):

...     return list(set(list1) & set(list2))

...

>>>

>>> common_elements([1,2,3,4,5,6], [3,5,7,9])

[3, 5]

>>>

>>> common_elements(['this','this','n','that'],['this','not','that','that'])

['this', 'that']

>>>

>>>

Note that result list could be different order with original list.

The previous answers all work to find the unique common elements, but will fail to account for repeated items in the lists. If you want the common elements to appear in the same number as they are found in common on the lists, you can use the following one-liner:

l2, common = l2[:], [ e for e in l1 if e in l2 and (l2.pop(l2.index(e)) or True)]

The or True part is only necessary if you expect any elements to evaluate to False.

you can use a simple list comprehension:

x=[1,2,3,4]

y=[3,4,5]

common = [i for i in x if i in y]

common: [3,4]

1) Method1
saving list1 is dictionary and then iterating each elem in list2

def findarrayhash(a,b):

    h1={k:1 for k in a}

    for val in b:

        if val in h1:

            print("common found",val)

            del h1[val]

        else:

            print("different found",val)

    for key in h1.iterkeys():

        print ("different found",key)

Finding Common and Different elements:

2) Method2
using set

def findarrayset(a,b):

    common = set(a)&set(b)

    diff=set(a)^set(b)

    print list(common)

    print list(diff) 

Hi, this is my propose (very simple)

import random



i = [1,4,10,22,44,6,12] #first random list, could be change in the future

j = [1,4,10,8,15,14] #second random list, could be change in the future

for x in i: 

    if x in j: #for any item 'x' from collection 'i', find the same item in collection of 'j'

        print(x) # print out the results

Here's a rather brute force method that I came up with. It's certainly not the most efficient but it's something.

The problem I found with some of the solutions here is that either it doesn't give repeated elements or it doesn't give the correct number of elements when the input order matters.

#finds common elements

def common(list1, list2):

    result = 

    intersect = list(set(list1).intersection(list2))



    #using the intersection, find the min

    count1 = 0

    count2 = 0

    for i in intersect: 

        for j in list1:

            if i == j:

                count1 += 1

        for k in list2: 

            if i == k:

                count2 += 1

        minCount = min(count2,count1)

        count1 = 0

        count2 = 0



        #append common factor that many times

        for j in range(minCount):

            result.append(i)



    return result

f_list=[1,2,3,4,5] # First list

s_list=[3,4,5,6,7,8] # Second list

# An empty list stores the common elements present in both the list

common_elements=



for i in f_list:

    # checking if each element of first list exists in second list

    if i in s_list:

        #if so add it in common elements list

        common_elements.append(i) 

print(common_elements)

a_list = range(1,10)

b_list = range(5, 25)

both = 



for i in b_list:

    for j in a_list:

        if i == j:

            both.append(i)

def common_member(a, b): 

    a_set = set(a) 

    b_set = set(b) 

    if (a_set & b_set): 

        print(a_set & b_set) 

    else: 

        print("No common elements") 

list_1=range(0,100)

list_2=range(0,100,5)

final_list=

for i in list_1:

    for j in list_2:

        if i==j:

            final_list.append(i)

print(set(final_list))

this is my proposition
i think its easier with sets than with a for loop

def unique_common_items(list1, list2):

"""Produce the set of *unique* common items in two lists.

   return list(set(list1) & set(list2))

Set is another way we can solve this

a = [3,2,4]

b = [2,3,5]

set(a)&set(b)

{2, 3}

Your problem is that you're returning from inside the for loop so you'll only get the first match. The solution is to move your return outside the loop.

def elementosEnComunEntre(lista1,lista2):



    elementosEnComun = set()



    for e1 in lista1:

         if(e1 in lista2):

             elementosEnComun.add(e1)



    return list(elementosEnComun)