python check two lists are equal using a not equal

Question

I am new to the python. I have two lists:

l1 = [1,1,1]
l2 = [1,1,1]

Now, I have to check whether these two are equal or not. I have a solution which I am currently using:

def comp(l1,l2):
    #condition we are checking [1,1,1] and  [1,1,1]
    try:
        a= set(l1)  #converting a list into set , as set will return the length on basis of unique elements.
        b= set(l2)
        return ((a == b) and (len(a) == 1) and (len(b) == 1))
    except TypeError:
        return False

Now, with this I am able to solve the task. But I am trying to do it elementwise.

l1 = [1,1,12]
l2 = [1,1,1]

Then I can treat the 12 as a 1 so I need to know which number is not matching. Can any one help me with this?

Answer 1

Assuming they have the same length, you could do:

{i: (v, l2[i]) for i, v in enumerate(l1) if v != l2[i]}

which returns

{2: (12, 1)}

for you example above. The keys are the indices, the values are tuples of the respective values in the lists.

Answer 2

If you need to know if the two list are exactly the same you can do l1 == l2 directly.

If you need to compare the lists element-wise (assuming they have the same length) and you only play with list of numbers, you can use numpy arrays:

import numpy as np

l1 = np.array([1,1,12])
l2 = np.array([1,1,1])

l1 == l2 # array([ True, True, False])

Answer 3

Assuming the lists are always the same length, You can do:

is_all_equel = True
for i in range(len(l1)):
    if l1[i] != l2[i]:
        print("index " + str(i) + " is " + str(l1[i]) + " in list 1, and " + str(l2[i]) + " in list 2")
        is_all_equel = False

In is_all_equel will be the answer if all the elements in the lists are equel

Answer 4

something like:

l1 = [1,1,12]
l2 = [1,1,1]
for elem in l1:
    if not elem in l2:
        print(elem)
        print(l1.index(elem))