To find duplicates in a python list, why doesn't this logic work?

Question

While I am aware that the best and simplest method to find duplicates in a list is by using Collections.Counter, I wish to know where does the following logic fail.

def uniquenumbercheck(listarg):
    for i in range(2,len(listarg)):
        for j in range(1,i-1):
            if(listarg[i]==listarg[j]):
                print("duplicate value appeared : "+str(listarg[i]))
                return
    print("all entered values are unique. ")

The program is running but failed to display the correct output.Could not figure out the mistake.For the example input of 1,1,0 integers to list, it says that they are unique.

Answer 1

Off-by-one error. In Python, indexes start at 0 , not 1 . So,

for i in range(2, len(listarg)):
    for j in range(1, i - 1):

should be:

for i in range(1, len(listarg)):
    for j in range(0, i):

Also, range(1, len(listarg)) can also be written as range(len(listarg)) .

Answer 2

i am assuming you are searching for duplicates within a list:

myl = [1, 2, 3, 5, 6, 1, 2]

for i in myl:
    if myl.count(i) > 1:
        print i

where the duplicates are printed out.