How to check if a list contains duplicate items?

Question

I want to validate a list to make sure that there are no duplicate items. My problem is that I don't know how to do this in the if statement. Is there a method or something in python that will return False if there are duplicates in the list?

Here is what I had in mind:

lst = ["1","2","3","3","4"]

if #lst contains no duplicates :
    print("success")
else:
    print("duplicate found")

Thanks in advance.

Possible duplicate of [Removing duplicates in lists](https://stackoverflow.com/questions/7961363/removing-duplicates-in-lists) — Pavel, Jan 11 '18 at 14:44
Make the list into a set using `set(lst)` and then compare the two lengths. If they are the same then there wont be any duplicates — wtreston, Jan 11 '18 at 14:44
Possible duplicate of [Check for duplicates in a flat list](https://stackoverflow.com/questions/1541797/check-for-duplicates-in-a-flat-list) — EoinS, Jan 11 '18 at 14:46

score 21 · Accepted Answer · answered Jan 11 '18 at 14:45

21

As said by Jkdc, convert it to a set and compare the length

lst = ["1","2","3","3","4"]

if len(set(lst)) == len(lst):
    print("success")
else:
    print("duplicate found")

answered Jan 11 '18 at 14:45

ThibThib

7,580
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3

This doesn't work when `lst` is a list of dictionaries. `set(lst)` then throws a `TypeError: unhashable type: 'dict'` – the21st Oct 10 '18 at 10:28

Poiz · Answer 2 · 2018-01-11T15:00:58.180

4

Exploit the Fact that Python set may not contain duplicates. The has_duplicate() Function takes care of determining if a list contains duplicates.

def has_duplicates(listObj):
    return len(listObj) != len(set(listObj))


print(has_duplicates([1, 2, 1, 1, 4, 7]))    ## PRINTS: True
print(has_duplicates([9, 2, 5, 4, 7]))       ## PRINTS: False

edited Jan 11 '18 at 15:00

answered Jan 11 '18 at 14:46

Poiz

7,519
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1

you can simplify to `return len(listObj) != len(set(listObj))`. – Ma0 Jan 11 '18 at 14:48

score 0 · Answer 3 · answered Jan 11 '18 at 15:15

Check this, the easiest way (at least for me)..

lst = ["1","2","3","3","4"]
status = True
for item in lst:
    if lst.count(item) > 1:
      # the count functions counts how many times the "item" is in lst
        status = False

if status == True:
    print("No duplicates")
else:
    print("Duplicates found")

score 0 · Answer 4 · answered Mar 11 '20 at 14:51

0

def check_duplicates(lst):
    seen = {}
    for item in lst:
        if seen.get(item):
            print("duplicate found")
            return
        else:
            seen[item] = True
    print("success")

answered Mar 11 '20 at 14:51

ahmed meraj

644
1
7
13

score 0 · Answer 5 · answered Apr 09 '20 at 00:52

0

def checkDuplicate():
    count = {}
    for item in lst:
            if item not in count:
                count[item] = 1
            else:
                return True
    return False

answered Apr 09 '20 at 00:52

Mahek Pavagadhi

1

4

While this code may solve the question, [including an explanation](https://meta.stackexchange.com/q/114762) of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes. Remember that you are answering the question for readers in the future, not just the person asking now. Please [edit] your answer to add explanations and give an indication of what limitations and assumptions apply. – Brian Apr 09 '20 at 01:05

How to check if a list contains duplicate items?

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