Count occurrences of an element in list

Question

I know that there are already a lot of questions about this specific topic, but I can't find a proper solution for my problem.

I have the input:

2, 20, 15, 16, 17, 3, 8, 10, 7

I want to see if there are 'double' numbers in my code. I have tried working with this code.

lijst = input('Give a list:  ')
teller = 0
for i in lijst.split(','):
    if lijst.count(i) != 1:
        teller += 1
print(teller != 0)

Normally I should get False, since there are no double numbers in the given list. However, I recieve True. I suggest that's because the 2 appears also in 20.

True

Does anyone know how to avoid this problem, so the the '2' isn't counted twice?

Chen A. · Accepted Answer · 2017-10-05 07:50:54Z

2

You can use collections.Counter which does exactly that

>>> data = [2, 20, 15, 16, 17, 3, 8, 10, 7]
>>> from collections import Counter
>>> Counter(data)
Counter({2: 1, 3: 1, 7: 1, 8: 1, 10: 1, 15: 1, 16: 1, 17: 1, 20: 1})
>>>

It counts number of occurences and returns a dict with keys indicates the item, and value is number of occurences.

If you just need to know if there are duplicates or not, regardless of which item is the duplicate, you can simply use Set over your list and check len() afterwards:

len(data) == len(set(data))

answered Oct 5, 2017 at 7:50

Chen A.

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Comments

Reblochon Masque · Accepted Answer · 2017-10-05 07:51:21Z

1

You can compare the length of the input with the length of the set of unique elements in input:

def has_repeated_elements(input):
    """returns True if input has repeated elements,
    False otherwise"""
    return len(set(input)) != len(input)

print(not has_repeated_elements(input))

answered Oct 5, 2017 at 7:51

Reblochon Masque

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