CheckIO - Home Challenge 1 : Non-unique Elements

checkio.1.01.Non-uniqueElements.py

# CheckIO - Home Challenge 1 : Non-unique Elements
# http://checkio.org

# You are given a non-empty list of integers (X). 
# For this task, you should return a list consisting of only the non-unique elements in this list. 
# To do so you will need to remove all unique elements (elements which are contained in a given list only once).
# When solving this task, do not change the order of the list. 
# Example: [1, 2, 3, 1, 3] 1 and 3 non-unique elements and result will be [1, 3, 1, 3].

def checkio(data):
    
    result = []
    for item in data:
        if (data.count(item) > 1):
            result.append(item)
    return result

#Some hints
#You can use list.count(element) method for counting.
#Create new list with non-unique elements
#or remove elements from original list (but it's bad practice for many real cases)
#Loop over original list


if __name__ == "__main__":
    #These "asserts" using only for self-checking and not necessary for auto-testing
    assert isinstance(checkio([1]), list), "The result must be a list"
    assert checkio([1, 2, 3, 1, 3]) == [1, 3, 1, 3], "1st example"
    assert checkio([1, 2, 3, 4, 5]) == [], "2nd example"
    assert checkio([5, 5, 5, 5, 5]) == [5, 5, 5, 5, 5], "3rd example"
    assert checkio([10, 9, 10, 10, 9, 8]) == [10, 9, 10, 10, 9], "4th example"