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First solution in Clear category for Non-unique Elements by cinekk
#Your optional code here
#You can import some modules or create additional functions
def checkio(data):
#Your code here
#It's main function. Don't remove this function
#It's used for auto-testing and must return a result for check.
numbersOfElements = {}
i = 0
for el in data:
if el in numbersOfElements:
numbersOfElements[el] += 1
else:
numbersOfElements[el] = 0
while i < len(data):
if numbersOfElements[data[i]] == 0:
del data[i]
i -= 1
i += 1
#replace this for solution
return data
print (checkio([1,2,3,1,3]))
#Some hints
#You can use list.count(element) method for counting.
#Create new list with non-unique elements
#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"
Oct. 18, 2016
