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This could be more effiecent? solution in Clear category for Non-unique Elements by chesse20
#Your optional code here
#You can import some modules or create additional functions
def checkio(data: list) -> list:
b_list = []
for index in range (0, len(data)): ## for entire list
if (data.count(data[index]) > 1): ##if the data point appears more than once
print("appending "+ str(data[index])) ##debug code
b_list.append(data[index]) ##put it in the new list to return
return b_list ##return the new list
#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
print(180 // 12)
print (184-169-8)
assert list(checkio([1, 2, 3, 1, 3])) == [1, 3, 1, 3], "1st example"
assert list(checkio([1, 2, 3, 4, 5])) == [], "2nd example"
assert list(checkio([5, 5, 5, 5, 5])) == [5, 5, 5, 5, 5], "3rd example"
assert list(checkio([10, 9, 10, 10, 9, 8])) == [10, 9, 10, 10, 9], "4th example"
print("It is all good. Let's check it now")
June 30, 2019
