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one-liner with alternate solution, list comprehension, list.count() solution in Clear category for Non-unique Elements by wdmwilcox
#Your optional code here
#You can import some modules or create additional functions
def checkio(data: list) -> list:
return [item for item in data if data.count(item) != 1]
'''Alternate (above is shorthand for below)'''
#new_list = list()
#for item in data:
# if data.count(item) != 1:
# new_list.append(item)
#return 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
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")
April 28, 2020
