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Non-unique solution in Clear category for Non-unique Elements by Seaclaid
def repet(list,num,k):
list_odp=[]
zero=[]
r=0
for z in range(len(list)):
if list[z]==num:
r=r+1
print('repet: ',r)
if r>1:
return num
else:
return zero
def checkio(data):
print(data)
temp=[]
odp=[]
for i in range(len(data)):
temp=repet(data,data[i],i)
if temp!=[]:
odp.append(temp)
print(odp)
return odp
#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"
Jan. 9, 2017
