给你一个非空的整数列表(X)。对于这个任务,你应该返回一个只包含这个列表中的非唯一元素的列表。为此,您需要删除所有唯一元素(仅包含在给定列表中的元素一次)。解决此任务时,请勿更改列表的顺序。例如:[1,2,3,1,3] 1和3个非唯一元素,结果将为[1,3,1,3]。
#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.
lista=[]
if 0 < len(data) and len(data) < 1000:
for i in data:
if data.count(i)>1:
lista.append(i)
return lista
#replace this for solution
return data
#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")
大神代码:
#Your optional code here
#You can import some modules or create additional functions
def checkio(data):
return ([x for x in data if data.count(x)>1])
#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")