python非独特元素代码实例

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]。

我的代码：

#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")