Question

在这段代码中，我试图将每次循环的集合的值与参数中传递的值（在本例中为a）进行比较。有趣的是它显示当我为每个循环使用a时，每个元素都是一个整数。如何在没有控制台错误的情况下获得整数与整数比较？

def remove(s,a,b):
    c=set()
    c=s
    for element in c:
        element=int(element)
        if(element<a or element>b):
            c.discard(element)
    return c

def main():
    remove({3, 17, -1, 4, 9, 2, 14}, 1, 10)

main()

输出：

    if(element<=a or element>=b):
TypeError: '>=' not supported between instances of 'int' and 'set'

Answer 1

您重新分配本地变量b：

def remove(s,a,b):
    b=set()  # now b is no longer the b you pass in, but an empty set
    b=s  # now it is the set s that you passed as an argument
    # ...    
    if(... element > b): # so the comparison must fail: int > set ??

使用集合理解的简短实现：

def remove(s, a, b):
    return {x for x in s if a <= x <= b}

>>> remove({3, 17, -1, 4, 9, 2, 14}, 1, 10)
{9, 2, 3, 4}

Answer 2

如果你想要int to int compare，那么把b作为s的列表。

def remove(s,a,b):
    b = list(s)
    for element in s:
        element=int(element)      
        if(element< a or element > b):
            b.remove(element)
    return b

def main():
    remove({3, 17, -1, 4, 9, 2, 14}, 1, 10)

main()

Answer 3

来吧，我们为什么不缩短代码？

试试这个：

def remove(s, a, b):
    return s.difference(filter(lambda x: not int(a) < int(x) < int(b), s))


def main():
    new_set = remove({3, 17, -1, 4, 9, 2, 14}, 1, 10)

    # {2, 3, 4, 9}
    print(new_set)


main()

如何更改集合中的特定元素？

3 个答案: