如何完全结束for循环python

Question

我有一个作业，无法找出真正的解决方案。

def triple(n):    #multiplies number with 3
    return (n*3)
def square(n):
    return (n**2)   #takes second power of number

for i in range(1,11):   #I asked to iterate 1 to 10
    triple_ = triple(i)
    square_ = square(i)
    if triple_ <= square_:   #it shouldn't print if square of the number is larger than the triple
        break                #it must END the loop
    print(f"triple({i})=={triple(i)} square({i})=={square(i)}")

triple(1)==3 square(1)==1

triple(2)==6 square(2)==4

我的代码停止打印了，但是他们不想要那样。他们想完全完成它。

edit1我对自己的陈述不实，对不起，它不应多次调用函数。他们每个人只有一次，我为此尝试过休息。

Answer 1

根据您的评论，如果条件全部错误：

def triple(n):    #multiplies number with 3
    return (n*3)
def square(n):
    return (n**2)   #takes second power of number

for i in range(1,11):   #I asked to iterate 1 to 10
    triple_ = triple(i)
    square_ = square(i)
    if triple_ > square_:   #it should only print if the square of the number is smaller than the triple
        print(f"triple({i})=={triple(i)} square({i})=={square(i)}")

中断将退出foor循环，您想要避免打印，这是一个完全不同的主题

Answer 2

尝试以下代码，

    def triple(n):    #multiplies number with 3
        return (n*3)
    def square(n):
        return (n**2)   #takes second power of number

    for i in range(1,11):   #I asked to iterate 1 to 10
        triple_ = triple(i)
        square_ = square(i)
        if triple_ < square_:   #it shouldnt print if square of the 
number is larger than the triple
            pass #it must END the loop
        else:
            print("Triple of " + str(i) + " is " + str(triple(i)) + " and its greater than or equal to its square " + str(square(i)))

在i = 3的情况下，平方是9，三元组也是9。因此，如果将<=替换为

此后停止打印，因为不满足任何条件。对于介于（1,10）之间的数字，三元组为正方形小于或等于三元组的唯一可能数字为1,2和3。