Question

我正在尝试用Python编写程序，以检查数字是否在给定范围内，我希望它在不存在数字的情况下显示一条消息。但是，我提供的以下代码无法打印这样的消息（它适用于前两个条件，但不适用于第三个条件）。如果有人指出我的代码有错误以及是否有更有效的编写方法，我将不胜感激。

def range(a,b,n):
    if b > a:
        if(n>=a) and (n<=b):
            print(n,"is within the range",a,"-",b)
    elif b < a:
        if(n>=b) and (n<=a):
            print(n,"is within the range",a,"-",b)
    else:
        if (n>a,b) or (n<a,b):
            print(n, "is not within the range", a,"-",b)

当我使用来测试代码时

range(2,8,300)

它应该打印300 is not within the range 2 - 8时不打印任何内容。

Answer 1

def range(a,b,n):
    if a < b:
        if n >= a:
            if n <= b:
                print(n,"is within the range",a,"-",b)
            else:
                print(n, "is not within the range", a,"-",b)    
        else:
            print(n, "is not within the range", a,"-",b)                
    if a > b:
        if n <= a:
            if n >= b:
                print(n,"is within the range",a,"-",b)
            else: 
                print(n, "is not within the range", a,"-",b)
        else:
            print(n, "is not within the range", a,"-",b)

这应该有效。

编辑：修改了代码。这还将测试涉及负数和零的情况。

Answer 2

public class InputStreamResource extends AbstractResource { //... @Override public boolean isOpen() { return true; } //... }是对的，因此它进入内部并且嵌套了if 8 > 2:条件失败并退出。因此，它永远不会进入if 300>=2 and 8<=300循环。

此外，else始终返回if (n>a,b) or (n<a,b)。由于 True 成为元组，并且始终为True。

这很好。

("anything..")

Answer 3

您需要进行间隔比较：

def range(a,b,n):
    if a <= n <= a:
            print(n,"is within the range",a,"-",b)
    else:
            print(n, "is not within the range", a,"-",b)


range(2,8,300)

输出：

300 is not within the range 2 - 8

或者：

   def check_range(a,b,n):    
   isRange = range(min(a,b),max(a,b))
   if n in isRange:
       print(n, "is within the range", a, "-", b)
   else:
       print(n, "is not within the range", a, "-", b)


check_range(5,2,4)

输出：

4 is within the range 5 - 2

PS。我不鼓励使用Python内置函数range() 作为您的自定义函数名称。

pyFiddle

Answer 4

您可以尝试以下方法：

def range(a, b, n):
    if a <= n <= b:
        print(n, "is within the range", a, "-", b)
    else:
        print(n, "is not within the range", a, "-", b)

更新：

def range(a, b, n):
    if a > b:
        if a>=n and b<=n:
            print(n, "is within the range", a, "-", b)
        else:
            print(n, "is not within the range", a, "-", b)
    else:
        if a <= n <= b:
            print(n, "is within the range", a, "-", b)
        else:
            print(n, "is not within the range", a, "-", b)

然后

>>> range(2, 8, 300)
300 is not within the range 2 - 8
>>> range(2, 8, 5)
5 is within the range 2 - 8

检查号码是否在给定范围内：无法按照指示打印消息

4 个答案: