Check whether a number is in a given range: fails to print message as instructed

Question

I am trying to write a program in Python to check whether a number is in a given range and I want it to print a message when it’s not. However, the following code I have come up with fails to print such a message (it works for the first two conditions, but not the third). I’d appreciate if someone could point out if there is a mistake in my code and also if there is a more efficient way to write it.

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)

When I test the code with

range(2,8,300)

it prints nothing when it should print 300 is not within the range 2 - 8.

Asked By: Madrid_datalady

||

Source

Answer 1

if 8 > 2: is true, so it goes inside and the nested if 300>=2 and 8<=300 condition
fails and exits. So, it never goes to else loop.

Also, if (n>a,b) or (n<a,b) this always returns True. Since ("anything..") becomes tuple and it’s always True.

This works fine.

def fun(x,y,z):
    x , y = min(x,y), max(x,y)
    if z >= x and z <= y:
         print(z," is within the range",x,"-",y)
    else:
         print(, "is not within the range", x,"-",y)

Answered By: Venkatesh Garnepudi

Answer 2

You need a interval comparison:

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)

OUTPUT:

300 is not within the range 2 - 8

OR alternatively:

   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)

OUTPUT:

4 is within the range 5 - 2

PS. I would not encourage using the Python built-in function range()
as your custom function name.

pyFiddle

Answered By: DirtyBit

Answer 3

You can try this:

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)

Update:

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)

Then

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

Answered By: shafik

Answer 4

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)

This should work.

EDIT: Modified the code. This will also test the cases involving negative numbers and zero.

Answered By: Ora Aff

Check whether a number is in a given range: fails to print message as instructed

Question:

Answers: