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How to print a statement if a user's calculated number is between two floats - Printable Version

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How to print a statement if a user's calculated number is between two floats - Bruizeh - Feb-10-2019

Hello, I have a problem. So I made a formula that calculates a User's BMI based on their height and weight, and gave it the variable userBmi. I would like it that if userBMI is between two floats, it prints a specific message, e.g. "You're Healthy" / "You're Overweight"

userBmi = round(weight / (height * height), 2)
 
        break
    except ValueError:
        print("Incorrect, Try again")

if userEth == "European":
    if userBmi < underWeight:
        print("Your BMI is", userBmi)
        print(under_msg)
    if userEth == "European":
        if userBmi < 18.5 && > 24.9:
            print("You're healthy")
This is just a snippet of the code I'm making, and the issue is the last 3 lines. It yields a syntax error, I don't know if I'm implementing the functions code or what, but any help you guys can give will be really appreciated.

PS: UserEth stands for the user's ethnicity, and a certain ethnicity determines whether the same BMI can be classified as UnderWeight or OverWeight for different ethnicity


RE: How to print a statement if a user's calculated number is between two floats - perfringo - Feb-10-2019

Your problem is on row # 12. You should write:

if 18.5 < userBmi < 24.9:

It is good practice to use 'growing pattern' i.e. values what are before are smaller. It it mentally easier to parse in which range number should be.


RE: How to print a statement if a user's calculated number is between two floats - DeaD_EyE - Feb-10-2019

if userBmi < 18.5 && > 24.9:
You are comparing 24.9 against what?
Also the && is not an operator and you should not use the bitwise operator &.

Wrong code:
userBmi < 18.5 & userBmi > 24.9
This raises a TypeError.

The & operator has a higher precedence as comparison operators.
The wrong code tries a bitwise and with the values 18.5 and userBmi.
Source: operator-precedence

Still wrong, but no Exception
(userBmi < 18.5) & (userBmi > 24.9)
There happens a bitwise and with two boolean.

Still wrong, but syntactically nicer
userBmi < 18.5 and userBmi > 24.9
This works, because the operator and has a lower precedence as a comparison operator.

Forget the greater than sign: Don't use the greater than sign in programming

But this is always False.

Better solution in my opinion:
if 18.5 < userBmi < 24.9:
This means 18.5 is smaller as userBmi, userBmi is smaller as 24.9.
18.5 -> False
18.6 -> True
24.8 -> True
24.9 -> False

The programmer who has written the article about the greater then sign, does not talk about python.
In Python you can chain comparisons.

The same solution where 24.9 and 18.5 are also True:
if 18.5 <= userBmi <= 24.9:
18.4 -> False
18.5 -> True
18.6 -> True
24.8 -> True
24.9 -> True
25.0 -> False


The same solution where 24.9 and 18.5 are also True, but this time inverted:
if not 18.5 <= userBmi <= 24.9:
18.4 -> True
18.5 -> False
18.6 -> False
24.8 -> False
24.9 -> False
25.0 -> True

I hope it helps. Please read the stuff. Operator precedence is inverted. Usually you get in the documentation of languages the highest operator precedence as first (descending order). In the python docs the precedence is listed from low to high (ascending order). It is very handy to chain the comparison. Last time I missed this feature in Structured Control Language?