How To Check If A Float Value Is Within A Certain Range And Has A Given Number Of Decimal Digits?

How to check if a float value is within a range (0.50,150.00) and has 2 decimal digits? For example, 15.22366 should be false (too many decimal digits). But 15.22 should be true. I

Solution 1:

Is that you are looking for?

def check(value):
    if 0.50 <= value <= 150 and round(value,2)==value:
        return True
    return False

---

Given your comment:

i input 15.22366 it is going to return true; that is why i specified the range; it should accept 15.22

Simply said, floating point values are imprecise. Many values don't have a precise representation. Say for example 1.40. It might be displayed "as it":

>>> f = 1.40
>>> print f
1.4

But this is an illusion. Python has rounded that value in order to nicely display it. The real value as referenced by the variable f is quite different:

>>> from decimal import Decimal
>>> Decimal(f)
Decimal('1.399999999999999911182158029987476766109466552734375')

According to your rule of having only 2 decimals, should f reference a valid value or not?

The easiest way to fix that issue is probably to use round(...,2) as I suggested in the code above. But this in only an heuristic -- only able to reject "largely wrong" values. See my point here:

>>> for v in [ 1.40,
...            1.405,
...            1.399999999999999911182158029987476766109466552734375,
...            1.39999999999999991118,
...            1.3999999999999991118]:
...     print check(v), v
...
True 1.4
False 1.405
True 1.4
True 1.4
False 1.4

Notice how the last few results might seems surprising at first. I hope my above explanations put some light on this.

---

As a final advice, for your needs as I guess them from your question, you should definitively consider using "decimal arithmetic". Python provides the decimal module for that purpose.

Solution 2:

float is the wrong data type to use for your case, Use Decimal instead.

Check python docs for issues and limitations. To quote from there (I've generalised the text in Italics)

Floating-point numbers are represented in computer hardware as base 2 (binary) fractions.

no matter how many base 2 digits you’re willing to use, some decimal value (like 0.1) cannot be represented exactly as a base 2 fraction.

Stop at any finite number of bits, and you get an approximation

On a typical machine running Python, there are 53 bits of precision available for a Python float, so the value stored internally when you enter a decimal number is the binary fraction which is close to, but not exactly equal to it.

The documentation for the built-in round() function says that it rounds to the nearest value, rounding ties away from zero.

And finally, it recommends

If you’re in a situation where you care which way your decimal halfway-cases are rounded, you should consider using the decimal module.

And this will hold for your case as well, as you are looking for a precision of 2 digits after decimal points, which float just can't guarantee.

---

EDIT Note: The answer below corresponds to original question related to random float generation

Seeing that you need 2 digits of sure shot precision, I would suggest generating integer random numbers in range [50, 15000] and dividing them by 100 to convert them to float yourself.

import random
random.randint(50, 15000)/100.0

Solution 3:

Why don't you just use round?

round(random.uniform(0.5, 150.0), 2)

Solution 4:

Probably what you want to do is not to change the value itself. As said by Cyber in the comment, even if your round a floating point number, it will always store the same precision. If you need to change the way it is printed:

n = random.uniform(0.5, 150)
print '%.2f' % n               # 58.03

Solution 5:

The easiest way is to first convert the decimal to string and split with '.' and check if the length of the character. If it is >2 then pass on. i.e. Convert use input number to check if it is in a given range.

a=15.22366
if len(str(a).split('.')[1])>2:
    if 0.50 <= value <= 150:
        <do your stuff>>