actually, this is not a bug and it is pretty common. its called a floating point “error” and in a nutshell, it has to do things with how decimal numbers are stored within a google sheets (even excel or any other app) more details can be found here: https://en.wikipedia.org/wiki/IEEE_754 to counter it you will need … Read more
The number 1.1 cannot be represented in finite form in binary. It looks like 1.00011001100110011… “Rounding errors” are just mathematically inevitable with simple floating-point arithmetic. If you want accuracy, use a Decimal number type. http://support.microsoft.com/kb/42980
I think it reflects more on your understanding of floating point types than on Python. See my article about floating point numbers (.NET-based, but still relevant) for the reasons behind this “inaccuracy”. If you need to keep the exact decimal representation, you should use the decimal module.
The question is confusingly worded. Let’s break it down into many smaller questions: Why is it that one tenth plus two tenths does not always equal three tenths in floating point arithmetic? Let me give you an analogy. Suppose we have a math system where all numbers are rounded off to exactly five decimal places. … Read more
The specific reason in your case is that the real number 0.94 cannot be represented exactly in a double precision floating point. When you type 0.94, the actual number stored is 0.939999999999999946709294817992486059665679931640625.
It is true. It is an inherent limitation of how floating point values are represented in memory in a finite number of bits. This program, for instance, prints “false”: public class Main { public static void main(String[] args) { double a = 0.7; double b = 0.9; double x = a + 0.1; double y … Read more
No, there is no best practice. Unfortunately, there cannot be, because almost all floating-point calculations introduce some rounding error, and the consequences of the errors are different for different applications. Typically, software will perform some calculations that ideally would yield some exact mathematical result x but, due to rounding errors (or other issues), produce an … Read more
Floating point numbers are an approximation, they cannot store decimal numbers exactly. Because they try to represent a very large range of numbers in only 64 bits, they must approximate to some extent. It is very important to be aware of this, because it results in some weird side-effects. For example, you might very reasonably … Read more
You can use truncatingRemainder and 1 as the divider. Returns the remainder of this value divided by the given value using truncating division. Apple doc Example: let myDouble1: Double = 12.25 let myDouble2: Double = 12.5 let myDouble3: Double = 12.75 let remainder1 = myDouble1.truncatingRemainder(dividingBy: 1) let remainder2 = myDouble2.truncatingRemainder(dividingBy: 1) let remainder3 = myDouble3.truncatingRemainder(dividingBy: … Read more
str(0.47000000000000003) give ‘0.47’ and float(‘0.47’) can be 0.46999999999999997. This is due to the way floating point number are represented (see this wikipedia article) Note: float(repr(0.47000000000000003)) or eval(repr(0.47000000000000003)) will give you the expected result, but you should use Decimal if you need precision.