As per spec: Constant expressions are always evaluated exactly; intermediate values and the constants themselves may require precision significantly larger than supported by any predeclared type in the language. Since 912 * 0.01 is a constant expression, it is evaluated exactly. Thus, writing fmt.Println(912 * 0.01) has the same effect as writing fmt.Println(9.12). When you … Read more
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
It is only display problem, see docs: #temporaly set display precision with pd.option_context(‘display.precision’, 10): print df 0 1 2 3 4 5 6 7 \ 0 895 2015-4-23 19 10000 LA 0.4677978806 0.477346934 0.4089938425 8 9 10 11 12 0 0.8224291972 0.8652525793 0.682994286 0.5139162227 NaN EDIT: (Thank you Mark Dickinson): Pandas uses a dedicated decimal-to-binary … Read more
floor will do as you ask. floor(45.8976 * 100) / 100; You won’t find a more direct function for this, since it’s a kind of odd thing to ask. Normally you’ll round mathematically correct. Out of curiosity – What do you need this for?
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 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.
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