![]() You need a way where you don’t have to rely on memorizing the rules for finding significant figures. So, you need a better way of calculating Significant Figures. Students often find it hard to memorize these rules and it can prove to be quite risky for them if they can’t recall the rules for Significant Figures at just the right time. This is because there are a lot of rules for Significant Figures that you need to memorize if you to accurately calculate them.Īlthough learning these rules is important, it is not so easy and can get quite confusing. N = 0 if Error =() else Sig_Digit(abs(Error), 1) import decimalĭef myrounding(value, roundstyle='ROUND_HALF_UP',sig = 3): Thus, I wrote a quick function (see code below) that can take value, rounding type, and desired significant digits into account. I ran into this as well but I needed control over the rounding type. if type(n) is int: return intround(n, sigfigs) This will be far less flexible, and will probably not play nicely with other numbers such as Decimal objects: > def roundall2(n, sigfigs): ![]() If you would like to create a function that handles any number, my preference would be to convert them both to strings and look for a decimal place to decide what to do: > def roundall1(n, sigfigs):Īnother option is to check for type. Or for integers: > def intround(n, sigfigs): It's fairly trivial to implement once they're a string though. They're not stored as base 10 in memory, so significant places isn't a natural thing to do. But it's fairly well handled for floating point numbers. I can't think of anything that would be able to handle this out of the box. Now we have some idea of how floats are represented we can use the decimal module to give us some more precision, showing us what's going on: from decimal import Decimalĭecimal(0x13333333333333) / 16**13 / 2**4 16**13 is because there are 13 hexadecimal digits after the decimal point, and 2**-4 is because hex exponents are base-2. To show how these numbers work a bit, we can get back to our starting point by doing: 0x13333333333333 / 16**13 * 2**-4 If you're not used to this format, a couple of useful references are the Python docs and the C standard. The reason for doing this is that the normal decimal representation often involves rounding and hence is not how the computer actually "sees" the number. To show in more detail, we ask Python to format the number in "hex" format: 0.075.hex() The nearest number to 0.075 that can actually be represented is slightly smaller, hence rounding comes out differently than you might naively expect.Īlso note that this applies to any use of non-decimal floating point arithmetic, e.g. This is due, as pointed out by "Novice C", to a combination of floating point arithmetic having both finite precision and a base-2 representation. My main reason for posting this answer are the comments complaining that "0.075" rounds to 0.07 rather than 0.08. ![]() To directly answer the question, here's my version using naming from the R function: import math
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