The simplest solution would be to round your numbers yourself (unless you can use something like decimal.Decimal, but this means you should forgo native doubles entirely, including literals) and reproduce MATLAB's mod that way, assuming that makes sense for your use cases. In computing, the modulo operation returns the remainder or signed remainder of a division. Assuming that you are using Dynare 5.2, copy the example model file C: to your working directory. So MATLAB goes out of its way to do some magic with the floating-point results. This suggests that when x/y is close to an integer then it's rounded first, rather than being truncated like in python. ![]() Within roundoff error of an integer, then n is that integer. If y is not an integer and the quotient x./y is Now here's help mod from MATLAB: MOD(x,y) returns x - floor(x./y).*y if y ~= 0, carefully computed toĪvoid rounding error. One can easily prove that these two numbers are indeed the same within double precision: > print(6/Decimal('0.05') - 6/Decimal(0.05)) Round each value in t to the nearest number of hours less than or equal to that value. The Department of Mathematics at the University of Central Arkansas is. Find the modulus after division by 1 0 for the polynomial x 3 - 2 x + 9 9 9. Tenure-Track Position in Mathematics Education. Round each value in a duration array to the nearest number of seconds less than or equal to that value. If the dividend is a polynomial expression, then mod returns a symbolic expression without evaluating the modulus. The first number is what you'd first get with 6/0.05, but the number 119.9999999999999933386618522 gets rounded to the nearest number representable with double precision, and this is 120. Round Duration Values Toward Negative Infinity. print(6/Decimal(0.05)) # exactly approximate Some proof: > from decimal import Decimal The Matlab Support Package for Arduino Hardware is a Matlab add-on that allows you to. However, it's ever so slightly smaller than 120, so explicit floor division will truncate that number to 119 before it could be normalized to 120.0. 0 (mod > ) pic hide this posting restore restore this posting. within the resolution of double precision) that it gets rounded to 120.0. The floating-point result of 6/0.05 is close enough to 120 (i.e. ![]() This is the core of the problem, in python: > 6/0.05 = 120 b mod (a,m) devuelve el resto después de dividir a por m, donde a es el dividendo y m es el divisor.
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