floating-accuracy

I have a function that computes a point in 3d spaced based on a value in range [0, 1] . The problem I'm facing is, that a binary floating-point number cannot represent exactly 1. The mathematical expression that is evaluated in the function is able to compute a value for t=1.0 , but the value will never be accepted by the function because it checks if for the range before computing. curves_error curves_bezier(curves_PointList* list, curves_Point* dest, curves_float t) { /* ... */ if (t < 0 || t > 1) return curves_invalid_args; /* ... */ return curves_no_error; } How can I, with this function,

This question already has an answer here: ruby: converting from float to integer in ruby produces strange results 3 answers When I add 0.1+0.2 I am getting 0.30000000000000004 but when I add the same number in ruby 1.8.7 I am getting the correct answer 0.3 . I get 0.3 by rounding but I just want to get 0.3 on ruby 1.9.2 by adding 0.1 and 0.2 You need bigdecimal for this to make work. (BigDecimal('0.1') + BigDecimal("0.2")).to_f See below link: http://redmine.ruby-lang.org/issues/4394 Your old ruby lied to you: $ ruby -v ruby 1.8.7 (2010-06-23 patchlevel 299) [x86_64-linux] $ irb irb(main):001

I reviewed the following posts beforehand. Is there a way to use DataFrame.isin() with an approximation factor or a tolerance value? Or is there another method that could? Filter dataframe rows if value in column is in a set list of values use a list of values to select rows from a pandas dataframe EX) df = DataFrame({'A' : [5,6,3.3,4], 'B' : [1,2,3.2, 5]}) In : df Out: A B 0 5 1 1 6 2 2 3.3 3.2 3 4 5 df[df['A'].isin([3, 6], tol=.5)] In : df Out: A B 1 6 2 2 3.3 3.2 You can do a similar thing with numpy's isclose : df[np.isclose(df['A'].values[:, None], [3, 6], atol=.5).any(axis=1)] Out: A B 1

This question already has an answer here: How to alter a float by its smallest increment (or close to it)? 7 answers A float (a.k.a. single) value is a 4-byte value, and supposed to represent any real-valued number. Because of the way it is formatted and the finite number of bytes it is made off, there is a minimum value and a maximum value it can represent, and it has a finite precision, depending on it's own value. I would like to know if there is a way to get the closest possible value above or below some reference value, given the finite precision of a float. With integers, this is trivial

I am seeking some more understanding about the "resolution" parameter of a numpy float (I guess any computer defined float for that matter). Consider the following script: import numpy as np a = np.finfo(10.1) print a I get an output which among other things prints out: precision=15 resolution= 1.0000000000000001e-15 max= 1.797(...)e+308 min= -max The numpy documentation specifies: "resolution: (floating point number of the appropriate type) The approximate decimal resolution of this type, i.e., 10**-precision." source resolution is derived from precision, but unfortunately this definition is

Can someone give me an example of a floating point number (double precision), that needs more than 16 significant decimal digits to represent it? I have found in this thread that sometimes you need up to 17 digits, but I am not able to find an example of such a number (16 seems enough to me). Can somebody clarify this? Thanks a lot! My other answer was dead wrong. #include <stdio.h> int main(int argc, char *argv[]) { unsigned long long n = 1ULL << 53; unsigned long long a = 2*(n-1); unsigned long long b = 2*(n-2); printf("%llu\n%llu\n%d\n", a, b, (double)a == (double)b); return 0; } Compile