Skip to main content

numerics - Is NumberForm double rounding numbers?



A number like 0.644696875 is represented internally as 0.6446968749999...:


N[FromDigits[RealDigits[0.644696875, 2], 2], $MachinePrecision]
(* 0.6446968749999999 *)

So if I ask NumberForm to print this number with 8 decimals, I would expect it to be 0.64469687 and not 0.64469688 because the digit after the 87 is a 4 which is less than 5. But it is not what we get with NumberForm:


NumberForm[0.644696875, {8, 8}]
(* 0.64469688 *)

So it looks like we have here two round operations when only one was requested:




  • First Rounding: From 0.6446968749999999 to 0.644696875

  • Second Rounding: From 0.644696875 to 0.64469688


I found this while comparing to Python, which doesn't double round. This leads to a result which I believe is correct:


ExternalEvaluate["Python", "'{:.8f}'.format(0.644696875)"]
(* 0.64469687 *)

Also notice that this floating point number is stored in the same way in both systems:


Divide @@ ExternalEvaluate["Python", "0.644696875.as_integer_ratio()"] == FromDigits[RealDigits[0.644696875, 2], 2]
(* True *)


Is Mathematica double rounding? Can this be avoided while still using machine numbers?


Motivation: I'm working on making a hash on an array of floating point numbers. This calculation should be the same from Mathematica and from Python. For this I need to be able to produce the same string representation of numbers in both system. This has proven more challenging than expected.


Update: I think Mathematica is double rounding just like Java does. Please see:



Update2: I asked support about this [CASE:4304365] and they said "It does appear that NumberForm is not behaving properly".




Comments

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....