Efficient Round edition with different rounding direction












1












$begingroup$


As pointed out in this post, Mathematica has a special version of Round that




Round rounds numbers of the form x.5 toward the nearest even integer.




A comment by David G suggest that why not have differnt options Direction → {"HalfDown","HalfUp","HalfEven","HalfOdd","Stochastic"}



These days I need a version of Round to HalfUp. I write a quite ugly and slow function as below



myRound[x_, d_] := Module[{},

  c1 = (1./d)*10;

  c2 = 1./d;

  theDigit = Last@IntegerDigits@IntegerPart[x*c1];

  If[theDigit >= 5,

   Internal`StringToDouble@ToString@N[(IntegerPart[x*c2] + 1)/c2],

   Internal`StringToDouble@ToString@N[(IntegerPart[x*c2])/c2]]]


speed test



In[267]:= 

myRound[#, 0.01] & /@ RandomReal[1., 1000000]; // AbsoluteTiming



Out[267]= {30.7072, Null}



In[268]:= 

Round[#, 0.01] & /@ RandomReal[1., 1000000]; // AbsoluteTiming



Out[268]= {0.285921, Null}


So I am wondering if someone on this site already have developed an efficient toolkit for round matters?






machine-precision precision-and-accuracy round







share|improve this question











$endgroup$












  • $begingroup$
    Floor[x+0.5]?
    $endgroup$
    – Szabolcs
    15 hours ago










  • $begingroup$
    @Szabolcs But Floor gives integer. Round can round at any digit
    $endgroup$
    – matheorem
    15 hours ago






  • 2




    $begingroup$
    Are your aware of RoundingRule?
    $endgroup$
    – Silvia
    15 hours ago






  • 1




    $begingroup$
    Could extend the method proposed by @Szabolcs: myRound[x_, d_] := d*Floor[x/d + 1/2]
    $endgroup$
    – Daniel Lichtblau
    13 hours ago






  • 1




    $begingroup$
    That's a result of using decimal values e.g. .01 that do not have exact binary equivalents. Could instead do myRound[8.121,1/100] and numericize afterward.
    $endgroup$
    – Daniel Lichtblau
    9 hours ago
















1












$begingroup$


As pointed out in this post, Mathematica has a special version of Round that




Round rounds numbers of the form x.5 toward the nearest even integer.




A comment by David G suggest that why not have differnt options Direction → {"HalfDown","HalfUp","HalfEven","HalfOdd","Stochastic"}



These days I need a version of Round to HalfUp. I write a quite ugly and slow function as below



myRound[x_, d_] := Module[{},

  c1 = (1./d)*10;

  c2 = 1./d;

  theDigit = Last@IntegerDigits@IntegerPart[x*c1];

  If[theDigit >= 5,

   Internal`StringToDouble@ToString@N[(IntegerPart[x*c2] + 1)/c2],

   Internal`StringToDouble@ToString@N[(IntegerPart[x*c2])/c2]]]


speed test



In[267]:= 

myRound[#, 0.01] & /@ RandomReal[1., 1000000]; // AbsoluteTiming



Out[267]= {30.7072, Null}



In[268]:= 

Round[#, 0.01] & /@ RandomReal[1., 1000000]; // AbsoluteTiming



Out[268]= {0.285921, Null}


So I am wondering if someone on this site already have developed an efficient toolkit for round matters?






machine-precision precision-and-accuracy round







share|improve this question











$endgroup$












  • $begingroup$
    Floor[x+0.5]?
    $endgroup$
    – Szabolcs
    15 hours ago










  • $begingroup$
    @Szabolcs But Floor gives integer. Round can round at any digit
    $endgroup$
    – matheorem
    15 hours ago






  • 2




    $begingroup$
    Are your aware of RoundingRule?
    $endgroup$
    – Silvia
    15 hours ago






  • 1




    $begingroup$
    Could extend the method proposed by @Szabolcs: myRound[x_, d_] := d*Floor[x/d + 1/2]
    $endgroup$
    – Daniel Lichtblau
    13 hours ago






  • 1




    $begingroup$
    That's a result of using decimal values e.g. .01 that do not have exact binary equivalents. Could instead do myRound[8.121,1/100] and numericize afterward.
    $endgroup$
    – Daniel Lichtblau
    9 hours ago














1












1








1





$begingroup$


As pointed out in this post, Mathematica has a special version of Round that




Round rounds numbers of the form x.5 toward the nearest even integer.




A comment by David G suggest that why not have differnt options Direction → {"HalfDown","HalfUp","HalfEven","HalfOdd","Stochastic"}



These days I need a version of Round to HalfUp. I write a quite ugly and slow function as below



myRound[x_, d_] := Module[{},

  c1 = (1./d)*10;

  c2 = 1./d;

  theDigit = Last@IntegerDigits@IntegerPart[x*c1];

  If[theDigit >= 5,

   Internal`StringToDouble@ToString@N[(IntegerPart[x*c2] + 1)/c2],

   Internal`StringToDouble@ToString@N[(IntegerPart[x*c2])/c2]]]


speed test



In[267]:= 

myRound[#, 0.01] & /@ RandomReal[1., 1000000]; // AbsoluteTiming



Out[267]= {30.7072, Null}



In[268]:= 

Round[#, 0.01] & /@ RandomReal[1., 1000000]; // AbsoluteTiming



Out[268]= {0.285921, Null}


So I am wondering if someone on this site already have developed an efficient toolkit for round matters?






machine-precision precision-and-accuracy round







share|improve this question











$endgroup$




As pointed out in this post, Mathematica has a special version of Round that




Round rounds numbers of the form x.5 toward the nearest even integer.




A comment by David G suggest that why not have differnt options Direction → {"HalfDown","HalfUp","HalfEven","HalfOdd","Stochastic"}



These days I need a version of Round to HalfUp. I write a quite ugly and slow function as below



myRound[x_, d_] := Module[{},

  c1 = (1./d)*10;

  c2 = 1./d;

  theDigit = Last@IntegerDigits@IntegerPart[x*c1];

  If[theDigit >= 5,

   Internal`StringToDouble@ToString@N[(IntegerPart[x*c2] + 1)/c2],

   Internal`StringToDouble@ToString@N[(IntegerPart[x*c2])/c2]]]


speed test



In[267]:= 

myRound[#, 0.01] & /@ RandomReal[1., 1000000]; // AbsoluteTiming



Out[267]= {30.7072, Null}



In[268]:= 

Round[#, 0.01] & /@ RandomReal[1., 1000000]; // AbsoluteTiming



Out[268]= {0.285921, Null}


So I am wondering if someone on this site already have developed an efficient toolkit for round matters?






machine-precision precision-and-accuracy round




machine-precision precision-and-accuracy round






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 15 hours ago







matheorem

















asked 15 hours ago









matheoremmatheorem

6,65743178




6,65743178












  • $begingroup$
    Floor[x+0.5]?
    $endgroup$
    – Szabolcs
    15 hours ago










  • $begingroup$
    @Szabolcs But Floor gives integer. Round can round at any digit
    $endgroup$
    – matheorem
    15 hours ago






  • 2




    $begingroup$
    Are your aware of RoundingRule?
    $endgroup$
    – Silvia
    15 hours ago






  • 1




    $begingroup$
    Could extend the method proposed by @Szabolcs: myRound[x_, d_] := d*Floor[x/d + 1/2]
    $endgroup$
    – Daniel Lichtblau
    13 hours ago






  • 1




    $begingroup$
    That's a result of using decimal values e.g. .01 that do not have exact binary equivalents. Could instead do myRound[8.121,1/100] and numericize afterward.
    $endgroup$
    – Daniel Lichtblau
    9 hours ago


















  • $begingroup$
    Floor[x+0.5]?
    $endgroup$
    – Szabolcs
    15 hours ago










  • $begingroup$
    @Szabolcs But Floor gives integer. Round can round at any digit
    $endgroup$
    – matheorem
    15 hours ago






  • 2




    $begingroup$
    Are your aware of RoundingRule?
    $endgroup$
    – Silvia
    15 hours ago






  • 1




    $begingroup$
    Could extend the method proposed by @Szabolcs: myRound[x_, d_] := d*Floor[x/d + 1/2]
    $endgroup$
    – Daniel Lichtblau
    13 hours ago






  • 1




    $begingroup$
    That's a result of using decimal values e.g. .01 that do not have exact binary equivalents. Could instead do myRound[8.121,1/100] and numericize afterward.
    $endgroup$
    – Daniel Lichtblau
    9 hours ago
















$begingroup$
Floor[x+0.5]?
$endgroup$
– Szabolcs
15 hours ago




$begingroup$
Floor[x+0.5]?
$endgroup$
– Szabolcs
15 hours ago












$begingroup$
@Szabolcs But Floor gives integer. Round can round at any digit
$endgroup$
– matheorem
15 hours ago




$begingroup$
@Szabolcs But Floor gives integer. Round can round at any digit
$endgroup$
– matheorem
15 hours ago




2




2




$begingroup$
Are your aware of RoundingRule?
$endgroup$
– Silvia
15 hours ago




$begingroup$
Are your aware of RoundingRule?
$endgroup$
– Silvia
15 hours ago




1




1




$begingroup$
Could extend the method proposed by @Szabolcs: myRound[x_, d_] := d*Floor[x/d + 1/2]
$endgroup$
– Daniel Lichtblau
13 hours ago




$begingroup$
Could extend the method proposed by @Szabolcs: myRound[x_, d_] := d*Floor[x/d + 1/2]
$endgroup$
– Daniel Lichtblau
13 hours ago




1




1




$begingroup$
That's a result of using decimal values e.g. .01 that do not have exact binary equivalents. Could instead do myRound[8.121,1/100] and numericize afterward.
$endgroup$
– Daniel Lichtblau
9 hours ago




$begingroup$
That's a result of using decimal values e.g. .01 that do not have exact binary equivalents. Could instead do myRound[8.121,1/100] and numericize afterward.
$endgroup$
– Daniel Lichtblau
9 hours ago










1 Answer
1






active

oldest

votes


















2












$begingroup$

I offer the following solution



r2[x_, a_] := x - Mod[x, a, -(a/2)]


We can verify that it has the desired result, using PiecewiseExpand



PiecewiseExpand[r2[x, a], -2 a < x < 2 a && a > 0]


Performance is only a little slower than the built-in Round



list = RandomReal[{0, 1}, 1000000];



AbsoluteTiming[Round[list, 0.1];]

(* {0.0079, Null} *)



AbsoluteTiming[r2[list, 0.1];]

(* {0.009414, Null} *)





share|improve this answer









$endgroup$













  • $begingroup$
    Wow, great solution. But there is a issue with hidden fractions as pointed out by mathematica.stackexchange.com/q/65298/4742 . But if we use InternalStringToDouble@ToString`, the performance will drop down.
    $endgroup$
    – matheorem
    14 hours ago










  • $begingroup$
    I figured out a trick, though I don't know why it works. The efficient solution to hidden fraction may be r2[x_, a_] := (IntegerPart[(x/a - Mod[x/a, 1, -(1/2)])]*10^12)/ 10.^(12 - 1/Log[a, 10])
    $endgroup$
    – matheorem
    14 hours ago






  • 3




    $begingroup$
    @matheorem Technically, 1.265 in binary floating point corresponds to the fraction 5697053528623677/4503599627370496 which less than 1265/1000. The issue is due to rounding decimal to binary. Thus it is a problem of inputting the number you meant, not a problem with r2.
    $endgroup$
    – Michael E2
    12 hours ago






  • 2




    $begingroup$
    @matheorem The problem with imprecise calculation is that unwanted errors creep in, such as what you're complaining about. :) How about this?: r2[x_, a_] := x (1 + Sign[x] $MachineEpsilon) - Mod[x (1 + Sign[x] $MachineEpsilon), a, -(a/2)]
    $endgroup$
    – Michael E2
    12 hours ago






  • 1




    $begingroup$
    @matheorem, I think you need to consider why you know the input is exactly 1.265 rather than a little more or a little less, and why the rounding is important to you. If you know the 3rd decimal place is exact, I suggest you multiply all your numbers by 1000, and work with integers.
    $endgroup$
    – mikado
    9 hours ago













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193412%2fefficient-round-edition-with-different-rounding-direction%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

I offer the following solution



r2[x_, a_] := x - Mod[x, a, -(a/2)]


We can verify that it has the desired result, using PiecewiseExpand



PiecewiseExpand[r2[x, a], -2 a < x < 2 a && a > 0]


Performance is only a little slower than the built-in Round



list = RandomReal[{0, 1}, 1000000];



AbsoluteTiming[Round[list, 0.1];]

(* {0.0079, Null} *)



AbsoluteTiming[r2[list, 0.1];]

(* {0.009414, Null} *)





share|improve this answer









$endgroup$













  • $begingroup$
    Wow, great solution. But there is a issue with hidden fractions as pointed out by mathematica.stackexchange.com/q/65298/4742 . But if we use InternalStringToDouble@ToString`, the performance will drop down.
    $endgroup$
    – matheorem
    14 hours ago










  • $begingroup$
    I figured out a trick, though I don't know why it works. The efficient solution to hidden fraction may be r2[x_, a_] := (IntegerPart[(x/a - Mod[x/a, 1, -(1/2)])]*10^12)/ 10.^(12 - 1/Log[a, 10])
    $endgroup$
    – matheorem
    14 hours ago






  • 3




    $begingroup$
    @matheorem Technically, 1.265 in binary floating point corresponds to the fraction 5697053528623677/4503599627370496 which less than 1265/1000. The issue is due to rounding decimal to binary. Thus it is a problem of inputting the number you meant, not a problem with r2.
    $endgroup$
    – Michael E2
    12 hours ago






  • 2




    $begingroup$
    @matheorem The problem with imprecise calculation is that unwanted errors creep in, such as what you're complaining about. :) How about this?: r2[x_, a_] := x (1 + Sign[x] $MachineEpsilon) - Mod[x (1 + Sign[x] $MachineEpsilon), a, -(a/2)]
    $endgroup$
    – Michael E2
    12 hours ago






  • 1




    $begingroup$
    @matheorem, I think you need to consider why you know the input is exactly 1.265 rather than a little more or a little less, and why the rounding is important to you. If you know the 3rd decimal place is exact, I suggest you multiply all your numbers by 1000, and work with integers.
    $endgroup$
    – mikado
    9 hours ago


















2












$begingroup$

I offer the following solution



r2[x_, a_] := x - Mod[x, a, -(a/2)]


We can verify that it has the desired result, using PiecewiseExpand



PiecewiseExpand[r2[x, a], -2 a < x < 2 a && a > 0]


Performance is only a little slower than the built-in Round



list = RandomReal[{0, 1}, 1000000];



AbsoluteTiming[Round[list, 0.1];]

(* {0.0079, Null} *)



AbsoluteTiming[r2[list, 0.1];]

(* {0.009414, Null} *)





share|improve this answer









$endgroup$













  • $begingroup$
    Wow, great solution. But there is a issue with hidden fractions as pointed out by mathematica.stackexchange.com/q/65298/4742 . But if we use InternalStringToDouble@ToString`, the performance will drop down.
    $endgroup$
    – matheorem
    14 hours ago










  • $begingroup$
    I figured out a trick, though I don't know why it works. The efficient solution to hidden fraction may be r2[x_, a_] := (IntegerPart[(x/a - Mod[x/a, 1, -(1/2)])]*10^12)/ 10.^(12 - 1/Log[a, 10])
    $endgroup$
    – matheorem
    14 hours ago






  • 3




    $begingroup$
    @matheorem Technically, 1.265 in binary floating point corresponds to the fraction 5697053528623677/4503599627370496 which less than 1265/1000. The issue is due to rounding decimal to binary. Thus it is a problem of inputting the number you meant, not a problem with r2.
    $endgroup$
    – Michael E2
    12 hours ago






  • 2




    $begingroup$
    @matheorem The problem with imprecise calculation is that unwanted errors creep in, such as what you're complaining about. :) How about this?: r2[x_, a_] := x (1 + Sign[x] $MachineEpsilon) - Mod[x (1 + Sign[x] $MachineEpsilon), a, -(a/2)]
    $endgroup$
    – Michael E2
    12 hours ago






  • 1




    $begingroup$
    @matheorem, I think you need to consider why you know the input is exactly 1.265 rather than a little more or a little less, and why the rounding is important to you. If you know the 3rd decimal place is exact, I suggest you multiply all your numbers by 1000, and work with integers.
    $endgroup$
    – mikado
    9 hours ago
















2












2








2





$begingroup$

I offer the following solution



r2[x_, a_] := x - Mod[x, a, -(a/2)]


We can verify that it has the desired result, using PiecewiseExpand



PiecewiseExpand[r2[x, a], -2 a < x < 2 a && a > 0]


Performance is only a little slower than the built-in Round



list = RandomReal[{0, 1}, 1000000];



AbsoluteTiming[Round[list, 0.1];]

(* {0.0079, Null} *)



AbsoluteTiming[r2[list, 0.1];]

(* {0.009414, Null} *)





share|improve this answer









$endgroup$



I offer the following solution



r2[x_, a_] := x - Mod[x, a, -(a/2)]


We can verify that it has the desired result, using PiecewiseExpand



PiecewiseExpand[r2[x, a], -2 a < x < 2 a && a > 0]


Performance is only a little slower than the built-in Round



list = RandomReal[{0, 1}, 1000000];



AbsoluteTiming[Round[list, 0.1];]

(* {0.0079, Null} *)



AbsoluteTiming[r2[list, 0.1];]

(* {0.009414, Null} *)






share|improve this answer












share|improve this answer



share|improve this answer










answered 15 hours ago









mikadomikado

6,6671929




6,6671929












  • $begingroup$
    Wow, great solution. But there is a issue with hidden fractions as pointed out by mathematica.stackexchange.com/q/65298/4742 . But if we use InternalStringToDouble@ToString`, the performance will drop down.
    $endgroup$
    – matheorem
    14 hours ago










  • $begingroup$
    I figured out a trick, though I don't know why it works. The efficient solution to hidden fraction may be r2[x_, a_] := (IntegerPart[(x/a - Mod[x/a, 1, -(1/2)])]*10^12)/ 10.^(12 - 1/Log[a, 10])
    $endgroup$
    – matheorem
    14 hours ago






  • 3




    $begingroup$
    @matheorem Technically, 1.265 in binary floating point corresponds to the fraction 5697053528623677/4503599627370496 which less than 1265/1000. The issue is due to rounding decimal to binary. Thus it is a problem of inputting the number you meant, not a problem with r2.
    $endgroup$
    – Michael E2
    12 hours ago






  • 2




    $begingroup$
    @matheorem The problem with imprecise calculation is that unwanted errors creep in, such as what you're complaining about. :) How about this?: r2[x_, a_] := x (1 + Sign[x] $MachineEpsilon) - Mod[x (1 + Sign[x] $MachineEpsilon), a, -(a/2)]
    $endgroup$
    – Michael E2
    12 hours ago






  • 1




    $begingroup$
    @matheorem, I think you need to consider why you know the input is exactly 1.265 rather than a little more or a little less, and why the rounding is important to you. If you know the 3rd decimal place is exact, I suggest you multiply all your numbers by 1000, and work with integers.
    $endgroup$
    – mikado
    9 hours ago




















  • $begingroup$
    Wow, great solution. But there is a issue with hidden fractions as pointed out by mathematica.stackexchange.com/q/65298/4742 . But if we use InternalStringToDouble@ToString`, the performance will drop down.
    $endgroup$
    – matheorem
    14 hours ago










  • $begingroup$
    I figured out a trick, though I don't know why it works. The efficient solution to hidden fraction may be r2[x_, a_] := (IntegerPart[(x/a - Mod[x/a, 1, -(1/2)])]*10^12)/ 10.^(12 - 1/Log[a, 10])
    $endgroup$
    – matheorem
    14 hours ago






  • 3




    $begingroup$
    @matheorem Technically, 1.265 in binary floating point corresponds to the fraction 5697053528623677/4503599627370496 which less than 1265/1000. The issue is due to rounding decimal to binary. Thus it is a problem of inputting the number you meant, not a problem with r2.
    $endgroup$
    – Michael E2
    12 hours ago






  • 2




    $begingroup$
    @matheorem The problem with imprecise calculation is that unwanted errors creep in, such as what you're complaining about. :) How about this?: r2[x_, a_] := x (1 + Sign[x] $MachineEpsilon) - Mod[x (1 + Sign[x] $MachineEpsilon), a, -(a/2)]
    $endgroup$
    – Michael E2
    12 hours ago






  • 1




    $begingroup$
    @matheorem, I think you need to consider why you know the input is exactly 1.265 rather than a little more or a little less, and why the rounding is important to you. If you know the 3rd decimal place is exact, I suggest you multiply all your numbers by 1000, and work with integers.
    $endgroup$
    – mikado
    9 hours ago


















$begingroup$
Wow, great solution. But there is a issue with hidden fractions as pointed out by mathematica.stackexchange.com/q/65298/4742 . But if we use InternalStringToDouble@ToString`, the performance will drop down.
$endgroup$
– matheorem
14 hours ago




$begingroup$
Wow, great solution. But there is a issue with hidden fractions as pointed out by mathematica.stackexchange.com/q/65298/4742 . But if we use InternalStringToDouble@ToString`, the performance will drop down.
$endgroup$
– matheorem
14 hours ago












$begingroup$
I figured out a trick, though I don't know why it works. The efficient solution to hidden fraction may be r2[x_, a_] := (IntegerPart[(x/a - Mod[x/a, 1, -(1/2)])]*10^12)/ 10.^(12 - 1/Log[a, 10])
$endgroup$
– matheorem
14 hours ago




$begingroup$
I figured out a trick, though I don't know why it works. The efficient solution to hidden fraction may be r2[x_, a_] := (IntegerPart[(x/a - Mod[x/a, 1, -(1/2)])]*10^12)/ 10.^(12 - 1/Log[a, 10])
$endgroup$
– matheorem
14 hours ago




3




3




$begingroup$
@matheorem Technically, 1.265 in binary floating point corresponds to the fraction 5697053528623677/4503599627370496 which less than 1265/1000. The issue is due to rounding decimal to binary. Thus it is a problem of inputting the number you meant, not a problem with r2.
$endgroup$
– Michael E2
12 hours ago




$begingroup$
@matheorem Technically, 1.265 in binary floating point corresponds to the fraction 5697053528623677/4503599627370496 which less than 1265/1000. The issue is due to rounding decimal to binary. Thus it is a problem of inputting the number you meant, not a problem with r2.
$endgroup$
– Michael E2
12 hours ago




2




2




$begingroup$
@matheorem The problem with imprecise calculation is that unwanted errors creep in, such as what you're complaining about. :) How about this?: r2[x_, a_] := x (1 + Sign[x] $MachineEpsilon) - Mod[x (1 + Sign[x] $MachineEpsilon), a, -(a/2)]
$endgroup$
– Michael E2
12 hours ago




$begingroup$
@matheorem The problem with imprecise calculation is that unwanted errors creep in, such as what you're complaining about. :) How about this?: r2[x_, a_] := x (1 + Sign[x] $MachineEpsilon) - Mod[x (1 + Sign[x] $MachineEpsilon), a, -(a/2)]
$endgroup$
– Michael E2
12 hours ago




1




1




$begingroup$
@matheorem, I think you need to consider why you know the input is exactly 1.265 rather than a little more or a little less, and why the rounding is important to you. If you know the 3rd decimal place is exact, I suggest you multiply all your numbers by 1000, and work with integers.
$endgroup$
– mikado
9 hours ago






$begingroup$
@matheorem, I think you need to consider why you know the input is exactly 1.265 rather than a little more or a little less, and why the rounding is important to you. If you know the 3rd decimal place is exact, I suggest you multiply all your numbers by 1000, and work with integers.
$endgroup$
– mikado
9 hours ago




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematica Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193412%2fefficient-round-edition-with-different-rounding-direction%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

數位音樂下載

Image
數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音
Read more

格利澤436b

Image
格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP
Read more

When can things happen in Etherscan, such as the picture below?

Image
2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  | 
Read more