Efficient Round edition with different rounding directionWhy round to even integers?How to prevent Round with hided fractionsVery different results from evaluating same expression with different precisionsProblems with rounding giving too many digits
If you attempt to grapple an opponent that you are hidden from, do they roll at disadvantage?
Curses work by shouting - How to avoid collateral damage?
Coordinate position not precise
Greatest common substring
Applicability of Single Responsibility Principle
Do there exist finite commutative rings with identity that are not Bézout rings?
Why are on-board computers allowed to change controls without notifying the pilots?
How can I get through very long and very dry, but also very useful technical documents when learning a new tool?
Understanding "audieritis" in Psalm 94
What to do with wrong results in talks?
Teaching indefinite integrals that require special-casing
Should my PhD thesis be submitted under my legal name?
The plural of 'stomach"
How was Earth single-handedly capable of creating 3 of the 4 gods of chaos?
Can I use my Chinese passport to enter China after I acquired another citizenship?
How does residential electricity work?
What would be the benefits of having both a state and local currencies?
Is expanding the research of a group into machine learning as a PhD student risky?
Time travel short story where a man arrives in the late 19th century in a time machine and then sends the machine back into the past
Trouble understanding overseas colleagues
(Bedrock Edition) Loading more than six chunks at once
Dot above capital letter not centred
Can a monster with multiattack use this ability if they are missing a limb?
Can I Retrieve Email Addresses from BCC?
Efficient Round edition with different rounding direction
Why round to even integers?How to prevent Round with hided fractionsVery different results from evaluating same expression with different precisionsProblems with rounding giving too many digits
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
asked Mar 17 at 11:54
matheoremmatheorem
6,65743178
$endgroup$
|
show 3 more comments
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
asked Mar 17 at 11:54
matheoremmatheorem
6,65743178
$endgroup$
$begingroup$
Floor[x+0.5]?
$endgroup$
– Szabolcs
Mar 17 at 11:58
$begingroup$
@Szabolcs But Floor gives integer. Round can round at any digit
$endgroup$
– matheorem
Mar 17 at 12:02
2
$begingroup$
Are your aware of RoundingRule?
$endgroup$
– Silvia
Mar 17 at 12:25
1
$begingroup$
Could extend the method proposed by @Szabolcs:myRound[x_, d_] := d*Floor[x/d + 1/2]
$endgroup$
– Daniel Lichtblau
Mar 17 at 13:58
1
$begingroup$
That's a result of using decimal values e.g. .01 that do not have exact binary equivalents. Could instead domyRound[8.121,1/100]and numericize afterward.
$endgroup$
– Daniel Lichtblau
Mar 17 at 18:36
|
show 3 more comments
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
asked Mar 17 at 11:54
matheoremmatheorem
6,65743178
$endgroup$
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
asked Mar 17 at 11:54
matheoremmatheorem
6,65743178
asked Mar 17 at 11:54
matheoremmatheorem
6,65743178
edited Mar 17 at 11:59
matheorem
asked Mar 17 at 11:54
matheoremmatheorem
6,65743178
asked Mar 17 at 11:54
matheoremmatheorem
6,65743178
asked Mar 17 at 11:54
matheoremmatheorem
6,65743178
6,65743178
$begingroup$
Floor[x+0.5]?
$endgroup$
– Szabolcs
Mar 17 at 11:58
$begingroup$
@Szabolcs But Floor gives integer. Round can round at any digit
$endgroup$
– matheorem
Mar 17 at 12:02
2
$begingroup$
Are your aware of RoundingRule?
$endgroup$
– Silvia
Mar 17 at 12:25
1
$begingroup$
Could extend the method proposed by @Szabolcs:myRound[x_, d_] := d*Floor[x/d + 1/2]
$endgroup$
– Daniel Lichtblau
Mar 17 at 13:58
1
$begingroup$
That's a result of using decimal values e.g. .01 that do not have exact binary equivalents. Could instead domyRound[8.121,1/100]and numericize afterward.
$endgroup$
– Daniel Lichtblau
Mar 17 at 18:36
|
show 3 more comments
$begingroup$
Floor[x+0.5]?
$endgroup$
– Szabolcs
Mar 17 at 11:58
$begingroup$
@Szabolcs But Floor gives integer. Round can round at any digit
$endgroup$
– matheorem
Mar 17 at 12:02
2
$begingroup$
Are your aware of RoundingRule?
$endgroup$
– Silvia
Mar 17 at 12:25
1
$begingroup$
Could extend the method proposed by @Szabolcs:myRound[x_, d_] := d*Floor[x/d + 1/2]
$endgroup$
– Daniel Lichtblau
Mar 17 at 13:58
1
$begingroup$
That's a result of using decimal values e.g. .01 that do not have exact binary equivalents. Could instead domyRound[8.121,1/100]and numericize afterward.
$endgroup$
– Daniel Lichtblau
Mar 17 at 18:36
$begingroup$
Floor[x+0.5]?$endgroup$
– Szabolcs
Mar 17 at 11:58
$begingroup$
Floor[x+0.5]?$endgroup$
– Szabolcs
Mar 17 at 11:58
$begingroup$
@Szabolcs But Floor gives integer. Round can round at any digit
$endgroup$
– matheorem
Mar 17 at 12:02
$begingroup$
@Szabolcs But Floor gives integer. Round can round at any digit
$endgroup$
– matheorem
Mar 17 at 12:02
2
2
$begingroup$
Are your aware of RoundingRule?
$endgroup$
– Silvia
Mar 17 at 12:25
$begingroup$
Are your aware of RoundingRule?
$endgroup$
– Silvia
Mar 17 at 12:25
1
1
$begingroup$
Could extend the method proposed by @Szabolcs:
myRound[x_, d_] := d*Floor[x/d + 1/2]$endgroup$
– Daniel Lichtblau
Mar 17 at 13:58
$begingroup$
Could extend the method proposed by @Szabolcs:
myRound[x_, d_] := d*Floor[x/d + 1/2]$endgroup$
– Daniel Lichtblau
Mar 17 at 13:58
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
Mar 17 at 18:36
$begingroup$
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
Mar 17 at 18:36
|
show 3 more comments
1 Answer
1
active
oldest
votes
3
$begingroup$
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 *)
answered Mar 17 at 12:33
mikadomikado
6,7471929
$endgroup$
$begingroup$
Wow, great solution. But there is a issue with hidden fractions as pointed out by mathematica.stackexchange.com/q/65298/4742 . But if we useInternalStringToDouble@ToString`, the performance will drop down.
$endgroup$
– matheorem
Mar 17 at 13:22
4
$begingroup$
@matheorem Technically,1.265in binary floating point corresponds to the fraction5697053528623677/4503599627370496which less than1265/1000. The issue is due to rounding decimal to binary. Thus it is a problem of inputting the number you meant, not a problem withr2[].
$endgroup$
– Michael E2
Mar 17 at 15:14
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
Mar 17 at 15:40
2
$begingroup$
@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
Mar 17 at 18:38
1
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
Mar 18 at 2:52
|
show 10 more comments
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
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
3
$begingroup$
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 *)
answered Mar 17 at 12:33
mikadomikado
6,7471929
$endgroup$
$begingroup$
Wow, great solution. But there is a issue with hidden fractions as pointed out by mathematica.stackexchange.com/q/65298/4742 . But if we useInternalStringToDouble@ToString`, the performance will drop down.
$endgroup$
– matheorem
Mar 17 at 13:22
4
$begingroup$
@matheorem Technically,1.265in binary floating point corresponds to the fraction5697053528623677/4503599627370496which less than1265/1000. The issue is due to rounding decimal to binary. Thus it is a problem of inputting the number you meant, not a problem withr2[].
$endgroup$
– Michael E2
Mar 17 at 15:14
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
Mar 17 at 15:40
2
$begingroup$
@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
Mar 17 at 18:38
1
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
Mar 18 at 2:52
|
show 10 more comments
3
$begingroup$
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 *)
answered Mar 17 at 12:33
mikadomikado
6,7471929
$endgroup$
$begingroup$
Wow, great solution. But there is a issue with hidden fractions as pointed out by mathematica.stackexchange.com/q/65298/4742 . But if we useInternalStringToDouble@ToString`, the performance will drop down.
$endgroup$
– matheorem
Mar 17 at 13:22
4
$begingroup$
@matheorem Technically,1.265in binary floating point corresponds to the fraction5697053528623677/4503599627370496which less than1265/1000. The issue is due to rounding decimal to binary. Thus it is a problem of inputting the number you meant, not a problem withr2[].
$endgroup$
– Michael E2
Mar 17 at 15:14
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
Mar 17 at 15:40
2
$begingroup$
@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
Mar 17 at 18:38
1
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
Mar 18 at 2:52
|
show 10 more comments
3
3
3
$begingroup$
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 *)
answered Mar 17 at 12:33
mikadomikado
6,7471929
$endgroup$
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 *)
answered Mar 17 at 12:33
mikadomikado
6,7471929
answered Mar 17 at 12:33
mikadomikado
6,7471929
answered Mar 17 at 12:33
mikadomikado
6,7471929
answered Mar 17 at 12:33
mikadomikado
6,7471929
6,7471929
$begingroup$
Wow, great solution. But there is a issue with hidden fractions as pointed out by mathematica.stackexchange.com/q/65298/4742 . But if we useInternalStringToDouble@ToString`, the performance will drop down.
$endgroup$
– matheorem
Mar 17 at 13:22
4
$begingroup$
@matheorem Technically,1.265in binary floating point corresponds to the fraction5697053528623677/4503599627370496which less than1265/1000. The issue is due to rounding decimal to binary. Thus it is a problem of inputting the number you meant, not a problem withr2[].
$endgroup$
– Michael E2
Mar 17 at 15:14
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
Mar 17 at 15:40
2
$begingroup$
@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
Mar 17 at 18:38
1
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
Mar 18 at 2:52
|
show 10 more comments
$begingroup$
Wow, great solution. But there is a issue with hidden fractions as pointed out by mathematica.stackexchange.com/q/65298/4742 . But if we useInternalStringToDouble@ToString`, the performance will drop down.
$endgroup$
– matheorem
Mar 17 at 13:22
4
$begingroup$
@matheorem Technically,1.265in binary floating point corresponds to the fraction5697053528623677/4503599627370496which less than1265/1000. The issue is due to rounding decimal to binary. Thus it is a problem of inputting the number you meant, not a problem withr2[].
$endgroup$
– Michael E2
Mar 17 at 15:14
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
Mar 17 at 15:40
2
$begingroup$
@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
Mar 17 at 18:38
1
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
Mar 18 at 2:52
$begingroup$
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
Mar 17 at 13:22
$begingroup$
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
Mar 17 at 13:22
4
4
$begingroup$
@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
Mar 17 at 15:14
$begingroup$
@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
Mar 17 at 15:14
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
Mar 17 at 15:40
$begingroup$
@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
Mar 17 at 15:40
2
2
$begingroup$
@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
Mar 17 at 18:38
$begingroup$
@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
Mar 17 at 18:38
1
1
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
Mar 18 at 2:52
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
Mar 18 at 2:52
|
show 10 more comments
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
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
Floor[x+0.5]?$endgroup$
– Szabolcs
Mar 17 at 11:58
$begingroup$
@Szabolcs But Floor gives integer. Round can round at any digit
$endgroup$
– matheorem
Mar 17 at 12:02
2
$begingroup$
Are your aware of RoundingRule?
$endgroup$
– Silvia
Mar 17 at 12:25
1
$begingroup$
Could extend the method proposed by @Szabolcs:
myRound[x_, d_] := d*Floor[x/d + 1/2]$endgroup$
– Daniel Lichtblau
Mar 17 at 13:58
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
Mar 17 at 18:36