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gerund and noun applications

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Matrix using tikz package

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Turning a hard to access nut?

What does Deadpool mean by "left the house in that shirt"?

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Tikz: place node leftmost of two nodes of different widths

Asserting that Atheism and Theism are both faith based positions

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Print a physical multiplication table

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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

share|improve this question

edited 12 hours ago

matheorem

asked 13 hours ago

matheoremmatheorem

6,65743178

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Could extend the method proposed by @Szabolcs: myRound[x_, d_] := d*Floor[x/d + 1/2]
  $endgroup$
  – Daniel Lichtblau
  10 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
  6 hours ago
  

  
  
  
  

 | 
show 2 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

share|improve this question

edited 12 hours ago

matheorem

asked 13 hours ago

matheoremmatheorem

6,65743178

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Could extend the method proposed by @Szabolcs: myRound[x_, d_] := d*Floor[x/d + 1/2]
  $endgroup$
  – Daniel Lichtblau
  10 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
  6 hours ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

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

edited 12 hours ago

matheorem

asked 13 hours ago

matheoremmatheorem

6,65743178

$endgroup$

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]]]

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

share|improve this question

edited 12 hours ago

matheorem

asked 13 hours ago

matheoremmatheorem

6,65743178

share|improve this question

edited 12 hours ago

matheorem

asked 13 hours ago

matheoremmatheorem

6,65743178

asked 13 hours ago

matheoremmatheorem

6,65743178

asked 13 hours ago

matheoremmatheorem

6,65743178

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

answered 12 hours ago

mikadomikado

6,6671929

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  11 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
  11 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
  9 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
  9 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
  6 hours ago
  
  

  
  
  
  

 | 
show 6 more comments

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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

answered 12 hours ago

mikadomikado

6,6671929

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  11 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
  11 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
  9 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
  9 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
  6 hours ago
  
  

  
  
  
  

 | 
show 6 more comments

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

answered 12 hours ago

mikadomikado

6,6671929

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  11 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
  11 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
  9 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
  9 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
  6 hours ago
  
  

  
  
  
  

 | 
show 6 more comments

$begingroup$

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

answered 12 hours ago

mikadomikado

6,6671929

$endgroup$

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

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

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

answered 12 hours ago

mikadomikado

6,6671929

answered 12 hours ago

mikadomikado

6,6671929

answered 12 hours ago

mikadomikado

6,6671929