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













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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




















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