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
asked 2 days ago
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 2 days ago
matheoremmatheorem
6,65743178
$endgroup$
$begingroup$
Floor[x+0.5]?
$endgroup$
– Szabolcs
2 days ago
$begingroup$
@Szabolcs But Floor gives integer. Round can round at any digit
$endgroup$
– matheorem
2 days ago
2
$begingroup$
Are your aware of RoundingRule?
$endgroup$
– Silvia
2 days ago
1
$begingroup$
Could extend the method proposed by @Szabolcs:myRound[x_, d_] := d*Floor[x/d + 1/2]
$endgroup$
– Daniel Lichtblau
2 days ago
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
2 days ago
|
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 2 days ago
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 2 days ago
matheoremmatheorem
6,65743178
asked 2 days ago
matheoremmatheorem
6,65743178
edited 2 days ago
matheorem
asked 2 days ago
matheoremmatheorem
6,65743178
asked 2 days ago
matheoremmatheorem
6,65743178
asked 2 days ago
matheoremmatheorem
6,65743178
6,65743178
$begingroup$
Floor[x+0.5]?
$endgroup$
– Szabolcs
2 days ago
$begingroup$
@Szabolcs But Floor gives integer. Round can round at any digit
$endgroup$
– matheorem
2 days ago
2
$begingroup$
Are your aware of RoundingRule?
$endgroup$
– Silvia
2 days ago
1
$begingroup$
Could extend the method proposed by @Szabolcs:myRound[x_, d_] := d*Floor[x/d + 1/2]
$endgroup$
– Daniel Lichtblau
2 days ago
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
2 days ago
|
show 3 more comments
$begingroup$
Floor[x+0.5]?
$endgroup$
– Szabolcs
2 days ago
$begingroup$
@Szabolcs But Floor gives integer. Round can round at any digit
$endgroup$
– matheorem
2 days ago
2
$begingroup$
Are your aware of RoundingRule?
$endgroup$
– Silvia
2 days ago
1
$begingroup$
Could extend the method proposed by @Szabolcs:myRound[x_, d_] := d*Floor[x/d + 1/2]
$endgroup$
– Daniel Lichtblau
2 days ago
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
2 days ago
$begingroup$
Floor[x+0.5]?$endgroup$
– Szabolcs
2 days ago
$begingroup$
Floor[x+0.5]?$endgroup$
– Szabolcs
2 days ago
$begingroup$
@Szabolcs But Floor gives integer. Round can round at any digit
$endgroup$
– matheorem
2 days ago
$begingroup$
@Szabolcs But Floor gives integer. Round can round at any digit
$endgroup$
– matheorem
2 days ago
2
2
$begingroup$
Are your aware of RoundingRule?
$endgroup$
– Silvia
2 days ago
$begingroup$
Are your aware of RoundingRule?
$endgroup$
– Silvia
2 days ago
1
1
$begingroup$
Could extend the method proposed by @Szabolcs:
myRound[x_, d_] := d*Floor[x/d + 1/2]$endgroup$
– Daniel Lichtblau
2 days ago
$begingroup$
Could extend the method proposed by @Szabolcs:
myRound[x_, d_] := d*Floor[x/d + 1/2]$endgroup$
– Daniel Lichtblau
2 days 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
2 days 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
2 days ago
|
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 2 days ago
mikadomikado
6,7171929
$endgroup$
$begingroup$
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
2 days ago
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
2 days 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
2 days ago
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
2 days ago
1
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
2 days ago
|
show 10 more comments
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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 2 days ago
mikadomikado
6,7171929
$endgroup$
$begingroup$
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
2 days ago
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
2 days 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
2 days ago
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
2 days ago
1
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
2 days ago
|
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 2 days ago
mikadomikado
6,7171929
$endgroup$
$begingroup$
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
2 days ago
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
2 days 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
2 days ago
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
2 days ago
1
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
2 days ago
|
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 2 days ago
mikadomikado
6,7171929
$endgroup$
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 2 days ago
mikadomikado
6,7171929
answered 2 days ago
mikadomikado
6,7171929
answered 2 days ago
mikadomikado
6,7171929
answered 2 days ago
mikadomikado
6,7171929
6,7171929
$begingroup$
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
2 days ago
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
2 days 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
2 days ago
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
2 days ago
1
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
2 days ago
|
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
2 days ago
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
2 days 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
2 days ago
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
2 days ago
1
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
2 days 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
2 days 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
2 days ago
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
2 days 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
2 days 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
2 days 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
2 days ago
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
2 days 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
2 days ago
1
1
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
2 days ago
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
2 days ago
|
show 10 more comments
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$begingroup$
Floor[x+0.5]?$endgroup$
– Szabolcs
2 days ago
$begingroup$
@Szabolcs But Floor gives integer. Round can round at any digit
$endgroup$
– matheorem
2 days ago
2
$begingroup$
Are your aware of RoundingRule?
$endgroup$
– Silvia
2 days ago
1
$begingroup$
Could extend the method proposed by @Szabolcs:
myRound[x_, d_] := d*Floor[x/d + 1/2]$endgroup$
– Daniel Lichtblau
2 days 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
2 days ago