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
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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
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
yesterday
|
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
yesterday
|
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
yesterday
|
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
yesterday
|
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
yesterday
$begingroup$
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
yesterday
$begingroup$
@matheorem That's what happens (automatically) when you use real (floating-point) numbers on all common CPUs.
$endgroup$
– Michael E2
yesterday
|
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