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 15 hours 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 15 hours ago
matheoremmatheorem
6,65743178
$endgroup$
$begingroup$
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 domyRound[8.121,1/100]and numericize afterward.
$endgroup$
– Daniel Lichtblau
9 hours 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 15 hours 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 15 hours ago
matheoremmatheorem
6,65743178
asked 15 hours ago
matheoremmatheorem
6,65743178
edited 15 hours ago
matheorem
asked 15 hours ago
matheoremmatheorem
6,65743178
asked 15 hours ago
matheoremmatheorem
6,65743178
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 domyRound[8.121,1/100]and numericize afterward.
$endgroup$
– Daniel Lichtblau
9 hours ago
|
show 3 more comments
$begingroup$
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 domyRound[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
|
show 3 more comments
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} *)
answered 15 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 useInternalStringToDouble@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 ber2[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,1.265in 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
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
|
show 9 more comments
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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} *)
answered 15 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 useInternalStringToDouble@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 ber2[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,1.265in 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
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
|
show 9 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} *)
answered 15 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 useInternalStringToDouble@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 ber2[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,1.265in 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
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
|
show 9 more comments
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} *)
answered 15 hours ago
mikadomikado
6,6671929
$endgroup$
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 15 hours ago
mikadomikado
6,6671929
answered 15 hours ago
mikadomikado
6,6671929
answered 15 hours ago
mikadomikado
6,6671929
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 useInternalStringToDouble@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 ber2[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,1.265in 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
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
|
show 9 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
14 hours ago
$begingroup$
I figured out a trick, though I don't know why it works. The efficient solution to hidden fraction may ber2[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,1.265in 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
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
|
show 9 more comments
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$begingroup$
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