how to round number to custom step (0.25, 20, 100...)

[ibiza]

Hi all,

I'd need a function that would rounds values like :
d = roundTo(64.2, 0.25) ' returns 64.25
d = roundTo(64.2, 10) ' returns 64
d = roundTo(64.2, 100) ' returns 100

well, that rounds a value to the given step...with rounding up if it is
it is right in the middle (e.g. 1.25 to 1.5 with a step of 0.25)

can anyone help with this one?

Thanks! :)

[vincent]

Try this (in C#). It has not been tested. It might not be the best but I
think it will work.
public double roundTo(double numToRound, double step)
{

if (step == 0)
return numToRound;

//Calc the floor value of numToRound
double floor = ((long) (numToRound / step)) * step;

//round up if more than half way of step
double round = floor;
double remainder = numToRound - floor;
if (remainder step / 2)
round += step;

return round;

}
"ibiza" wrote:

Hi all,

I'd need a function that would rounds values like :
d = roundTo(64.2, 0.25) ' returns 64.25
d = roundTo(64.2, 10) ' returns 64
d = roundTo(64.2, 100) ' returns 100

well, that rounds a value to the given step...with rounding up if it is
it is right in the middle (e.g. 1.25 to 1.5 with a step of 0.25)

can anyone help with this one?

Thanks! :)

[vincent]

I found a bug. The comparision in the if statement should be >= instead of >
if (remainder >= step / 2)
round += step;

"vincent" wrote:

Try this (in C#). It has not been tested. It might not be the best but I
think it will work.
public double roundTo(double numToRound, double step)
{

if (step == 0)
return numToRound;

//Calc the floor value of numToRound
double floor = ((long) (numToRound / step)) * step;

//round up if more than half way of step
double round = floor;
double remainder = numToRound - floor;
if (remainder step / 2)
round += step;

return round;

}
"ibiza" wrote:

Hi all,

I'd need a function that would rounds values like :
d = roundTo(64.2, 0.25) ' returns 64.25
d = roundTo(64.2, 10) ' returns 64
d = roundTo(64.2, 100) ' returns 100

well, that rounds a value to the given step...with rounding up if it is
it is right in the middle (e.g. 1.25 to 1.5 with a step of 0.25)

can anyone help with this one?

Thanks! :)

[ibiza]

well, thank you very much! very appreciated ^_^
works fine!

vincent wrote:

I found a bug. The comparision in the if statement should be >= instead of >
if (remainder >= step / 2)
round += step;

"vincent" wrote:

Try this (in C#). It has not been tested. It might not be the best but I
think it will work.
public double roundTo(double numToRound, double step)
{

if (step == 0)
return numToRound;

//Calc the floor value of numToRound
double floor = ((long) (numToRound / step)) * step;

//round up if more than half way of step
double round = floor;
double remainder = numToRound - floor;
if (remainder step / 2)
round += step;

return round;

}
"ibiza" wrote:

Hi all,
>
I'd need a function that would rounds values like :
d = roundTo(64.2, 0.25) ' returns 64.25
d = roundTo(64.2, 10) ' returns 64
d = roundTo(64.2, 100) ' returns 100
>
well, that rounds a value to the given step...with rounding up if it is
it is right in the middle (e.g. 1.25 to 1.5 with a step of 0.25)
>
can anyone help with this one?
>
Thanks! :)
>
>

[Göran Andersson]

ibiza wrote:

Hi all,

I'd need a function that would rounds values like :
d = roundTo(64.2, 0.25) ' returns 64.25
d = roundTo(64.2, 10) ' returns 64
d = roundTo(64.2, 100) ' returns 100

well, that rounds a value to the given step...with rounding up if it is
it is right in the middle (e.g. 1.25 to 1.5 with a step of 0.25)

can anyone help with this one?

Thanks! :)

Simply:

Shared Function RoundTo(value As Double, step As Double) As Double
Return Math.Round(value / step) * step
End Function

This will of course return the value 60 in the second case you listed,
not 64. I believe that's what you really intended.

[ibiza]

wow, even simpler :)

and you got it right, the second case should have returned 60, my
mistake ;)

Göran Andersson wrote:

ibiza wrote:

Hi all,

I'd need a function that would rounds values like :
d = roundTo(64.2, 0.25) ' returns 64.25
d = roundTo(64.2, 10) ' returns 64
d = roundTo(64.2, 100) ' returns 100

well, that rounds a value to the given step...with rounding up if it is
it is right in the middle (e.g. 1.25 to 1.5 with a step of 0.25)

can anyone help with this one?

Thanks! :)

Simply:

Shared Function RoundTo(value As Double, step As Double) As Double
Return Math.Round(value / step) * step
End Function

This will of course return the value 60 in the second case you listed,
not 64. I believe that's what you really intended.