Rounding Question

[fosch]

To anyone C# savvy enough to help me out...

I'm trying to round a number to the hundreds place. Yes, that's hundreds and not hundred-ths place. My example below, shows exactly what I'm talking about.

myNumber = 213.4444444

(1) Math.Round(myNumber, 2) = 213.44 <-- hundred-ths
(2) Math.Round(myNumber, 1) = 213.4 <--ten-ths
(3) Math.Round(myNumber, 0) = 213
(4) Math.Round(myNumber, -1) = 210 <--tens
(5) Math.Round(myNumber, -2) = 200 <--hundreds


Ultimately, I'm trying to arrive at (5), as I am writing a position sizing algorithm that I want to always result in an even multiple of 100 shares.

Is there a function that will round to the "left" of the decimal? I know in VB that the code woud be Round(myNumber, -2). However, in C# the Round() function does not take negative numbers.

Help.

[NinjaTrader_Austin]

fosch, I'll post the reply here that I emailed you for future reference and in case anyone else has the same question.

This is a function that you will need to place on the same level as OnBarUpdate() (see screenshot for details):

static double Round(double value, int digits)
{
    if ((digits < -15) || (digits > 15))
        throw new ArgumentOutOfRangeException("digits", "Rounding digits must be between -15 and 15, inclusive.");

    if (digits >= 0)
        return Math.Round(value, digits);

    double n = Math.Pow(10, -digits);
    return Math.Round(value / n, 0) * n;
}

protected override void OnBarUpdate()
{
    if (CurrentBar == 0)
    {
        try
        {
            double number = 213.4444444;
            double newNumber = Round(number, -2);
            Print(newNumber);
            // Prints "200"
        }
        catch (Exception ex)
        {
            Print(ex.ToString());
        }
    }
}

The page with the rest of the information (and where I pulled the code from) can be found here.

[NinjaTrader_Art]

To anyone C# savvy enough to help me out...

I'm trying to round a number to the hundreds place. Yes, that's hundreds and not hundred-ths place. My example below, shows exactly what I'm talking about.

myNumber = 213.4444444

(1) Math.Round(myNumber, 2) = 213.44 <-- hundred-ths
(2) Math.Round(myNumber, 1) = 213.4 <--ten-ths
(3) Math.Round(myNumber, 0) = 213
(4) Math.Round(myNumber, -1) = 210 <--tens
(5) Math.Round(myNumber, -2) = 200 <--hundreds


Ultimately, I'm trying to arrive at (5), as I am writing a position sizing algorithm that I want to always result in an even multiple of 100 shares.

Is there a function that will round to the "left" of the decimal? I know in VB that the code woud be Round(myNumber, -2). However, in C# the Round() function does not take negative numbers.

Help.

fosch,

double x = Math.Round(myNumber * 0.01, 0) * 100;

That should do the trick.

In position sizing you may want the result to be 200 if myNumber is 287. Rounding would give the result of 300.

You might use:
double x = Math.Floor(myNumber * 0.01) * 100;

[fosch]

Austin, I received your e-mail right after I made this post. You should've received an e-mail back from me already thanking you. I tried your code out, and it words very nicely. Very robust.

Art, I love the way that you just broke it down to simple number manipulation.

Thank you both for your replies. Now, if I can only figure out which one to use.