Find the highest, the second highest and third highest

[derekfan]

Hi All,

Is there any function or code that can determine the highest , second highest and third highest price over a specified numbers of bars ?

Any help is appreciated.

Thank you.

Derek

[ACM]

Derek
Save the values that you want to rank in an array and then sort the array (following image illustrates one way to do this)



Also search the web for javascript sort and you will find many other examples on how to accomplish this
Alex

Hi All,

Is there any function or code that can determine the highest , second highest and third highest price over a specified numbers of bars ?

Any help is appreciated.

Thank you.

Derek

[waynecd]

Just modified existing code to sort the aH array. Note: it just outputs the 3 highest values of the last 20 bars (user defined default). You can modify the script to do whatever you need with those values.

Try:

/*****************************************************************
http://share.esignal.com/groupcontents.jsp?folder=&groupid=165
Provided By : eSignal. (c) Copyright 2003
Study:          x Bars Highest High and Lowest Low

Notes:
*This study draws two lines marking the Highest High and Lowest Low
 within the most recent X number of bars from the current bar.
*Enter the number of bars to be included for the HH and LL via
 the "Edit Studies" option.
 
Formula Parameters:     Defaults:
Number of Bars          20
Line Thickness          1
High Bar Color          blue
Low Bar Color           blue
*****************************************************************/
//debugClear();


function preMain() {
    setPriceStudy(true);
    //setStudyTitle("x Bars HH LL ");
    setShowCursorLabel(false);
    setShowTitleParameters(false);
    
    var fp1 = new FunctionParameter("xBars", FunctionParameter.NUMBER);
    fp1.setName("Number of Bars");
    fp1.setLowerLimit(1);
    fp1.setDefault(20);

    var fp2 = new FunctionParameter("nThickness", FunctionParameter.NUMBER);
    fp2.setName("Line Thickness");
    fp2.setLowerLimit(1);
    fp2.setDefault(1);
    
    var fp3 = new FunctionParameter("cH", FunctionParameter.COLOR);
    fp3.setName("High Bar Color");
    fp3.setDefault(Color.blue);

    var fp4 = new FunctionParameter("cL", FunctionParameter.COLOR);
    fp4.setName("Low Bar Color");
    fp4.setDefault(Color.blue);
}

var nBars = 20;
var vHigh = null;
var vLow = null;
var aH = null;
var aL = null;
var bEdit = true;

function main(xBars, nThickness, cH, cL) {
    var nIndex = getCurrentBarIndex();
    var nState = getBarState();
    
    if (xBars != null) nBars = Math.round(xBars);
    if (nIndex < -nBars+1) return;

    if (bEdit == true) {
        aH = null;
        aL = null;
        vHigh = null;
        vLow = null;
        bEdit = false;
    }
    
    if (aH == null || aL == null) {
        aH = new Array(nBars);
        aL = new Array(nBars);
    }
    
    if (nState == BARSTATE_NEWBAR && vHigh != null && vLow != null) {
        aH.pop();
        aH.unshift(vHigh);
        aL.pop();
        aL.unshift(vLow);
    }
    
    var h = high();
    aH[0] = h;
    var l = low();
    aL[0] = l;
    
    if (vHigh == null) vHigh = h;
    if (vLow == null) vLow = l;
    
    if (nIndex == 0 && aH[nBars-1] != null && aL[nBars-1] != null) {
        var i = 0;
        for (i = 0; i < nBars; ++i) {
            vHigh = Math.max(vHigh, aH[i]);
            vLow = Math.min(vLow, aL[i]);
        }
        drawLineRelative(-nBars+1, vHigh, 0, vHigh, PS_SOLID, nThickness, cH, "high");
        drawLineRelative(-nBars+1, vLow, 0, vLow, PS_SOLID, nThickness, cL, "low");
    }
    //ADDED CODE START
    if(isLastBarOnChart()){
        aH.sort(function(a, b) {
            return a - b;
        });
        debugPrintln("103: "+aH[aH.length-1]+"\t"+aH[aH.length-2]+"\t"+aH[aH.length-3]);
    }
    //ADDED CODE ENDS
    return;
} 

Wayne

PS: just saw reply by ACM but since I coded the sort already, I'll leave the post.

[ACM]

Derek
FWIW the example provided by waynecd will actually return incorrect results as new bars are added to the chart because once the array is sorted the highest value is removed on each subsequent new bar (and reversing the sort order simply creates the opposite problem)
To correct this you could populate a new array using the values from the existing one and sort its values so as not to affect the original array (see illustration enclosed below). While at it you might as well reverse the sort as I show in my prior post as it is more intuitive/easier IMHO to look for elements [0],[1] etc rather than for [array.length-1], [array.length-2], etc.
Alternatively you could just use the example I provided in the first place [or something more along those lines].
Alex