Trendline in FLOT

Question

I included jquery.flot.trendline.js. From Here

It is my code

$.plot($(\"#placeholder\"), seriesdata, {
        series: {
     trendlin         


        

Answer 1

That plugin is a no-go. It requires modifications to the flot source to work and in my opinion isn't very well done. The simpliest approach would be to just add the trendline yourself as an additional series. The math is not difficult...

  // calc slope and intercept
  // then use resulting y = mx + b to create trendline
  lineFit = function(points){
    sI = slopeAndIntercept(points);
    if (sI){
      // we have slope/intercept, get points on fit line
      var N = points.length;
      var rV = [];
      rV.push([points[0][0], sI.slope * points[0][0] + sI.intercept]);
      rV.push([points[N-1][0], sI.slope * points[N-1][0] + sI.intercept]);
      return rV;
    }
    return [];
  }

  // simple linear regression
  slopeAndIntercept = function(points){
    var rV = {},
        N = points.length,
        sumX = 0, 
        sumY = 0,
        sumXx = 0,
        sumYy = 0,
        sumXy = 0;

    // can't fit with 0 or 1 point
    if (N < 2){
      return rV;
    }    

    for (var i = 0; i < N; i++){
      var x = points[i][0],
          y = points[i][1];
      sumX += x;
      sumY += y;
      sumXx += (x*x);
      sumYy += (y*y);
      sumXy += (x*y);
    }

    // calc slope and intercept
    rV['slope'] = ((N * sumXy) - (sumX * sumY)) / (N * sumXx - (sumX*sumX));
    rV['intercept'] = (sumY - rV['slope'] * sumX) / N;
    rV['rSquared'] = Math.abs((rV['slope'] * (sumXy - (sumX * sumY) / N)) / (sumYy - ((sumY * sumY) / N)));

    return rV;
  }

You can then call this as:

  lineFitSeries = lineFit(someSeries);

And add the lineFitSeries as another series to flot...

Here's a working example.