CODEWITHSUNDEEP
javascriptmathd3.js
jcnesci
jcnesci

Reputation: 91

d3.js moving average with previous and next data values

I'm new to D3 and trying to do a moving average of previous and next values on my data in order to smooth it out.

Currently, I have it working using the 2 previous values + the current value. It does work but 1) how would I also use the next values, and 2) what if I wanted to use the 15 previous and 15 next values? (it would be crazy to have 30 individual vars for storing all of them)

I'm used to traditional javascript but lost as to how to traverse the data this way in D3. Hope someone can enlighten me, thanks.

See the whole code on bl.ocks.org: http://bl.ocks.org/jcnesci/7439277

Or just the data parsing code here:

d3.json("by_date_mod.json", function(error, data) {

// Setup each row of data by formatting the Date for X, and by converting to a number for Y.
data = data.rows;
data.forEach(function(d) {
  d.key = parseDate(String(d.key));
  d.value = +d.value;
});

x.domain(d3.extent(data, function(d) { return d.key; }));
y.domain([0, d3.max(data, function(d) { return d.value; })]);

// Setup the moving average calculation.
// Currently is a hacky way of doing it by manually storing and using the previous 3 values for averaging.
// Looking for another way to address previous values so we can make the averaging window much larger (like 15 previous values).
var prevPrevVal = 0;
var prevVal = 0;
var curVal = 0
var movingAverageLine = d3.svg.line()
.x(function(d,i) { return x(d.key); })
.y(function(d,i) {
    if (i == 0) {
        prevPrevVal  = y(d.value);
        prevVal = y(d.value);
        curVal =  y(d.value);
    } else if (i == 1) {
        prevPrevVal = prevVal;
        prevVal = curVal;
        curVal = (prevVal + y(d.value)) / 2.0;
    } else {
        prevPrevVal = prevVal;
        prevVal = curVal;
        curVal = (prevPrevVal + prevVal + y(d.value)) / 3.0;
    }
    return curVal;
})
.interpolate("basis");

// Draw the moving average version of the data, as a line.
graph1.append("path")
  .attr("class", "average")
  .attr("d", movingAverageLine(data));

// Draw the raw data as an area.
graph1.append("path")
    .datum(data)
    .attr("class", "area")
    .attr("d", area);

// Draw the X-axis of the graph.
graph1.append("g")
    .attr("class", "x axis")
    .attr("transform", "translate(0," + height + ")")
    .call(xAxis);

// Draw the Y-axis of the graph.
graph1.append("g")
    .attr("class", "y axis")
    .call(yAxis)
  .append("text")
    .attr("transform", "rotate(-90)")
    .attr("y", 6)
    .attr("dy", ".71em")
    .style("text-anchor", "end")
    .text("Value");
});

Upvotes: 3

Views: 3487

Answers (2)

musically_ut
musically_ut

Reputation: 34288

You need a function to calculate the moving average:

var movingWindowAvg = function (arr, step) {  // Window size = 2 * step + 1
    return arr.map(function (_, idx) { 
        var wnd = arr.slice(idx - step, idx + step + 1); 
        var result = d3.sum(wnd) / wnd.length;

        // Check for isNaN, the javascript way
        result = (result == result) ? result : _;

        return result; 
    });
};

var avgData = movingWindowAvg(avg, 7); // 15 step moving window.

Note that this functions fudges the values a bit at the borders of the original array, when a complete window cannot be extracted.

Update: If the result is NaN, convert the result to the present number in the beginning. Checking result == result is the recommended way of testing for NaNs in Javascript.

Upvotes: 5

williaster
williaster

Reputation: 93

If you really don't need a variable size window, this cumulative average might be a faster option without the slicing overhead:

    function cumAvg(objects, accessor) {
      return objects.reduce(
          function(avgs, currObj, i) { 
            if (i == 1) {
              return [ accessor(currObj) ];
            } else {
              var lastAvg = avgs[i - 2]; // reduce idxs are 1-based, arrays are 0
              avgs.push( lastAvg + ( (accessor(currObj) - lastAvg) / i) );
              return avgs;
            }
          }
    }

Upvotes: 1

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