在d3.js中绘制滚动/移动平均值 [英] Plot rolling/moving average in d3.js

问题描述

寻找一种方法来绘制d3中的滚动/移动平均值，而无需事先处理数据.因此，我想通过平均每个数据点和其后的两个点来平滑该行.我的代码是这样的

Looking for a way to plot rolling/moving average in d3 without having to manipulate the data in advance. So I want to smooth out the line by averaging each data point with the two after it. My code is like this

var data = [3, 66, 2, 76, 5, 20, 1, 3, 8, 90, 2, 5, 70];

var w = 20,
    h = 80;

var x = d3.scale.linear()
    .domain([0, 1])
    .range([0, w]);
var y = d3.scale.linear()
    .domain([0, 100])
    .rangeRound([h, 0]);

var chart = d3.select("body").append("svg")
    .attr("class", "chart")
    .attr("width", w * data.length -1)
    .attr("height", h);

var line = d3.svg.line()
    .x(function(d,i) { return x(i); })
    .y(function(d) { return y(d); })


var movingAverageLine = d3.svg.line()
    .x(function(d,i) { return x(i); })
    .y(function(d) { return y(d); })

chart.append("svg:path").attr("d", line(data));
chart.append("svg:path").attr("d", movingAverageLine(data));

我可以指定movingAverageLine来计算以下数据点的平均值吗?我想不出在该函数中访问它们的方法.

Can I specify movingAverageLine to calculate the average of the following data points? I can't think of a way to access them in that function.

我已经在jsfiddle上建立了一个示例. http://jsfiddle.net/tjjjohnson/XXFrg/2/#run

I have set up an example on jsfiddle. http://jsfiddle.net/tjjjohnson/XXFrg/2/#run

推荐答案

先前的解决方案会导致累积移动平均值.

A previous solution results in a cumulative moving average.

我修改了/users/2255689/john-oconnor>约翰·奥康纳来提供 n 移动平均值，方法是将自定义插值函数传递给d3.svg.line():

I modified the fiddle made by John O'Connor to provide an n-moving average by passing a custom interpolation function to d3.svg.line():

movingAvg = function(n) {
    return function (points) {
        points = points.map(function(each, index, array) {
            var to = index + n - 1;
            var subSeq, sum;
            if (to < points.length) {
                subSeq = array.slice(index, to + 1);
                sum = subSeq.reduce(function(a,b) { 
                    return [a[0] + b[0], a[1] + b[1]]; 
                });
                return sum.map(function(each) { return each / n; });
            }
            return undefined;
        });
        points = points.filter(function(each) { return typeof each !== 'undefined' })
        // Note that one could re-interpolate the points 
        // to form a basis curve (I think...)
        return points.join("L");
    }
}
var movingAverageLine = d3.svg.line()
    .x(function(d,i) { return x(i); })
    .y(function(d,i) { return y(d); })
    .interpolate(movingAvg(6));

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