在深入研究了Highcharts API之后，我最终找到了一种通过配置来实现这一点的方法。每个轴都有一个名为tickPositioner的配置选项，您可以为该选项提供一个返回数组的函数。该数组包含您希望刻度出现在轴上的确切值。下面是我的新tickPositioner配置，它在Y轴上放置了5个刻度。零正好在中间，最大值在两个极端：

yAxis: {

    tickPositioner: function () {

        var maxDeviation = Math.ceil(Math.max(Math.abs(this.dataMax), Math.abs(this.dataMin)));
        var halfMaxDeviation = Math.ceil(maxDeviation / 2);

        return [-maxDeviation, -halfMaxDeviation, 0, halfMaxDeviation, maxDeviation];
    },

    ...
}

我知道这是一个老职位，但我想我会张贴我的解决方案无论如何（这是灵感来自一个macserv建议以上在接受的答案），因为它可能会帮助其他人谁正在寻找一个类似的解决方案：

tickPositioner: function (min, max) {
            var maxDeviation = Math.ceil(Math.max(Math.abs(this.dataMax), Math.abs(this.dataMin)));
            return this.getLinearTickPositions(this.tickInterval, -maxDeviation, maxDeviation);
        }

您可以使用getExtremes和setExtremes方法执行此操作

• http://api.highcharts.com/highcharts#Axis.getExtremes%28%29
• http://api.highcharts.com/highcharts#Axis.setExtremes%28%29

例如：
http://jsfiddle.net/jlbriggs/j3NTM/1/

var ext = chart.yAxis[0].getExtremes();

这是我的解决方案。它的好处是你可以维护tickInterval。

tickPositioner(min, max) {
    let { tickPositions, tickInterval } = this;
    tickPositions = _.map(tickPositions, (tickPos) => Math.abs(tickPos));
    tickPositions = tickPositions.sort((a, b) => (b - a));

    const maxTickPosition = _.first(tickPositions);
    let minTickPosition = maxTickPosition * -1;

    let newTickPositions = [];
    while (minTickPosition <= maxTickPosition) {
      newTickPositions.push(minTickPosition);
      minTickPosition += tickInterval;
    }

    return newTickPositions;
  }

以防有人搜查，
还有一个选择。我最终陷入了类似的境地。下面是我的解决方案：

tickPositioner: function () {
    var dataMin,
    dataMax = this.dataMax;
    var positivePositions = [], negativePositions = [];

    if(this.dataMin<0) dataMin = this.dataMin*-1;
    if(this.dataMax<0) dataMax = this.dataMax*-1;

    for (var i = 0; i <= (dataMin)+10; i+=10) {
        negativePositions.push(i*-1)
    }

    negativePositions.reverse().pop();

    for (var i = 0; i <= (dataMax)+10; i+=10) {
        positivePositions.push(i)
    }

    return negativePositions.concat(positivePositions);

},

http://jsfiddle.net/j3NTM/21/

这是一个老问题，但最近我也遇到了同样的问题，这里是我的解决方案，可以概括：

const TICK_PRECISION = 2;
const AXIS_MAX_EXPAND_RATE = 1.2;

function setAxisTicks(axis, tickCount) {
    // first you calc the max from the data, then multiply with 1.1 or 1.2 
    // which can expand the max a little, in order to leave some space from the bottom/top to the max value. 
    // toPrecision decide the significant number.
    let maxDeviation = (Math.max(Math.abs(axis.dataMax), Math.abs(axis.dataMin)) * AXIS_MAX_EXPAND_RATE).toPrecision(TICK_PRECISION);

    // in case it is not a whole number
    let wholeMaxDeviation = maxDeviation * 10**TICK_PRECISION;

    // halfCount will be the tick counts on each side of 0
    let halfCount = Math.floor(tickCount / 2);

    // look for the nearest larger number which can mod the halfCount
    while (wholeMaxDeviation % halfCount != 0) {
        wholeMaxDeviation++;
    }

    // calc the unit tick amount, remember to divide by the precision
    let unitTick = (wholeMaxDeviation / halfCount) / 10**TICK_PRECISION;

    // finally get all ticks
    let tickPositions = [];
    for (let i = -halfCount; i <= halfCount; i++) {
        // there are problems with the precision when multiply a float, make sure no anything like 1.6666666667 in your result
        let tick = parseFloat((unitTick * i).toFixed(TICK_PRECISION));
        tickPositions.push(tick);
    }
    return tickPositions;
}

因此，在您的图表轴tickPositioner中，您可以添加：

tickPositioner: function () {
    return setAxisTicks(this, 7);
},